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Proposition 2.7 Suppose \( f \in {H}_{0}^{1}\left( \Omega \right) \) . There exists a unique function \( u \) from \( \left( {0, + \infty }\right) \) to \( {H}_{0}^{1}\left( \Omega \right) \), differentiable in \( \left( {0, + \infty }\right) \) and satisfying the following conditions:\n\n\[ - {u}^{\prime }\left( t\right) = {\Delta u}\left( t\right) \text{for all}t > 0\text{.}\]\n\n\[ - \mathop{\lim }\limits_{{t \rightarrow 0}}u\left( t\right) = f\text{in}{H}_{0}^{1}\left( \Omega \right) \text{.} \]
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Proof. Suppose \( u \) satisfies the conditions of the statement. Then\n\n\[ \frac{d}{dt}{\left( \parallel u\left( t\right) {\parallel }_{{L}^{2}}\right) }^{2} = 2\operatorname{Re}{\left( {u}^{\prime }\left( t\right) \mid u\left( t\right) \right) }_{{L}^{2}} = - 2{\left( \parallel u\left( t\right) {\parallel }_{{H}_{0}^{1}}\right) }^{2}, \]\n\nby Proposition 2.1. It follows that the function \( t \mapsto {\left( \parallel u\left( t\right) {\parallel }_{{L}^{2}}\right) }^{2} \) is decreasing and, in particular, that\n\n\[ \parallel u\left( t\right) {\parallel }_{{L}^{2}} \leq \parallel f{\parallel }_{{L}^{2}}\;\text{ for all }t > 0. \]\n\nConsequently, if \( f = 0 \), we have \( u\left( t\right) = 0 \) for all \( t > 0 \), which proves uniqueness.\n\nRegarding existence, it suffices to check that the given formula is good. This is easy if we take into account that \( {\mu }_{n} \leq {\mu }_{0} < 0 \) for every \( n \) .
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Yes
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Proposition 2.8 Suppose \( f, g \in {H}_{0}^{1}\left( \Omega \right) \) . There exists at most one function \( u \) from \( \mathbb{R} \) to \( {H}_{0}^{1}\left( \Omega \right) \), twice differentiable on \( \mathbb{R} \) and satisfying these conditions:\n\n\[ \n- {u}^{\prime \prime }\left( t\right) = {\Delta u}\left( t\right) \text{for all}t \in \mathbb{R}.\n\]\n\n\[ \n\text{-}u\left( 0\right) = f\text{and}{u}^{\prime }\left( 0\right) = g.\n\]\n\nIf the sequences \( {\left( {\mu }_{n}{\left( f \mid {u}_{n}\right) }_{{H}_{0}^{1}}\right) }_{n \in \mathbb{N}} \) and \( {\left( \sqrt{-{\mu }_{n}}{\left( g \mid {u}_{n}\right) }_{{H}_{0}^{1}}\right) }_{n \in \mathbb{N}} \) lie in \( {\ell }^{2} \) , such a function \( u \) exists and is given by\n\n\[ \nu\left( t\right) = \mathop{\sum }\limits_{{n = 0}}^{{+\infty }}\left( {\cos \left( {\sqrt{-{\mu }_{n}}t}\right) {\left( f \mid {u}_{n}\right) }_{{H}_{0}^{1}} + \frac{1}{\sqrt{-{\mu }_{n}}}\sin \left( {\sqrt{-{\mu }_{n}}t}\right) {\left( g \mid {u}_{n}\right) }_{{H}_{0}^{1}}}\right) {u}_{n}\n\]\n\nfor all \( t \in \mathbb{R} \), the series being convergent in \( {H}_{0}^{1}\left( \Omega \right) .
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Proof. Let \( u \) satisfy the conditions of the statement. By Proposition 2.1,\n\n\[ \n\frac{d}{dt}{\left( {\begin{Vmatrix}{u}^{\prime }\left( t\right) \end{Vmatrix}}_{{L}^{2}}\right) }^{2} = 2\operatorname{Re}{\left( {u}^{\prime \prime }\left( t\right) \mid {u}^{\prime }\left( t\right) \right) }_{{L}^{2}}\n\]\n\n\[ \n= - 2\operatorname{Re}{\left( u\left( t\right) \mid {u}^{\prime }\left( t\right) \right) }_{{H}_{0}^{1}} = - \frac{d}{dt}{\left( \parallel u\left( t\right) {\parallel }_{{H}_{0}^{1}}\right) }^{2}.\n\]\n\nIt follows that the expression \( {\left( \parallel {u}^{\prime }\left( t\right) {\parallel }_{{L}^{2}}\right) }^{2} + {\left( \parallel u\left( t\right) {\parallel }_{{H}_{0}^{1}}\right) }^{2} \) does not depend on \( t \) . In particular, if \( f = g = 0 \), we have \( u\left( t\right) = 0 \) for \( t \in \mathbb{R} \), which proves uniqueness.\n\nThe proof of existence, as in the previous example, is straightforward.
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Yes
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Lemma 1.1 If a rectangle is the almost disjoint union of finitely many other rectangles, say \( R = \mathop{\bigcup }\limits_{{k = 1}}^{N}{R}_{k} \), then\n\n\[ \left| R\right| = \mathop{\sum }\limits_{{k = 1}}^{N}\left| {R}_{k}\right| \]
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Proof. We consider the grid formed by extending indefinitely the sides of all rectangles \( {R}_{1},\ldots ,{R}_{N} \) . This construction yields finitely many rectangles \( {\widetilde{R}}_{1},\ldots ,{\widetilde{R}}_{M} \), and a partition \( {J}_{1},\ldots ,{J}_{N} \) of the integers between 1 and \( M \), such that the unions\n\n\[ R = \mathop{\bigcup }\limits_{{j = 1}}^{M}{\widetilde{R}}_{j}\;\text{ and }\;{R}_{k} = \mathop{\bigcup }\limits_{{j \in {J}_{k}}}{\widetilde{R}}_{j},\;\text{ for }k = 1,\ldots, N \]\n\nare almost disjoint (see the illustration in Figure 2).\n\n\n\nFigure 2. The grid formed by the rectangles \( {R}_{k} \)\n\nFor the rectangle \( R \), for example, we see that \( \left| R\right| = \mathop{\sum }\limits_{{j = 1}}^{M}\left| {\widetilde{R}}_{j}\right| \), since the grid actually partitions the sides of \( R \) and each \( {\widetilde{R}}_{j} \) consists of taking products of the intervals in these partitions. Thus when adding the volumes of the \( {\widetilde{R}}_{j} \) we are summing the corresponding products of lengths of the intervals that arise. Since this also holds for the other rectangles \( {R}_{1},\ldots ,{R}_{N} \), we conclude that\n\n\[ \left| R\right| = \mathop{\sum }\limits_{{j = 1}}^{M}\left| {\widetilde{R}}_{j}\right| = \mathop{\sum }\limits_{{k = 1}}^{N}\mathop{\sum }\limits_{{j \in {J}_{k}}}\left| {\widetilde{R}}_{j}\right| = \mathop{\sum }\limits_{{k = 1}}^{N}\left| {R}_{k}\right| . \]
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Yes
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Theorem 1.3 Every open subset \( \mathcal{O} \) of \( \mathbb{R} \) can be writen uniquely as a countable union of disjoint open intervals.
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Proof. For each \( x \in \mathcal{O} \), let \( {I}_{x} \) denote the largest open interval containing \( x \) and contained in \( \mathcal{O} \). More precisely, since \( \mathcal{O} \) is open, \( x \) is contained in some small (non-trivial) interval, and therefore if\n\n\[ \n{a}_{x} = \inf \{ a < x : \left( {a, x}\right) \subset \mathcal{O}\} \;\text{ and }\;{b}_{x} = \sup \{ b > x : \left( {x, b}\right) \subset \mathcal{O}\} \n\]\n\nwe must have \( {a}_{x} < x < {b}_{x} \) (with possibly infinite values for \( {a}_{x} \) and \( {b}_{x} \) ). If we now let \( {I}_{x} = \left( {{a}_{x},{b}_{x}}\right) \), then by construction we have \( x \in {I}_{x} \) as well as \( {I}_{x} \subset \mathcal{O} \). Hence\n\n\[ \n\mathcal{O} = \mathop{\bigcup }\limits_{{x \in \mathcal{O}}}{I}_{x} \n\]\n\nNow suppose that two intervals \( {I}_{x} \) and \( {I}_{y} \) intersect. Then their union (which is also an open interval) is contained in \( \mathcal{O} \) and contains \( x \). Since \( {I}_{x} \) is maximal, we must have \( \left( {{I}_{x} \cup {I}_{y}}\right) \subset {I}_{x} \), and similarly \( \left( {{I}_{x} \cup {I}_{y}}\right) \subset {I}_{y} \). This can happen only if \( {I}_{x} = {I}_{y} \); therefore, any two distinct intervals in the collection \( \mathcal{I} = {\left\{ {I}_{x}\right\} }_{x \in \mathcal{O}} \) must be disjoint. The proof will be complete once we have shown that there are only countably many distinct intervals in the collection \( \mathcal{I} \). This, however, is easy to see, since every open interval \( {I}_{x} \) contains a rational number. Since different intervals are disjoint, they must contain distinct rationals, and therefore \( \mathcal{I} \) is countable, as desired.
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Yes
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Theorem 1.4 Every open subset \( \mathcal{O} \) of \( {\mathbb{R}}^{d}, d \geq 1 \), can be written as a countable union of almost disjoint closed cubes.
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Proof. We must construct a countable collection \( \mathcal{Q} \) of closed cubes whose interiors are disjoint, and so that \( \mathcal{O} = \mathop{\bigcup }\limits_{{Q \in \mathcal{Q}}}Q \) . As a first step, consider the grid in \( {\mathbb{R}}^{d} \) formed by taking all closed cubes of side length 1 whose vertices have integer coordinates. In other words, we consider the natural grid of lines parallel to the axes, that is, the grid generated by the lattice \( {\mathbb{Z}}^{d} \) . We shall also use the grids formed by cubes of side length \( {2}^{-N} \) obtained by successively bisecting the original grid. We either accept or reject cubes in the initial grid as part of \( \mathcal{Q} \) according to the following rule: if \( Q \) is entirely contained in \( \mathcal{O} \) then we accept \( Q \) ; if \( Q \) intersects both \( \mathcal{O} \) and \( {\mathcal{O}}^{c} \) then we tentatively accept it; and if \( Q \) is entirely contained in \( {\mathcal{O}}^{c} \) then we reject it. As a second step, we bisect the tentatively accepted cubes into \( {2}^{d} \) cubes with side length \( 1/2 \) . We then repeat our procedure, by accepting the smaller cubes if they are completely contained in \( \mathcal{O} \), tentatively accepting them if they intersect both \( \mathcal{O} \) and \( {\mathcal{O}}^{c} \), and rejecting them if they are contained in \( {\mathcal{O}}^{c} \) . Figure 3 illustrates these steps for an open set in \( {\mathbb{R}}^{2} \) . This procedure is then repeated indefinitely, and (by construction) the resulting collection \( \mathcal{Q} \) of all accepted cubes is countable and consists of almost disjoint cubes. To see why their union is all of \( \mathcal{O} \), we note that given \( x \in \mathcal{O} \) there exists a cube of side length \( {2}^{-N} \) (obtained from successive bisections of the original grid) that contains \( x \) and that is entirely contained in \( \mathcal{O} \) . Either this cube has been accepted, or it is contained in a cube that has been previously accepted. This shows that the union of all cubes in \( \mathcal{Q} \) covers \( \mathcal{O} \) .
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Yes
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Property 2 If \( {m}_{ * }\left( E\right) = 0 \), then \( E \) is measurable. In particular, if \( F \) is a subset of a set of exterior measure 0, then \( F \) is measurable.
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By Observation 3 of the exterior measure, for every \( \epsilon > 0 \) there exists an open set \( \mathcal{O} \) with \( E \subset \mathcal{O} \) and \( {m}_{ * }\left( \mathcal{O}\right) \leq \epsilon \) . Since \( \left( {\mathcal{O} - E}\right) \subset \mathcal{O} \) , monotonicity implies \( {m}_{ * }\left( {\mathcal{O} - E}\right) \leq \epsilon \), as desired.
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No
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Property 3 A countable union of measurable sets is measurable.
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Suppose \( E = \mathop{\bigcup }\limits_{{j = 1}}^{\infty }{E}_{j} \), where each \( {E}_{j} \) is measurable. Given \( \epsilon > 0 \), we may choose for each \( j \) an open set \( {\mathcal{O}}_{j} \) with \( {E}_{j} \subset {\mathcal{O}}_{j} \) and \( {m}_{ * }\left( {{\mathcal{O}}_{j} - {E}_{j}}\right) \leq \epsilon /{2}^{j} \) . Then the union \( \mathcal{O} = \mathop{\bigcup }\limits_{{j = 1}}^{\infty }{\mathcal{O}}_{j} \) is open, \( E \subset \mathcal{O} \), and \( \left( {\mathcal{O} - E}\right) \subset \mathop{\bigcup }\limits_{{j = 1}}^{\infty }\left( {{\mathcal{O}}_{j} - {E}_{j}}\right) \), so monotonicity and sub-additivity of the exterior measure imply\n\n\[ \n{m}_{ * }\left( {\mathcal{O} - E}\right) \leq \mathop{\sum }\limits_{{j = 1}}^{\infty }{m}_{ * }\left( {{\mathcal{O}}_{j} - {E}_{j}}\right) \leq \epsilon \n\]
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Yes
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Lemma 3.1 If \( F \) is closed, \( K \) is compact, and these sets are disjoint, then \( d\left( {F, K}\right) > 0 \) .
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Proof. Since \( F \) is closed, for each point \( x \in K \), there exists \( {\delta }_{x} > 0 \) so that \( d\left( {x, F}\right) > 3{\delta }_{x} \) . Since \( \mathop{\bigcup }\limits_{{x \in K}}{B}_{2{\delta }_{x}}\left( x\right) \) covers \( K \), and \( K \) is compact, we may find a subcover, which we denote by \( \mathop{\bigcup }\limits_{{j = 1}}^{N}{B}_{2{\delta }_{j}}\left( {x}_{j}\right) \) . If we let \( \delta = \) \( \min \left( {{\delta }_{1},\ldots ,{\delta }_{N}}\right) \), then we must have \( d\left( {K, F}\right) \geq \delta > 0 \) . Indeed, if \( x \in K \) and \( y \in F \), then for some \( j \) we have \( \left| {{x}_{j} - x}\right| \leq 2{\delta }_{j} \), and by construction \( \left| {y - {x}_{j}}\right| \geq 3{\delta }_{j} \) . Therefore\n\n\[ \left| {y - x}\right| \geq \left| {y - {x}_{j}}\right| - \left| {{x}_{j} - x}\right| \geq 3{\delta }_{j} - 2{\delta }_{j} \geq \delta \]\n\nand the lemma is proved.
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Yes
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Property 5 The complement of a measurable set is measurable.
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If \( E \) is measurable, then for every positive integer \( n \) we may choose an open set \( {\mathcal{O}}_{n} \) with \( E \subset {\mathcal{O}}_{n} \) and \( {m}_{ * }\left( {{\mathcal{O}}_{n} - E}\right) \leq 1/n \) . The complement \( {\mathcal{O}}_{n}^{c} \) is closed, hence measurable, which implies that the union \( S = \mathop{\bigcup }\limits_{{n = 1}}^{\infty }{\mathcal{O}}_{n}^{c} \) is also measurable by Property 3. Now we simply note that \( S \subset {E}^{c} \), and\n\n\[ \left( {{E}^{c} - S}\right) \subset \left( {{\mathcal{O}}_{n} - E}\right) \]\n\nsuch that \( {m}_{ * }\left( {{E}^{c} - S}\right) \leq 1/n \) for all \( n \) . Therefore, \( {m}_{ * }\left( {{E}^{c} - S}\right) = 0 \), and \( {E}^{c} - S \) is measurable by Property 2. Therefore \( {E}^{c} \) is measurable since it is the union of two measurable sets, namely \( S \) and \( \left( {{E}^{c} - S}\right) \) .
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Yes
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Property 6 A countable intersection of measurable sets is measurable.
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This follows from Properties 3 and 5, since\n\n\[ \mathop{\bigcap }\limits_{{j = 1}}^{\infty }{E}_{j} = {\left( \mathop{\bigcup }\limits_{{j = 1}}^{\infty }{E}_{j}^{c}\right) }^{c} \]\n\n
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Yes
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Theorem 3.2 If \( {E}_{1},{E}_{2},\ldots \), are disjoint measurable sets, and \( E = \) \( \mathop{\bigcup }\limits_{{j = 1}}^{\infty }{E}_{j} \), then\n\n\[ m\left( E\right) = \mathop{\sum }\limits_{{j = 1}}^{\infty }m\left( {E}_{j}\right) \]
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Proof. First, we assume further that each \( {E}_{j} \) is bounded. Then, for each \( j \), by applying the definition of measurability to \( {E}_{j}^{c} \), we can choose a closed subset \( {F}_{j} \) of \( {E}_{j} \) with \( {m}_{ * }\left( {{E}_{j} - {F}_{j}}\right) \leq \epsilon /{2}^{j} \) . For each fixed \( N \) , the sets \( {F}_{1},\ldots ,{F}_{N} \) are compact and disjoint, so that \( m\left( {\mathop{\bigcup }\limits_{{j = 1}}^{N}{F}_{j}}\right) = \) \( \mathop{\sum }\limits_{{j = 1}}^{N}m\left( {F}_{j}\right) \) . Since \( \mathop{\bigcup }\limits_{{j = 1}}^{N}{F}_{j} \subset E \), we must have\n\n\[ m\left( E\right) \geq \mathop{\sum }\limits_{{j = 1}}^{N}m\left( {F}_{j}\right) \geq \mathop{\sum }\limits_{{j = 1}}^{N}m\left( {E}_{j}\right) - \epsilon . \]\n\nLetting \( N \) tend to infinity, since \( \epsilon \) was arbitrary we find that\n\n\[ m\left( E\right) \geq \mathop{\sum }\limits_{{j = 1}}^{\infty }m\left( {E}_{j}\right) \]\n\nSince the reverse inequality always holds (by sub-additivity in Observation 2), this concludes the proof when each \( {E}_{j} \) is bounded.\n\nIn the general case, we select any sequence of cubes \( {\left\{ {Q}_{k}\right\} }_{k = 1}^{\infty } \) that increases to \( {\mathbb{R}}^{d} \), in the sense that \( {Q}_{k} \subset {Q}_{k + 1} \) for all \( k \geq 1 \) and \( \mathop{\bigcup }\limits_{{k = 1}}^{\infty }{Q}_{k} = \) \( {\mathbb{R}}^{d} \) . We then let \( {S}_{1} = {Q}_{1} \) and \( {S}_{k} = {Q}_{k} - {Q}_{k - 1} \) for \( k \geq 2 \) . If we define measurable sets by \( {E}_{j, k} = {E}_{j} \cap {S}_{k} \), then\n\n\[ E = \mathop{\bigcup }\limits_{{j, k}}{E}_{j, k} \]\n\nThe union above is disjoint and every \( {E}_{j, k} \) is bounded. Moreover \( {E}_{j} = \) \( \mathop{\bigcup }\limits_{{k = 1}}^{\infty }{E}_{j, k} \), and this union is also disjoint. Putting these facts together, and using what has already been proved, we obtain\n\n\[ m\left( E\right) = \mathop{\sum }\limits_{{j, k}}m\left( {E}_{j, k}\right) = \mathop{\sum }\limits_{j}\mathop{\sum }\limits_{k}m\left( {E}_{j, k}\right) = \mathop{\sum }\limits_{j}m\left( {E}_{j}\right) ,\]\n\nas claimed.
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Yes
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Corollary 3.3 Suppose \( {E}_{1},{E}_{2},\ldots \) are measurable subsets of \( {\mathbb{R}}^{d} \). (i) If \( {E}_{k} \nearrow E \), then \( m\left( E\right) = \mathop{\lim }\limits_{{N \rightarrow \infty }}m\left( {E}_{N}\right) \).
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Proof. For the first part, let \( {G}_{1} = {E}_{1},{G}_{2} = {E}_{2} - {E}_{1} \), and in general \( {G}_{k} = {E}_{k} - {E}_{k - 1} \) for \( k \geq 2 \). By their construction, the sets \( {G}_{k} \) are measurable, disjoint, and \( E = \mathop{\bigcup }\limits_{{k = 1}}^{\infty }{G}_{k} \). Hence \[ m\left( E\right) = \mathop{\sum }\limits_{{k = 1}}^{\infty }m\left( {G}_{k}\right) = \mathop{\lim }\limits_{{N \rightarrow \infty }}\mathop{\sum }\limits_{{k = 1}}^{N}m\left( {G}_{k}\right) = \mathop{\lim }\limits_{{N \rightarrow \infty }}m\left( {\mathop{\bigcup }\limits_{{k = 1}}^{N}{G}_{k}}\right) , \] and since \( \mathop{\bigcup }\limits_{{k = 1}}^{N}{G}_{k} = {E}_{N} \) we get the desired limit.
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Yes
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Theorem 3.4 Suppose \( E \) is a measurable subset of \( {\mathbb{R}}^{d} \) . Then, for every \( \epsilon > 0 \) :\n\n(i) There exists an open set \( \mathcal{O} \) with \( E \subset \mathcal{O} \) and \( m\left( {\mathcal{O} - E}\right) \leq \epsilon \) .\n\n(ii) There exists a closed set \( F \) with \( F \subset E \) and \( m\left( {E - F}\right) \leq \epsilon \) .\n\n(iii) If \( m\left( E\right) \) is finite, there exists a compact set \( K \) with \( K \subset E \) and \( m\left( {E - K}\right) \leq \epsilon \)\n\n(iv) If \( m\left( E\right) \) is finite, there exists a finite union \( F = \mathop{\bigcup }\limits_{{j = 1}}^{N}{Q}_{j} \) of closed cubes such that\n\n\[ m\left( {E\bigtriangleup F}\right) \leq \epsilon \]
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Proof. Part (i) is just the definition of measurability. For the second part, we know that \( {E}^{c} \) is measurable, so there exists an open set \( \mathcal{O} \) with \( {E}^{c} \subset \mathcal{O} \) and \( m\left( {\mathcal{O} - {E}^{c}}\right) \leq \epsilon \) . If we let \( F = {\mathcal{O}}^{c} \), then \( F \) is closed, \( F \subset E \) , and \( E - F = \mathcal{O} - {E}^{c} \) . Hence \( m\left( {E - F}\right) \leq \epsilon \) as desired.\n\nFor (iii), we first pick a closed set \( F \) so that \( F \subset E \) and \( m\left( {E - F}\right) \leq \) \( \epsilon /2 \) . For each \( n \), we let \( {B}_{n} \) denote the ball centered at the origin of radius\n\n\( n \), and define compact sets \( {K}_{n} = F \cap {B}_{n} \) . Then \( E - {K}_{n} \) is a sequence of measurable sets that decreases to \( E - F \), and since \( m\left( E\right) < \infty \), we conclude that for all large \( n \) one has \( m\left( {E - {K}_{n}}\right) \leq \epsilon \) .\n\nFor the last part, choose a family of closed cubes \( {\left\{ {Q}_{j}\right\} }_{j = 1}^{\infty } \) so that\n\n\[ E \subset \mathop{\bigcup }\limits_{{j = 1}}^{\infty }{Q}_{j}\;\text{ and }\;\mathop{\sum }\limits_{{j = 1}}^{\infty }\left| {Q}_{j}\right| \leq m\left( E\right) + \epsilon /2 \]\n\nSince \( m\left( E\right) < \infty \), the series converges and there exists \( N > 0 \) such that \( \mathop{\sum }\limits_{{j = N + 1}}^{\infty }\left| {Q}_{j}\right| < \epsilon /2 \) . If \( F = \mathop{\bigcup }\limits_{{j = 1}}^{N}{Q}_{j} \), then\n\n\[ m\left( {E\bigtriangleup F}\right) = m\left( {E - F}\right) + m\left( {F - E}\right) \]\n\n\[ \leq m\left( {\mathop{\bigcup }\limits_{{j = N + 1}}^{\infty }{Q}_{j}}\right) + m\left( {\mathop{\bigcup }\limits_{{j = 1}}^{\infty }{Q}_{j} - E}\right) \]\n\n\[ \leq \mathop{\sum }\limits_{{j = N + 1}}^{\infty }\left| {Q}_{j}\right| + \mathop{\sum }\limits_{{j = 1}}^{\infty }\left| {Q}_{j}\right| - m\left( E\right) \]\n\n\[ \leq \epsilon \text{.} \]
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Yes
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Corollary 3.5 A subset \( E \) of \( {\mathbb{R}}^{d} \) is measurable\n\n(i) if and only if \( E \) differs from a \( {G}_{\delta } \) by a set of measure zero,\n\n(ii) if and only if \( E \) differs from an \( {F}_{\sigma } \) by a set of measure zero.
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Proof. Clearly \( E \) is measurable whenever it satisfies either (i) or (ii), since the \( {F}_{\sigma },{G}_{\delta } \), and sets of measure zero are measurable.\n\nConversely, if \( E \) is measurable, then for each integer \( n \geq 1 \) we may select an open set \( {\mathcal{O}}_{n} \) that contains \( E \), and such that \( m\left( {{\mathcal{O}}_{n} - E}\right) \leq 1/n \) . Then \( S = \mathop{\bigcap }\limits_{{n = 1}}^{\infty }{\mathcal{O}}_{n} \) is a \( {G}_{\delta } \) that contains \( E \), and \( \left( {S - E}\right) \subset \left( {{\mathcal{O}}_{n} - E}\right) \) for all \( n \) . Therefore \( m\left( {S - E}\right) \leq 1/n \) for all \( n \) ; hence \( S - E \) has exterior measure zero, and is therefore measurable.\n\nFor the second implication, we simply apply part (ii) of Theorem 3.4 with \( \epsilon = 1/n \), and take the union of the resulting closed sets.
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Yes
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Theorem 3.6 The set \( \mathcal{N} \) is not measurable.
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The proof is by contradiction, so we assume that \( \mathcal{N} \) is measurable. Let \( {\left\{ {r}_{k}\right\} }_{k = 1}^{\infty } \) be an enumeration of all the rationals in \( \left\lbrack {-1,1}\right\rbrack \), and consider the translates\n\n\[ \n{\mathcal{N}}_{k} = \mathcal{N} + {r}_{k} \n\]\n\nWe claim that the sets \( {\mathcal{N}}_{k} \) are disjoint, and\n\n(4)\n\n\[ \n\left\lbrack {0,1}\right\rbrack \subset \mathop{\bigcup }\limits_{{k = 1}}^{\infty }{\mathcal{N}}_{k} \subset \left\lbrack {-1,2}\right\rbrack \n\]\n\nTo see why these sets are disjoint, suppose that the intersection \( {\mathcal{N}}_{k} \cap {\mathcal{N}}_{{k}^{\prime }} \) is non-empty. Then there exist rationals \( {r}_{k} \neq {r}_{k}^{\prime } \) and \( \alpha \) and \( \beta \) with \( {x}_{\alpha } + {r}_{k} = {x}_{\beta } + {r}_{{k}^{\prime }} \) ; hence\n\n\[ \n{x}_{\alpha } - {x}_{\beta } = {r}_{{k}^{\prime }} - {r}_{k} \n\]\n\nConsequently \( \alpha \neq \beta \) and \( {x}_{\alpha } - {x}_{\beta } \) is rational; hence \( {x}_{\alpha } \sim {x}_{\beta } \), which contradicts the fact that \( \mathcal{N} \) contains only one representative of each equivalence class.\n\nThe second inclusion is straightforward since each \( {\mathcal{N}}_{k} \) is contained in \( \left\lbrack {-1,2}\right\rbrack \) by construction. Finally, if \( x \in \left\lbrack {0,1}\right\rbrack \), then \( x \sim {x}_{\alpha } \) for some \( \alpha \), and therefore \( x - {x}_{\alpha } = {r}_{k} \) for some \( k \) . Hence \( x \in {\mathcal{N}}_{k} \), and the first inclusion holds.\n\nNow we may conclude the proof of the theorem. If \( \mathcal{N} \) were measurable, then so would be \( {\mathcal{N}}_{k} \) for all \( k \), and since the union \( \mathop{\bigcup }\limits_{{k = 1}}^{\infty }{\mathcal{N}}_{k} \) is disjoint, the inclusions in (4) yield\n\n\[ \n1 \leq \mathop{\sum }\limits_{{k = 1}}^{\infty }m\left( {\mathcal{N}}_{k}\right) \leq 3 \n\]\n\nSince \( {\mathcal{N}}_{k} \) is a translate of \( \mathcal{N} \), we must have \( m\left( {\mathcal{N}}_{k}\right) = m\left( \mathcal{N}\right) \) for all \( k \) . Consequently,\n\n\[ \n1 \leq \mathop{\sum }\limits_{{k = 1}}^{\infty }m\left( \mathcal{N}\right) \leq 3 \n\]\n\nThis is the desired contradiction, since neither \( m\left( \mathcal{N}\right) = 0 \) nor \( m\left( \mathcal{N}\right) > 0 \) is possible.
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Yes
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If \( f \) is measurable and finite-valued, and \( \Phi \) is continuous, then \( \Phi \circ f \) is measurable.
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In fact, \( \Phi \) is continuous, so \( {\Phi }^{-1}\left( \left( {-\infty, a}\right) \right) \) is an open set \( \mathcal{O} \), and hence \( {\left( \Phi \circ f\right) }^{-1}\left( \left( {-\infty, a}\right) \right) = {f}^{-1}\left( \mathcal{O}\right) \) is measurable.
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Yes
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Property 3 Suppose \( {\left\{ {f}_{n}\right\} }_{n = 1}^{\infty } \) is a sequence of measurable functions. Then\n\n\[ \mathop{\sup }\limits_{n}{f}_{n}\left( x\right) ,\;\mathop{\inf }\limits_{n}{f}_{n}\left( x\right) ,\;\mathop{\limsup }\limits_{{n \rightarrow \infty }},{f}_{n}\left( x\right) \;\text{ and }\;\mathop{\liminf }\limits_{{n \rightarrow \infty }}{f}_{n}\left( x\right) \]\n\nare measurable.
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Proving that \( \mathop{\sup }\limits_{n}{f}_{n} \) is measurable requires noting that \( \left\{ {\mathop{\sup }\limits_{n}{f}_{n} > a}\right\} = \) \( \mathop{\bigcup }\limits_{n}\left\{ {{f}_{n} > a}\right\} \) . This also yields the result for \( \mathop{\inf }\limits_{n}{f}_{n}\left( x\right) \), since this quantity equals \( - \mathop{\sup }\limits_{n}\left( {-{f}_{n}\left( x\right) }\right) \) .\n\nThe result for the limsup and liminf also follows from the two observations\n\n\[ \mathop{\limsup }\limits_{{n \rightarrow \infty }}{f}_{n}\left( x\right) = \mathop{\inf }\limits_{k}\left\{ {\mathop{\sup }\limits_{{n \geq k}}{f}_{n}}\right\} \;\text{ and }\;\mathop{\liminf }\limits_{{n \rightarrow \infty }}{f}_{n}\left( x\right) = \mathop{\sup }\limits_{k}\left\{ {\mathop{\inf }\limits_{{n \geq k}}{f}_{n}}\right\} . \]
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Yes
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Property 4 If \( {\left\{ {f}_{n}\right\} }_{n = 1}^{\infty } \) is a collection of measurable functions, and\n\n\[ \mathop{\lim }\limits_{{n \rightarrow \infty }}{f}_{n}\left( x\right) = f\left( x\right) \]\n\nthen \( f \) is measurable.
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Since \( f\left( x\right) = \mathop{\limsup }\limits_{{n \rightarrow \infty }}{f}_{n}\left( x\right) = \mathop{\liminf }\limits_{{n \rightarrow \infty }}{f}_{n}\left( x\right) \), this property is a consequence of property 3.
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Yes
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Property 5 If \( f \) and \( g \) are measurable, then\n\n(i) The integer powers \( {f}^{k}, k \geq 1 \) are measurable.\n\n(ii) \( f + g \) and \( {fg} \) are measurable if both \( f \) and \( g \) are finite-valued.
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For (i) we simply note that if \( k \) is odd, then \( \left\{ {{f}^{k} > a}\right\} = \left\{ {f > {a}^{1/k}}\right\} \), and if \( k \) is even and \( a \geq 0 \), then \( \left\{ {{f}^{k} > a}\right\} = \left\{ {f > {a}^{1/k}}\right\} \cup \left\{ {f < - {a}^{1/k}}\right\} \).\n\nFor (ii), we first see that \( f + g \) is measurable because\n\n\[ \{ f + g > a\} = \mathop{\bigcup }\limits_{{r \in \mathbb{Q}}}\{ f > a - r\} \cap \{ g > r\} \]\n\nwith \( \mathbb{Q} \) denoting the rationals.\n\nFinally, \( {fg} \) is measurable because of the previous results and the fact that\n\n\[ {fg} = \frac{1}{4}\left\lbrack {{\left( f + g\right) }^{2} - {\left( f - g\right) }^{2}}\right\rbrack \]
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Yes
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Theorem 4.1 Suppose \( f \) is a non-negative measurable function on \( {\mathbb{R}}^{d} \) . Then there exists an increasing sequence of non-negative simple functions \( {\left\{ {\varphi }_{k}\right\} }_{k = 1}^{\infty } \) that converges pointwise to \( f \), namely,\n\n\[ \n{\varphi }_{k}\left( x\right) \leq {\varphi }_{k + 1}\left( x\right) \;\text{ and }\;\mathop{\lim }\limits_{{k \rightarrow \infty }}{\varphi }_{k}\left( x\right) = f\left( x\right) ,\text{ for all }x.\n\]
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Proof. We begin first with a truncation. For \( N \geq 1 \), let \( {Q}_{N} \) denote the cube centered at the origin and of side length \( N \) . Then we define\n\n\[ \n{F}_{N}\left( x\right) = \left\{ \begin{matrix} f\left( x\right) & \text{ if }x \in {Q}_{N}\text{ and }f\left( x\right) \leq N \\ N & \text{ if }x \in {Q}_{N}\text{ and }f\left( x\right) > N \\ 0 & \text{ otherwise. } \end{matrix}\right.\n\]\n\nThen, \( {F}_{N}\left( x\right) \rightarrow f\left( x\right) \) as \( N \) tends to infinity for all \( x \) . Now, we partition the range of \( {F}_{N} \), namely \( \left\lbrack {0, N}\right\rbrack \), as follows. For fixed \( N, M \geq 1 \), we define\n\n\[ \n{E}_{\ell, M} = \left\{ {x \in {Q}_{N} : \frac{\ell }{M} < {F}_{N}\left( x\right) \leq \frac{\ell + 1}{M}}\right\} ,\;\text{ for }0 \leq \ell < {NM}.\n\]\n\nThen we may form\n\n\[ \n{F}_{N, M}\left( x\right) = \mathop{\sum }\limits_{\ell }\frac{\ell }{M}{\chi }_{{E}_{\ell, M}}\left( x\right)\n\]\n\nEach \( {F}_{N, M} \) is a simple function that satisfies \( 0 \leq {F}_{N}\left( x\right) - {F}_{N, M}\left( x\right) \leq \) \( 1/M \) for all \( x \) . If we now choose \( N = M = {2}^{k} \) with \( k \geq 1 \) integral, and let \( {\varphi }_{k} = {F}_{{2}^{k},{2}^{k}} \), then we see that \( 0 \leq {F}_{M}\left( x\right) - {\varphi }_{k}\left( x\right) \leq 1/{2}^{k} \) for all \( x \) , \( \left\{ {\varphi }_{k}\right\} \) is increasing, and this sequence satisfies all the desired properties.
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Yes
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Theorem 4.2 Suppose \( f \) is measurable on \( {\mathbb{R}}^{d} \) . Then there exists a sequence of simple functions \( {\left\{ {\varphi }_{k}\right\} }_{k = 1}^{\infty } \) that satisfies\n\n\[ \left| {{\varphi }_{k}\left( x\right) }\right| \leq \left| {{\varphi }_{k + 1}\left( x\right) }\right| \;\text{ and }\;\mathop{\lim }\limits_{{k \rightarrow \infty }}{\varphi }_{k}\left( x\right) = f\left( x\right) ,\text{ for all }x. \]\n\nIn particular, we have \( \left| {{\varphi }_{k}\left( x\right) }\right| \leq \left| {f\left( x\right) }\right| \) for all \( x \) and \( k \) .
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Proof. We use the following decomposition of the function \( f : f\left( x\right) = \) \( {f}^{ + }\left( x\right) - {f}^{ - }\left( x\right) \), where\n\n\[ {f}^{ + }\left( x\right) = \max \left( {f\left( x\right) ,0}\right) \;\text{ and }\;{f}^{ - }\left( x\right) = \max \left( {-f\left( x\right) ,0}\right) . \]\n\nSince both \( {f}^{ + } \) and \( {f}^{ - } \) are non-negative, the previous theorem yields two increasing sequences of non-negative simple functions \( {\left\{ {\varphi }_{k}^{\left( 1\right) }\left( x\right) \right\} }_{k = 1}^{\infty } \) and \( {\left\{ {\varphi }_{k}^{\left( 2\right) }\left( x\right) \right\} }_{k = 1}^{\infty } \) which converge pointwise to \( {f}^{ + } \) and \( {f}^{ - } \), respectively. Then, if we let\n\n\[ {\varphi }_{k}\left( x\right) = {\varphi }_{k}^{\left( 1\right) }\left( x\right) - {\varphi }_{k}^{\left( 2\right) }\left( x\right) \]\n\nwe see that \( {\varphi }_{k}\left( x\right) \) converges to \( f\left( x\right) \) for all \( x \) . Finally, the sequence \( \left\{ \left| {\varphi }_{k}\right| \right\} \) is increasing because the definition of \( {f}^{ + },{f}^{ - } \) and the properties of \( {\varphi }_{k}^{\left( 1\right) } \) and \( {\varphi }_{k}^{\left( 2\right) } \) imply that\n\n\[ \left| {{\varphi }_{k}\left( x\right) }\right| = {\varphi }_{k}^{\left( 1\right) }\left( x\right) + {\varphi }_{k}^{\left( 2\right) }\left( x\right) \]\n\nWe may now go one step further, and approximate by step functions. Here, in general, the convergence may hold only almost everywhere.
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Yes
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Theorem 4.3 Suppose \( f \) is measurable on \( {\mathbb{R}}^{d} \) . Then there exists a sequence of step functions \( {\left\{ {\psi }_{k}\right\} }_{k = 1}^{\infty } \) that converges pointwise to \( f\left( x\right) \) for almost every \( x \) .
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Proof. By the previous result, it suffices to show that if \( E \) is a measurable set with finite measure, then \( f = {\chi }_{E} \) can be approximated by step functions. To this end, we recall part (iv) of Theorem 3.4, which states that for every \( \epsilon \) there exist cubes \( {Q}_{1},\ldots ,{Q}_{N} \) such that \( m\left( {E\bigtriangleup \mathop{\bigcup }\limits_{{j = 1}}^{N}{Q}_{j}}\right) \leq \epsilon \) . By considering the grid formed by extending the sides of these cubes, we see that there exist almost disjoint rectangles \( {\widetilde{R}}_{1},\ldots ,{\widetilde{R}}_{M} \) such that \( \mathop{\bigcup }\limits_{{j = 1}}^{N}{Q}_{j} = \mathop{\bigcup }\limits_{{j = 1}}^{M}{\widetilde{R}}_{j} \) . By taking rectangles \( {R}_{j} \) contained in \( {\widetilde{R}}_{j} \), and slightly smaller in size, we find a collection of disjoint rectangles that satisfy \( m\left( {E\bigtriangleup \mathop{\bigcup }\limits_{{j = 1}}^{M}{R}_{j}}\right) \leq {2\epsilon } \) . Therefore\n\n\[ f\left( x\right) = \mathop{\sum }\limits_{{j = 1}}^{M}{\chi }_{{R}_{j}}\left( x\right) \]\n\nexcept possibly on a set of measure \( \leq {2\epsilon } \) . Consequently, for every \( k \geq 1 \) , there exists a step function \( {\psi }_{k}\left( x\right) \) such that if\n\n\[ {E}_{k} = \left\{ {x : f\left( x\right) \neq {\psi }_{k}\left( x\right) }\right\} \]\n\nthen \( m\left( {E}_{k}\right) \leq {2}^{-k} \) . If we let \( {F}_{K} = \mathop{\bigcup }\limits_{{j = K + 1}}^{\infty }{E}_{j} \) and \( F = \mathop{\bigcap }\limits_{{K = 1}}^{\infty }{F}_{K} \), then \( m\left( F\right) = 0 \) since \( m\left( {F}_{K}\right) \leq {2}^{-K} \), and \( {\psi }_{k}\left( x\right) \rightarrow f\left( x\right) \) for all \( x \) in the complement of \( F \), which is the desired result.
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Yes
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Theorem 4.4 (Egorov) Suppose \( {\left\{ {f}_{k}\right\} }_{k = 1}^{\infty } \) is a sequence of measurable functions defined on a measurable set \( E \) with \( m\left( E\right) < \infty \), and assume that \( {f}_{k} \rightarrow f \) a.e on \( E \) . Given \( \epsilon > 0 \), we can find a closed set \( {A}_{\epsilon } \subset E \) such that \( m\left( {E - {A}_{\epsilon }}\right) \leq \epsilon \) and \( {f}_{k} \rightarrow f \) uniformly on \( {A}_{\epsilon } \) .
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Proof. We may assume without loss of generality that \( {f}_{k}\left( x\right) \rightarrow f\left( x\right) \) for every \( x \in E \) . For each pair of non-negative integers \( n \) and \( k \), let\n\n\[ \n{E}_{k}^{n} = \left\{ {x \in E : \left| {{f}_{j}\left( x\right) - f\left( x\right) }\right| < 1/n,\text{ for all }j > k}\right\} .\n\]\n\nNow fix \( n \) and note that \( {E}_{k}^{n} \subset {E}_{k + 1}^{n} \), and \( {E}_{k}^{n} \nearrow E \) as \( k \) tends to infinity. By Corollary 3.3, we find that there exists \( {k}_{n} \) such that \( m\left( {E - {E}_{{k}_{n}}^{n}}\right) < \) \( 1/{2}^{n} \) . By construction, we then have\n\n\[ \n\left| {{f}_{j}\left( x\right) - f\left( x\right) }\right| < 1/n\;\text{ whenever }j > {k}_{n}\text{ and }x \in {E}_{{k}_{n}}^{n}.\n\]\n\nWe choose \( N \) so that \( \mathop{\sum }\limits_{{n = N}}^{\infty }{2}^{-n} < \epsilon /2 \), and let\n\n\[ \n{\widetilde{A}}_{\epsilon } = \mathop{\bigcap }\limits_{{n \geq N}}{E}_{{k}_{n}}^{n}\n\]\n\nWe first observe that\n\n\[ \nm\left( {E - {\widetilde{A}}_{\epsilon }}\right) \leq \mathop{\sum }\limits_{{n = N}}^{\infty }m\left( {E - {E}_{{k}_{n}}^{n}}\right) < \epsilon /2.\n\]\n\nNext, if \( \delta > 0 \), we choose \( n \geq N \) such that \( 1/n < \delta \), and note that \( x \in \) \( {\widetilde{A}}_{\epsilon } \) implies \( x \in {E}_{{k}_{n}}^{n} \) . We see therefore that \( \left| {{f}_{j}\left( x\right) - f\left( x\right) }\right| < \delta \) whenever \( j > {k}_{n} \) . Hence \( {f}_{k} \) converges uniformly to \( f \) on \( {\widetilde{A}}_{\epsilon } \) .\n\nFinally, using Theorem 3.4 choose a closed subset \( {A}_{\epsilon } \subset {\widetilde{A}}_{\epsilon } \) with \( m\left( {{\widetilde{A}}_{\epsilon } - }\right. \) \( \left. {A}_{\epsilon }\right) < \epsilon /2 \) . As a result, we have \( m\left( {E - {A}_{\epsilon }}\right) < \epsilon \) and the theorem is proved.
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Yes
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Theorem 4.5 (Lusin) Suppose \( f \) is measurable and finite valued on \( E \) with \( E \) of finite measure. Then for every \( \epsilon > 0 \) there exists a closed set \( {F}_{\epsilon } \), with\n\n\[ \n{F}_{\epsilon } \subset E,\;\text{ and }\;m\left( {E - {F}_{\epsilon }}\right) \leq \epsilon \n\]\n\nand such that \( {\left. f\right| }_{{F}_{\epsilon }} \) is continuous.
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Proof. Let \( {f}_{n} \) be a sequence of step functions so that \( {f}_{n} \rightarrow f \) a.e. Then we may find sets \( {E}_{n} \) so that \( m\left( {E}_{n}\right) < 1/{2}^{n} \) and \( {f}_{n} \) is continuous outside \( {E}_{n} \) . By Egorov’s theorem, we may find a set \( {A}_{\epsilon /3} \) on which \( {f}_{n} \rightarrow f \) uniformly and \( m\left( {E - {A}_{\epsilon /3}}\right) \leq \epsilon /3 \) . Then we consider\n\n\[ \n{F}^{\prime } = {A}_{\epsilon /3} - \mathop{\bigcup }\limits_{{n \geq N}}{E}_{n} \n\]\n\nfor \( N \) so large that \( \mathop{\sum }\limits_{{n > N}}1/{2}^{n} < \epsilon /3 \) . Now for every \( n \geq N \) the function \( {f}_{n} \) is continuous on \( {F}^{\prime } \) ; thus \( f \) (being the uniform limit of \( \left\{ {f}_{n}\right\} \) ) is also continuous on \( {F}^{\prime } \) . To finish the proof, we merely need to approximate the set \( {F}^{\prime } \) by a closed set \( {F}_{\epsilon } \subset {F}^{\prime } \) such that \( m\left( {{F}^{\prime } - {F}_{\epsilon }}\right) < \epsilon /3 \) .
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Yes
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Proposition 1.1 The integral of simple functions defined above satisfies the following properties:\n\n(i) Independence of the representation. If \( \varphi = \mathop{\sum }\limits_{{k = 1}}^{N}{a}_{k}{\chi }_{{E}_{k}} \) is any representation of \( \varphi \), then\n\n\[ \int \varphi = \mathop{\sum }\limits_{{k = 1}}^{N}{a}_{k}m\left( {E}_{k}\right) \]
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Proof. The only conclusion that is a little tricky is the first, which asserts that the integral of a simple function can be calculated by using any of its decompositions as a linear combination of characteristic functions.\n\nSuppose that \( \varphi = \mathop{\sum }\limits_{{k = 1}}^{N}{a}_{k}{\chi }_{{E}_{k}} \), where we assume that the sets \( {E}_{k} \) are disjoint, but we do not suppose that the numbers \( {a}_{k} \) are distinct and nonzero. For each distinct non-zero value \( a \) among the \( \left\{ {a}_{k}\right\} \) we define \( {E}_{a}^{\prime } = \) \( \bigcup {E}_{k} \), where the union is taken over those indices \( k \) such that \( {a}_{k} = a \) . Note then that the sets \( {E}_{a}^{\prime } \) are disjoint, and \( m\left( {E}_{a}^{\prime }\right) = \sum m\left( {E}_{k}\right) \), where\n\nthe sum is taken over the same set of \( k \)’s. Then clearly \( \varphi = \sum a{\chi }_{{E}_{a}^{\prime }} \) , where the sum is over the distinct non-zero values of \( \left\{ {a}_{k}\right\} \) . Thus\n\n\[ \int \varphi = \sum \operatorname{am}\left( {E}_{a}^{\prime }\right) = \mathop{\sum }\limits_{{k = 1}}^{N}{a}_{k}m\left( {E}_{k}\right) . \]
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Yes
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Lemma 1.2 Let \( f \) be a bounded function supported on a set \( E \) of finite measure. If \( {\left\{ {\varphi }_{n}\right\} }_{n = 1}^{\infty } \) is any sequence of simple functions bounded by \( M \) , supported on \( E \), and with \( {\varphi }_{n}\left( x\right) \rightarrow f\left( x\right) \) for a.e. \( x \), then:\n\n(i) The limit \( \mathop{\lim }\limits_{{n \rightarrow \infty }}\int {\varphi }_{n} \) exists.\n\n(ii) If \( f = 0 \) a.e., then the limit \( \mathop{\lim }\limits_{{n \rightarrow \infty }}\int {\varphi }_{n} \) equals 0 .
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Proof. The assertions of the lemma would be nearly obvious if we had that \( {\varphi }_{n} \) converges to \( f \) uniformly on \( E \) . Instead, we recall one of Littlewood's principles, which states that the convergence of a sequence of measurable functions is \
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No
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Proposition 1.3 Suppose \( f \) and \( g \) are bounded functions supported on sets of finite measure. Then the following properties hold.\n\n(i) Linearity. If \( a, b \in \mathbb{R} \), then\n\n\[ \int \left( {{af} + {bg}}\right) = a\int f + b\int g. \]\n\n(ii) Additivity. If \( E \) and \( F \) are disjoint subsets of \( {\mathbb{R}}^{d} \), then\n\n\[ {\int }_{E \cup F}f = {\int }_{E}f + {\int }_{F}f \]\n\n(iii) Monotonicity. If \( f \leq g \), then\n\n\[ \int f \leq \int g \]\n\n(iv) Triangle inequality. \( \left| f\right| \) is also bounded, supported on a set of finite measure, and\n\n\[ \left| {\int f}\right| \leq \int \left| f\right| \]
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All these properties follow by using approximations by simple functions, and the properties of the integral of simple functions given in Proposition 1.1.
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No
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Theorem 1.4 (Bounded convergence theorem) Suppose that \( \left\{ {f}_{n}\right\} \) is a sequence of measurable functions that are all bounded by \( M \), are supported on a set \( E \) of finite measure, and \( {f}_{n}\left( x\right) \rightarrow f\left( x\right) \) a.e. \( x \) as \( n \rightarrow \) \( \infty \) . Then \( f \) is measurable, bounded, supported on \( E \) for a.e. \( x \), and\n\n\[ \n\int \left| {{f}_{n} - f}\right| \rightarrow 0\;\text{ as }n \rightarrow \infty .\n\]\n\nConsequently,\n\n\[ \n\int {f}_{n} \rightarrow \int f\;\text{ as }n \rightarrow \infty \n\]
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Proof. From the assumptions one sees at once that \( f \) is bounded by \( M \) almost everywhere and vanishes outside \( E \), except possibly on a set of measure zero. Clearly, the triangle inequality for the integral implies that it suffices to prove that \( \int \left| {{f}_{n} - f}\right| \rightarrow 0 \) as \( n \) tends to infinity.\n\nThe proof is a reprise of the argument in Lemma 1.2. Given \( \epsilon > 0 \), we may find, by Egorov’s theorem, a measurable subset \( {A}_{\epsilon } \) of \( E \) such that \( m\left( {E - {A}_{\epsilon }}\right) \leq \epsilon \) and \( {f}_{n} \rightarrow f \) uniformly on \( {A}_{\epsilon } \) . Then, we know that for all sufficiently large \( n \) we have \( \left| {{f}_{n}\left( x\right) - f\left( x\right) }\right| \leq \epsilon \) for all \( x \in {A}_{\epsilon } \) . Putting these facts together yields\n\n\[ \n\int \left| {{f}_{n}\left( x\right) - f\left( x\right) }\right| {dx} \leq {\int }_{{A}_{\epsilon }}\left| {{f}_{n}\left( x\right) - f\left( x\right) }\right| {dx} + {\int }_{E - {A}_{\epsilon }}\left| {{f}_{n}\left( x\right) - f\left( x\right) }\right| {dx}\n\]\n\n\[ \n\leq {\epsilon m}\left( E\right) + {2Mm}\left( {E - {A}_{\epsilon }}\right)\n\]\n\nfor all large \( n \) . Since \( \epsilon \) is arbitrary, the proof of the theorem is complete.
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Yes
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Proposition 1.6 The integral of non-negative measurable functions enjoys the following properties:\n\n(i) Linearity. If \( f, g \geq 0 \), and \( a, b \) are positive real numbers, then\n\n\[ \n\int \left( {{af} + {bg}}\right) = a\int f + b\int g.\n\]
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Proof. Of the first four assertions, only (i) is not an immediate consequence of the definitions, and to prove it we argue as follows. We take \( a = b = 1 \) and note that if \( \varphi \leq f \) and \( \psi \leq g \), where both \( \varphi \) and \( \psi \) are bounded and supported on sets of finite measure, then \( \varphi + \psi \leq f + g \) , and \( \varphi + \psi \) is also bounded and supported on a set of finite measure. Consequently\n\n\[ \n\int f + \int g \leq \int \left( {f + g}\right)\n\]\n\nTo prove the reverse inequality, suppose \( \eta \) is bounded and supported on a set of finite measure, and \( \eta \leq f + g \) . If we define \( {\eta }_{1}\left( x\right) = \min \left( {f\left( x\right) ,\eta \left( x\right) }\right) \) and \( {\eta }_{2} = \eta - {\eta }_{1} \), we note that\n\n\[ \n{\eta }_{1} \leq f\;\text{ and }\;{\eta }_{2} \leq g.\n\]\n\nMoreover both \( {\eta }_{1},{\eta }_{2} \) are bounded and supported on sets of finite measure. Hence\n\n\[ \n\int \eta = \int \left( {{\eta }_{1} + {\eta }_{2}}\right) = \int {\eta }_{1} + \int {\eta }_{2} \leq \int f + \int g.\n\]\n\nTaking the supremum over \( \eta \) yields the required inequality.
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Yes
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Lemma 1.7 (Fatou) Suppose \( \\left\\{ {f}_{n}\\right\\} \) is a sequence of measurable functions with \( {f}_{n} \\geq 0 \) . If \( \\mathop{\\lim }\\limits_{{n \\rightarrow \\infty }}{f}_{n}\\left( x\\right) = f\\left( x\\right) \) for a.e. \( x \), then
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Proof. Suppose \( 0 \\leq g \\leq f \), where \( g \) is bounded and supported on a set \( E \) of finite measure. If we set \( {g}_{n}\\left( x\\right) = \\min \\left( {g\\left( x\\right) ,{f}_{n}\\left( x\\right) }\\right) \), then \( {g}_{n} \) is measurable, supported on \( E \), and \( {g}_{n}\\left( x\\right) \\rightarrow g\\left( x\\right) \) a.e., so by the bounded convergence theorem\n\n\[ \n\\int {g}_{n} \\rightarrow \\int g \n\]\n\nBy construction, we also have \( {g}_{n} \\leq {f}_{n} \), so that \( \\int {g}_{n} \\leq \\int {f}_{n} \), and therefore\n\n\[ \n\\int g \\leq \\mathop{\\liminf }\\limits_{{n \\rightarrow \\infty }}\\int {f}_{n} \n\]\n\nTaking the supremum over all \( g \) yields the desired inequality.
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Yes
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Corollary 1.8 Suppose \( f \) is a non-negative measurable function, and \( \left\{ {f}_{n}\right\} \) a sequence of non-negative measurable functions with \( {f}_{n}\left( x\right) \leq f\left( x\right) \) and \( {f}_{n}\left( x\right) \rightarrow f\left( x\right) \) for almost every \( x \) . Then\n\n\[ \mathop{\lim }\limits_{{n \rightarrow \infty }}\int {f}_{n} = \int f \]
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Proof. Since \( {f}_{n}\left( x\right) \leq f\left( x\right) \) a.e \( x \), we necessarily have \( \int {f}_{n} \leq \int f \) for all \( n \) ; hence\n\n\[ \mathop{\limsup }\limits_{{n \rightarrow \infty }}\int {f}_{n} \leq \int f \]\n\n## This inequality combined with Fatou's lemma proves the desired limit.
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No
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Corollary 1.10 Consider a series \( \mathop{\sum }\limits_{{k = 1}}^{\infty }{a}_{k}\left( x\right) \), where \( {a}_{k}\left( x\right) \geq 0 \) is measurable for every \( k \geq 1 \) . Then\n\n\[ \n\int \mathop{\sum }\limits_{{k = 1}}^{\infty }{a}_{k}\left( x\right) {dx} = \mathop{\sum }\limits_{{k = 1}}^{\infty }\int {a}_{k}\left( x\right) {dx}.\n\]\n\nIf \( \mathop{\sum }\limits_{{k = 1}}^{\infty }\int {a}_{k}\left( x\right) {dx} \) is finite, then the series \( \mathop{\sum }\limits_{{k = 1}}^{\infty }{a}_{k}\left( x\right) \) converges for a.e. \( x \) .
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Proof. Let \( {f}_{n}\left( x\right) = \mathop{\sum }\limits_{{k = 1}}^{n}{a}_{k}\left( x\right) \) and \( f\left( x\right) = \mathop{\sum }\limits_{{k = 1}}^{\infty }{a}_{k}\left( x\right) \) . The functions \( {f}_{n} \) are measurable, \( {f}_{n}\left( x\right) \leq {f}_{n + 1}\left( x\right) \), and \( {f}_{n}\left( x\right) \rightarrow f\left( x\right) \) as \( n \) tends to infinity. Since\n\n\[ \n\int {f}_{n} = \mathop{\sum }\limits_{{k = 1}}^{n}\int {a}_{k}\left( x\right) {dx}\n\]\n\nthe monotone convergence theorem implies\n\n\[ \n\mathop{\sum }\limits_{{k = 1}}^{\infty }\int {a}_{k}\left( x\right) {dx} = \int \mathop{\sum }\limits_{{k = 1}}^{\infty }{a}_{k}\left( x\right) {dx}.\n\]\n\nIf \( \sum \int {a}_{k} < \infty \), then the above implies that \( \mathop{\sum }\limits_{{k = 1}}^{\infty }{a}_{k}\left( x\right) \) is integrable, and by our earlier observation, we conclude that \( \mathop{\sum }\limits_{{k = 1}}^{\infty }{a}_{k}\left( x\right) \) is finite almost everywhere.
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Yes
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Proposition 1.12 Suppose \( f \) is integrable on \( {\mathbb{R}}^{d} \) . Then for every \( \epsilon > 0 \) :\n\n(i) There exists a set of finite measure \( B \) (a ball, for example) such that\n\n\[{\int }_{{B}^{c}}\left| f\right| < \epsilon\]\n\n(ii) There is a \( \delta > 0 \) such that\n\n\[{\int }_{E}\left| f\right| < \epsilon \;\text{ whenever }m\left( E\right) < \delta .
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Proof. By replacing \( f \) with \( \left| f\right| \) we may assume without loss of generality that \( f \geq 0 \) .\n\nFor the first part, let \( {B}_{N} \) denote the ball of radius \( N \) centered at the origin, and note that if \( {f}_{N}\left( x\right) = f\left( x\right) {\chi }_{{B}_{N}}\left( x\right) \), then \( {f}_{N} \geq 0 \) is measurable, \( {f}_{N}\left( x\right) \leq {f}_{N + 1}\left( x\right) \), and \( \mathop{\lim }\limits_{{N \rightarrow \infty }}{f}_{N}\left( x\right) = f\left( x\right) \) . By the monotone convergence theorem, we must have\n\n\[ \mathop{\lim }\limits_{{N \rightarrow \infty }}\int {f}_{N} = \int f \]\n\nIn particular, for some large \( N \) ,\n\n\[ 0 \leq \int f - \int f{\chi }_{{B}_{N}} < \epsilon \]\n\nand since \( 1 - {\chi }_{{B}_{N}} = {\chi }_{{B}_{N}^{c}} \), this implies \( {\int }_{{B}_{N}^{c}}f < \epsilon \), as we set out to prove.\n\nFor the second part, assuming again that \( f \geq 0 \), we let \( {f}_{N}\left( x\right) = f\left( x\right) {\chi }_{{E}_{N}} \) where\n\n\[ {E}_{N} = \{ x : f\left( x\right) \leq N\} . \]\n\nOnce again, \( {f}_{N} \geq 0 \) is measurable, \( {f}_{N}\left( x\right) \leq {f}_{N + 1}\left( x\right) \), and given \( \epsilon > 0 \) there exists (by the monotone convergence theorem) an integer \( N > 0 \) such that\n\n\[ \int \left( {f - {f}_{N}}\right) < \frac{\epsilon }{2} \]\n\nWe now pick \( \delta > 0 \) so that \( {N\delta } < \epsilon /2 \) . If \( m\left( E\right) < \delta \), then\n\n\[ {\int }_{E}f = {\int }_{E}\left( {f - {f}_{N}}\right) + {\int }_{E}{f}_{N} \]\n\n\[ \leq \int \left( {f - {f}_{N}}\right) + {\int }_{E}{f}_{N} \]\n\n\[ \leq \int \left( {f - {f}_{N}}\right) + {Nm}\left( E\right) \]\n\n\[ \leq \frac{\epsilon }{2} + \frac{\epsilon }{2} = \epsilon \]\n\nThis concludes the proof of the proposition.
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Yes
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Theorem 1.13 Suppose \( \left\{ {f}_{n}\right\} \) is a sequence of measurable functions such that \( {f}_{n}\left( x\right) \rightarrow f\left( x\right) \) a.e. \( x \), as \( n \) tends to infinity. If \( \left| {{f}_{n}\left( x\right) }\right| \leq g\left( x\right) \), where \( g \) is integrable, then\n\n\[ \n\int \left| {{f}_{n} - f}\right| \rightarrow 0\;\text{ as }n \rightarrow \infty \n\]\n\nand consequently\n\n\[ \n\int {f}_{n} \rightarrow \int f\;\text{ as }n \rightarrow \infty \n\]
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Proof. For each \( N \geq 0 \) let \( {E}_{N} = \{ x : \left| x\right| \leq N, g\left( x\right) \leq N\} \) . Given \( \epsilon > 0 \), we may argue as in the first part of the previous lemma, to see that there exists \( N \) so that \( {\int }_{{E}_{N}^{c}}g < \epsilon \) . Then the functions \( {f}_{n}{\chi }_{{E}_{N}} \) are bounded (by \( N \) ) and supported on a set of finite measure, so that by the bounded convergence theorem, we have\n\n\[ \n{\int }_{{E}_{N}}\left| {{f}_{n} - f}\right| < \epsilon ,\;\text{ for all large }n.\n\]\n\nHence, we obtain the estimate\n\n\[ \n\int \left| {{f}_{n} - f}\right| = {\int }_{{E}_{N}}\left| {{f}_{n} - f}\right| + {\int }_{{E}_{N}^{c}}\left| {{f}_{n} - f}\right| \n\]\n\n\[ \n\leq {\int }_{{E}_{N}}\left| {{f}_{n} - f}\right| + 2{\int }_{{E}_{N}^{c}}g \n\]\n\n\[ \n\leq \epsilon + {2\epsilon } = {3\epsilon } \n\]\n\nfor all large \( n \) . This proves the theorem.
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Yes
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Theorem 2.4 The following families of functions are dense in \( {L}^{1}\left( {\mathbb{R}}^{d}\right) \) :\n\n(i) The simple functions.\n\n(ii) The step functions.\n\n(iii) The continuous functions of compact support.
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Proof. Let \( f \) be an integrable function on \( {\mathbb{R}}^{d} \) . First, we may assume that \( f \) is real-valued, because we may approximate its real and imaginary parts independently. If this is the case, we may then write \( f = {f}^{ + } - {f}^{ - } \) , where \( {f}^{ + },{f}^{ - } \geq 0 \), and it now suffices to prove the theorem when \( f \geq 0 \) .\n\nFor (i), Theorem 4.1 in Chapter 1 guarantees the existence of a sequence \( \left\{ {\varphi }_{k}\right\} \) of non-negative simple functions that increase to \( f \) pointwise. By the dominated convergence theorem (or even simply the monotone convergence theorem) we then have\n\n\[ \n{\begin{Vmatrix}f - {\varphi }_{k}\end{Vmatrix}}_{{L}^{1}} \rightarrow 0\;\text{ as }k \rightarrow \infty .\n\]\n\nThus there are simple functions that are arbitrarily close to \( f \) in the \( {L}^{1} \) norm.\n\nFor (ii), we first note that by (i) it suffices to approximate simple functions by step functions. Then, we recall that a simple function is a finite linear combination of characteristic functions of sets of finite measure, so it suffices to show that if \( E \) is such a set, then there is a step function \( \psi \) so that \( {\begin{Vmatrix}{\chi }_{E} - \psi \end{Vmatrix}}_{{L}^{1}} \) is small. However, we now recall that this argument was already carried out in the proof of Theorem 4.3, Chapter 1. Indeed, there it is shown that there is an almost disjoint family of rectangles \( \left\{ {R}_{j}\right\} \) with \( m\left( {E\bigtriangleup \mathop{\bigcup }\limits_{{j = 1}}^{M}{R}_{j}}\right) \leq {2\epsilon } \) . Thus \( {\chi }_{E} \) and \( \psi = \) \( \mathop{\sum }\limits_{j}{\chi }_{{R}_{j}} \) differ at most on a set of measure \( {2\epsilon } \), and as a result we find that \( {\begin{Vmatrix}{\chi }_{E} - \psi \end{Vmatrix}}_{{L}^{1}} < {2\epsilon } \).\n\nBy (ii), it suffices to establish (iii) when \( f \) is the characteristic function of a rectangle. In the one-dimensional case, where \( f \) is the characteristic function of an interval \( \left\lbrack {a, b}\right\rbrack \), we may choose a continuous piecewise linear function \( g \) defined by\n\n\[ \ng\left( x\right) = \left\{ \begin{array}{ll} 1 & \text{ if }a \leq x \leq b \\ 0 & \text{ if }x \leq a - \epsilon \text{ or }x \geq b + \epsilon \end{array}\right.\n\]\n\nand with \( g \) linear on the intervals \( \left\lbrack {a - \epsilon, a}\right\rbrack \) and \( \left\lbrack {b, b + \epsilon }\right\rbrack \) . Then \( \parallel f - \) \( g{\parallel }_{{L}^{1}} < {2\epsilon } \) . In \( d \) dimensions, it suffices to note that the characteristic function of a rectangle is the product of characteristic functions of intervals. Then, the desired continuous function of compact support is simply the product of functions like \( g \) defined above.
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Yes
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Proposition 2.5 Suppose \( f \in {L}^{1}\left( {\mathbb{R}}^{d}\right) \) . Then\n\n\[ \n{\begin{Vmatrix}{f}_{h} - f\end{Vmatrix}}_{{L}^{1}} \rightarrow 0\;\text{ as }h \rightarrow 0.\n\]
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The proof is a simple consequence of the approximation of integrable functions by continuous functions of compact support as given in Theorem 2.4. In fact for any \( \epsilon > 0 \), we can find such a function \( g \) so that \( \parallel f - g\parallel < \epsilon \) . Now\n\n\[ \n{f}_{h} - f = \left( {{g}_{h} - g}\right) + \left( {{f}_{h} - {g}_{h}}\right) - \left( {f - g}\right) .\n\]\n\nHowever, \( \begin{Vmatrix}{{f}_{h} - {g}_{h}}\end{Vmatrix} = \parallel f - g\parallel < \epsilon \), while since \( g \) is continuous and has compact support we have that clearly\n\n\[ \n\begin{Vmatrix}{{g}_{h} - g}\end{Vmatrix} = {\int }_{{\mathbb{R}}^{d}}\left| {g\left( {x - h}\right) - g\left( x\right) }\right| {dx} \rightarrow 0\;\text{ as }h \rightarrow 0.\n\]\n\nSo if \( \left| h\right| < \delta \), where \( \delta \) is sufficiently small, then \( \begin{Vmatrix}{{g}_{h} - g}\end{Vmatrix} < \epsilon \), and as a result \( \begin{Vmatrix}{{f}_{h} - f}\end{Vmatrix} < {3\epsilon } \), whenever \( \left| h\right| < \delta \) .
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Yes
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Theorem 3.1 Suppose \( f\left( {x, y}\right) \) is integrable on \( {\mathbb{R}}^{{d}_{1}} \times {\mathbb{R}}^{{d}_{2}} \) . Then for almost every \( y \in {\mathbb{R}}^{{d}_{2}} \) :\n\n(i) The slice \( {f}^{y} \) is integrable on \( {\mathbb{R}}^{{d}_{1}} \) .\n\n(ii) The function defined by \( {\int }_{{\mathbb{R}}^{{d}_{1}}}{f}^{y}\left( x\right) {dx} \) is integrable on \( {\mathbb{R}}^{{d}_{2}} \) .\n\nMoreover:\n\n(iii) \( {\int }_{{\mathbb{R}}^{{d}_{2}}}\left( {{\int }_{{\mathbb{R}}^{{d}_{1}}}f\left( {x, y}\right) {dx}}\right) {dy} = {\int }_{{\mathbb{R}}^{d}}f \) .
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We first note that we may assume that \( f \) is real-valued, since the theorem then applies to the real and imaginary parts of a complex-valued function. The proof of Fubini's theorem which we give next consists of a sequence of six steps. We begin by letting \( \mathcal{F} \) denote the set of integrable functions on \( {\mathbb{R}}^{d} \) which satisfy all three conclusions in the theorem, and set out to prove that \( {L}^{1}\left( {\mathbb{R}}^{d}\right) \subset \mathcal{F} \) .\n\nWe proceed by first showing that \( \mathcal{F} \) is closed under operations such as linear combinations (Step 1) and limits (Step 2). Then we begin to construct families of functions in \( \mathcal{F} \) . Since any integrable function is the \
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No
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Theorem 3.2 Suppose \( f\left( {x, y}\right) \) is a non-negative measurable function on \( {\mathbb{R}}^{{d}_{1}} \times {\mathbb{R}}^{{d}_{2}} \) . Then for almost every \( y \in {\mathbb{R}}^{{d}_{2}} \) :\n\n(i) The slice \( {f}^{y} \) is measurable on \( {\mathbb{R}}^{{d}_{1}} \) .\n\n(ii) The function defined by \( {\int }_{{\mathbb{R}}^{{d}_{1}}}{f}^{y}\left( x\right) {dx} \) is measurable on \( {\mathbb{R}}^{{d}_{2}} \) .\n\nMoreover:\n\n(iii) \( {\int }_{{\mathbb{R}}^{{d}_{2}}}\left( {{\int }_{{\mathbb{R}}^{{d}_{1}}}f\left( {x, y}\right) {dx}}\right) {dy} = {\int }_{{\mathbb{R}}^{d}}f\left( {x, y}\right) {dxdy} \) in the extended sense.
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Proof of Theorem 3.2. Consider the truncations\n\n\[ \n{f}_{k}\left( {x, y}\right) = \left\{ \begin{matrix} f\left( {x, y}\right) & \text{ if }\left| \left( {x, y}\right) \right| < k\text{ and }f\left( {x, y}\right) < k, \\ 0 & \text{ otherwise. } \end{matrix}\right. \]\n\nEach \( {f}_{k} \) is integrable, and by part (i) in Fubini’s theorem there exists a set \( {E}_{k} \subset {\mathbb{R}}^{{d}_{2}} \) of measure 0 such that the slice \( {f}_{k}^{y}\left( x\right) \) is measurable for all \( y \in {E}_{k}^{c} \) . Then, if we set \( E = \mathop{\bigcup }\limits_{k}{E}_{k} \), we find that \( {f}^{y}\left( x\right) \) is measurable for all \( y \in {E}^{c} \) and all \( k \) . Moreover, \( m\left( E\right) = 0 \) . Since \( {f}_{k}^{y} \nearrow {f}^{y} \), the monotone convergence theorem implies that if \( y \notin E \), then\n\n\[ \n{\int }_{{\mathbb{R}}^{{d}_{1}}}{f}_{k}\left( {x, y}\right) {dx} \nearrow {\int }_{{\mathbb{R}}^{{d}_{1}}}f\left( {x, y}\right) {dx}\;\text{ as }k \rightarrow \infty . \]\n\nAgain by Fubini’s theorem, \( {\int }_{{\mathbb{R}}^{{d}_{1}}}{f}_{k}\left( {x, y}\right) {dx} \) is measurable for all \( y \in {E}^{c} \) , hence so is \( {\int }_{{\mathbb{R}}^{{d}_{1}}}f\left( {x, y}\right) {dx} \) . Another application of the monotone convergence theorem then gives\n\n(11)\n\n\[ \n{\int }_{{\mathbb{R}}^{{d}_{2}}}\left( {{\int }_{{\mathbb{R}}^{{d}_{1}}}{f}_{k}\left( {x, y}\right) {dx}}\right) {dy} \rightarrow {\int }_{{\mathbb{R}}^{{d}_{2}}}\left( {{\int }_{{\mathbb{R}}^{{d}_{1}}}f\left( {x, y}\right) {dx}}\right) {dy}. \]\n\nBy part (iii) in Fubini's theorem we know that\n\n(12)\n\n\[ \n{\int }_{{\mathbb{R}}^{{d}_{2}}}\left( {{\int }_{{\mathbb{R}}^{{d}_{1}}}{f}_{k}\left( {x, y}\right) {dx}}\right) {dy} = {\int }_{{\mathbb{R}}^{d}}{f}_{k}. \]\n\nA final application of the monotone convergence theorem directly to \( {f}_{k} \) also gives\n\n(13)\n\n\[ \n{\int }_{{\mathbb{R}}^{d}}{f}_{k} \rightarrow {\int }_{{\mathbb{R}}^{d}}f \]\n\nCombining (11), (12), and (13) completes the proof of Theorem 3.2.
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Yes
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Corollary 3.3 If \( E \) is a measurable set in \( {\mathbb{R}}^{{d}_{1}} \times {\mathbb{R}}^{{d}_{2}} \), then for almost every \( y \in {\mathbb{R}}^{{d}_{2}} \) the slice\n\n\[ \n{E}^{y} = \left\{ {x \in {\mathbb{R}}^{{d}_{1}} : \left( {x, y}\right) \in E}\right\}\n\]\n\nis a measurable subset of \( {\mathbb{R}}^{{d}_{1}} \) . Moreover \( m\left( {E}^{y}\right) \) is a measurable function of \( y \) and\n\n\[ \nm\left( E\right) = {\int }_{{\mathbb{R}}^{{d}_{2}}}m\left( {E}^{y}\right) {dy}\n\]\n
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This is an immediate consequence of the first part of Theorem 3.2 applied to the function \( {\chi }_{E} \) . Clearly a symmetric result holds for the \( x \) -slices in \( {\mathbb{R}}^{{d}_{2}} \) .
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Yes
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Proposition 3.4 If \( E = {E}_{1} \times {E}_{2} \) is a measurable subset of \( {\mathbb{R}}^{d} \), and \( {m}_{ * }\left( {E}_{2}\right) > 0 \), then \( {E}_{1} \) is measurable.
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Proof. By Corollary 3.3, we know that for a.e. \( y \in {\mathbb{R}}^{{d}_{2}} \), the slice function\n\n\[ \n{\left( {\chi }_{{E}_{1} \times {E}_{2}}\right) }^{y}\left( x\right) = {\chi }_{{E}_{1}}\left( x\right) {\chi }_{{E}_{2}}\left( y\right) \n\]\n\nis measurable as a function of \( x \) . In fact, we claim that there is some \( y \in {E}_{2} \) such that the above slice function is measurable in \( x \) ; for such a \( y \) we would have \( {\chi }_{{E}_{1} \times {E}_{2}}\left( {x, y}\right) = {\chi }_{{E}_{1}}\left( x\right) \), and this would imply that \( {E}_{1} \) is measurable.\n\nTo prove the existence of such a \( y \), we use the assumption that \( {m}_{ * }\left( {E}_{2}\right) > \) 0 . Indeed, let \( F \) denote the set of \( y \in {\mathbb{R}}^{{d}_{2}} \) such that the slice \( {E}^{y} \) is measurable. Then \( m\left( {F}^{c}\right) = 0 \) (by the previous corollary). However, \( {E}_{2} \cap F \) is not empty because \( {m}_{ * }\left( {{E}_{2} \cap F}\right) > 0 \) . To see this, note that \( {E}_{2} = \left( {{E}_{2} \cap F}\right) \bigcup \left( {{E}_{2} \cap {F}^{c}}\right) \), hence\n\n\[ \n0 < {m}_{ * }\left( {E}_{2}\right) \leq {m}_{ * }\left( {{E}_{2} \cap F}\right) + {m}_{ * }\left( {{E}_{2} \cap {F}^{c}}\right) = {m}_{ * }\left( {{E}_{2} \cap F}\right) , \n\]\n\nbecause \( {E}_{2} \cap {F}^{c} \) is a subset of a set of measure zero.
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Yes
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Lemma 3.5 If \( {E}_{1} \subset {\mathbb{R}}^{{d}_{1}} \) and \( {E}_{2} \subset {\mathbb{R}}^{{d}_{2}} \), then\n\n\[ \n{m}_{ * }\left( {{E}_{1} \times {E}_{2}}\right) \leq {m}_{ * }\left( {E}_{1}\right) {m}_{ * }\left( {E}_{2}\right) \n\]\n\nwith the understanding that if one of the sets \( {E}_{j} \) has exterior measure zero, then \( {m}_{ * }\left( {{E}_{1} \times {E}_{2}}\right) = 0 \) .
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Proof. Let \( \epsilon > 0 \) . By definition, we can find cubes \( {\left\{ {Q}_{k}\right\} }_{k = 1}^{\infty } \) in \( {\mathbb{R}}^{{d}_{1}} \) and \( {\left\{ {Q}_{\ell }^{\prime }\right\} }_{\ell = 1}^{\infty } \) in \( {\mathbb{R}}^{{d}_{2}} \) such that\n\n\[ \n{E}_{1} \subset \mathop{\bigcup }\limits_{{k = 1}}^{\infty }{Q}_{k},\;\text{ and }\;{E}_{2} \subset \mathop{\bigcup }\limits_{{\ell = 1}}^{\infty }{Q}_{\ell }^{\prime } \n\]\n\nand\n\n\[ \n\mathop{\sum }\limits_{{k = 1}}^{\infty }\left| {Q}_{k}\right| \leq {m}_{ * }\left( {E}_{1}\right) + \epsilon \;\text{ and }\;\mathop{\sum }\limits_{{\ell = 1}}^{\infty }\left| {Q}_{\ell }^{\prime }\right| \leq {m}_{ * }\left( {E}_{2}\right) + \epsilon . \n\]\n\nSince \( {E}_{1} \times {E}_{2} \subset \mathop{\bigcup }\limits_{{k,\ell = 1}}^{\infty }{Q}_{k} \times {Q}_{\ell }^{\prime } \), the sub-additivity of the exterior measure yields\n\n\[ \n{m}_{ * }\left( {{E}_{1} \times {E}_{2}}\right) \leq \mathop{\sum }\limits_{{k,\ell = 1}}^{\infty }\left| {{Q}_{k} \times {Q}_{\ell }^{\prime }}\right| \n\]\n\n\[ \n= \left( {\mathop{\sum }\limits_{{k = 1}}^{\infty }\left| {Q}_{k}\right| }\right) \left( {\mathop{\sum }\limits_{{\ell = 1}}^{\infty }\left| {Q}_{\ell }^{\prime }\right| }\right) \n\]\n\n\[ \n\leq \left( {{m}_{ * }\left( {E}_{1}\right) + \epsilon }\right) \left( {{m}_{ * }\left( {E}_{2}\right) + \epsilon }\right) \n\]\n\nIf neither \( {E}_{1} \) nor \( {E}_{2} \) has exterior measure 0, then from the above we find\n\n\[ \n{m}_{ * }\left( {{E}_{1} \times {E}_{2}}\right) \leq {m}_{ * }\left( {E}_{1}\right) {m}_{ * }\left( {E}_{2}\right) + O\left( \epsilon \right) \n\]\n\nand since \( \epsilon \) is arbitrary, we must have \( {m}_{ * }\left( {{E}_{1} \times {E}_{2}}\right) \leq {m}_{ * }\left( {E}_{1}\right) {m}_{ * }\left( {E}_{2}\right) \) .\n\nIf for instance \( {m}_{ * }\left( {E}_{1}\right) = 0 \), consider for each positive integer \( j \) the set \( {E}_{2}^{j} = {E}_{2} \cap \left\{ {y \in {\mathbb{R}}^{{d}_{2}} : \left| y\right| \leq j}\right\} \) . Then, by the above argument, we find that \( {m}_{ * }\left( {{E}_{1} \times {E}_{2}^{j}}\right) = 0 \) . Since \( \left( {{E}_{1} \times {E}_{2}^{j}}\right) \nearrow \left( {{E}_{1} \times {E}_{2}}\right) \) as \( j \rightarrow \infty \), we conclude that \( {m}_{ * }\left( {{E}_{1} \times {E}_{2}}\right) = 0 \) .
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Yes
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Proposition 3.6 Suppose \( {E}_{1} \) and \( {E}_{2} \) are measurable subsets of \( {\mathbb{R}}^{{d}_{1}} \) and \( {\mathbb{R}}^{{d}_{2}} \), respectively. Then \( E = {E}_{1} \times {E}_{2} \) is a measurable subset of \( {\mathbb{R}}^{d} \) . Moreover,\n\n\[ m\left( E\right) = m\left( {E}_{1}\right) m\left( {E}_{2}\right) \]\n\nwith the understanding that if one of the sets \( {E}_{j} \) has measure zero, then \( m\left( E\right) = 0 \) .
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Proof. It suffices to prove that \( E \) is measurable, because then the assertion about \( m\left( E\right) \) follows from Corollary 3.3. Since each set \( {E}_{j} \) is measurable, there exist sets \( {G}_{j} \subset {\mathbb{R}}^{{d}_{j}} \) of type \( {G}_{\delta } \), with \( {G}_{j} \supset {E}_{j} \) and \( {m}_{ * }\left( {{G}_{j} - {E}_{j}}\right) = 0 \) for each \( j = 1,2 \) . (See Corollary 3.5 in Chapter 1.) Clearly, \( G = {G}_{1} \times {G}_{2} \) is measurable in \( {\mathbb{R}}^{{d}_{1}} \times {\mathbb{R}}^{{d}_{2}} \) and\n\n\[ \left( {{G}_{1} \times {G}_{2}}\right) - \left( {{E}_{1} \times {E}_{2}}\right) \subset \left( {\left( {{G}_{1} - {E}_{1}}\right) \times {G}_{2}}\right) \cup \left( {{G}_{1} \times \left( {{G}_{2} - {E}_{2}}\right) }\right) .\n\nBy the lemma we conclude that \( {m}_{ * }\left( {G - E}\right) = 0 \), hence \( E \) is measurable.
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Yes
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Corollary 3.7 Suppose \( f \) is a measurable function on \( {\mathbb{R}}^{{d}_{1}} \) . Then the function \( \widetilde{f} \) defined by \( \widetilde{f}\left( {x, y}\right) = f\left( x\right) \) is measurable on \( {\mathbb{R}}^{{d}_{1}} \times {\mathbb{R}}^{{d}_{2}} \) .
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Proof. To see this, we may assume that \( f \) is real-valued, and recall first that if \( a \in \mathbb{R} \) and \( {E}_{1} = \left\{ {x \in {\mathbb{R}}^{{d}_{1}} : f\left( x\right) < a}\right\} \), then \( {E}_{1} \) is measurable by definition. Since\n\n\[ \left\{ {\left( {x, y}\right) \in {\mathbb{R}}^{{d}_{1}} \times {\mathbb{R}}^{{d}_{2}} : \widetilde{f}\left( {x, y}\right) < a}\right\} = {E}_{1} \times {\mathbb{R}}^{{d}_{2}}, \]\n\nthe previous proposition shows that \( \{ \widetilde{f}\left( {x, y}\right) < a\} \) is measurable for each \( a \in \mathbb{R} \) . Thus \( \widetilde{f}\left( {x, y}\right) \) is a measurable function on \( {\mathbb{R}}^{{d}_{1}} \times {\mathbb{R}}^{{d}_{2}} \), as desired.
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Yes
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Corollary 3.8 Suppose \( f\left( x\right) \) is a non-negative function on \( {\mathbb{R}}^{d} \), and let\n\n\[ \mathcal{A} = \left\{ {\left( {x, y}\right) \in {\mathbb{R}}^{d} \times \mathbb{R} : 0 \leq y \leq f\left( x\right) }\right\} \]\n\nThen:\n\n(i) \( f \) is measurable on \( {\mathbb{R}}^{d} \) if and only if \( \mathcal{A} \) is measurable in \( {\mathbb{R}}^{d + 1} \) .\n\n(ii) If the conditions in (i) hold, then\n\n\[ {\int }_{{\mathbb{R}}^{d}}f\left( x\right) {dx} = m\left( \mathcal{A}\right) \]
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Proof. If \( f \) is measurable on \( {\mathbb{R}}^{d} \), then the previous proposition guarantees that the function\n\n\[ F\left( {x, y}\right) = y - f\left( x\right) \]\nis measurable on \( {\mathbb{R}}^{d + 1} \), so we immediately see that \( \mathcal{A} = \{ y \geq 0\} \cap \{ F \leq \) \( 0\} \) is measurable.\n\nConversely, suppose that \( \mathcal{A} \) is measurable. We note that for each \( x \in {\mathbb{R}}^{{d}_{1}} \) the slice \( {\mathcal{A}}_{x} = \{ y \in \mathbb{R} : \left( {x, y}\right) \in \mathcal{A}\} \) is a closed segment, namely \( {\mathcal{A}}_{x} = \left\lbrack {0, f\left( x\right) }\right\rbrack \) . Consequently Corollary 3.3 (with the roles of \( x \) and \( y \) interchanged) yields the measurability of \( m\left( {\mathcal{A}}_{x}\right) = f\left( x\right) \) . Moreover\n\n\[ m\left( \mathcal{A}\right) = \int {\chi }_{\mathcal{A}}\left( {x, y}\right) {dxdy} = {\int }_{{\mathbb{R}}^{{d}_{1}}}m\left( {\mathcal{A}}_{x}\right) {dx} = {\int }_{{\mathbb{R}}^{{d}_{1}}}f\left( x\right) {dx} \]\n\nas was to be shown.
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Yes
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Proposition 3.9 If \( f \) is a measurable function on \( {\mathbb{R}}^{d} \), then the function \( \widetilde{f}\left( {x, y}\right) = f\left( {x - y}\right) \) is measurable on \( {\mathbb{R}}^{d} \times {\mathbb{R}}^{d} \) .
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By picking \( E = \left\{ {z \in {\mathbb{R}}^{d} : f\left( z\right) < a}\right\} \), we see that it suffices to prove that whenever \( E \) is a measurable subset of \( {\mathbb{R}}^{d} \), then \( \widetilde{E} = \{ \left( {x, y}\right) : x - y \in \) \( E\} \) is a measurable subset of \( {\mathbb{R}}^{d} \times {\mathbb{R}}^{d} \) . Note first that if \( \mathcal{O} \) is an open set, then \( \widetilde{\mathcal{O}} \) is also open. Taking countable intersections shows that if \( E \) is a \( {G}_{\delta } \) set, then so is \( \widetilde{E} \) . Assume now that \( m\left( {\widetilde{E}}_{k}\right) = 0 \) for each \( k \), where \( {\widetilde{E}}_{k} = \widetilde{E} \cap {B}_{k} \) and \( {B}_{k} = \{ \left| y\right| < k\} \) . Again, take \( \mathcal{O} \) to be open in \( {\mathbb{R}}^{d} \), and let us calculate \( m\left( {\widetilde{\mathcal{O}} \cap {B}_{k}}\right) \) . We have that \( {\chi }_{\widetilde{\mathcal{O}} \cap {B}_{k}} = {\chi }_{\mathcal{O}}\left( {x - y}\right) {\chi }_{{B}_{k}}\left( y\right) \) . Hence\n\n\[ m\left( {\widetilde{\mathcal{O}} \cap {B}_{k}}\right) = \int {\chi }_{\mathcal{O}}\left( {x - y}\right) {\chi }_{{B}_{k}}\left( y\right) {dydx} \]\n\n\[ = \int \left( {\int {\chi }_{\mathcal{O}}\left( {x - y}\right) {dx}}\right) {\chi }_{{B}_{k}}\left( y\right) {dy} \]\n\n\[ = m\left( \mathcal{O}\right) m\left( {B}_{k}\right) \]\n\nby the translation-invariance of the measure. Now if \( m\left( E\right) = 0 \), there is a sequence of open sets \( {\mathcal{O}}_{n} \) such that \( E \subset {\mathcal{O}}_{n} \) and \( m\left( {\mathcal{O}}_{n}\right) \rightarrow 0 \) . It follows from the above that \( {\widetilde{E}}_{k} \subset {\widetilde{\mathcal{O}}}_{n} \cap {B}_{k} \) and \( m\left( {{\widetilde{\mathcal{O}}}_{n} \cap {B}_{k}}\right) \rightarrow 0 \) in \( n \) for each fixed \( k \) . This shows \( m\left( {\widetilde{E}}_{k}\right) = 0 \), and hence \( m\left( \widetilde{E}\right) = 0 \) . The proof of the proposition is concluded once we recall that any measurable set \( E \) can be written as the difference of a \( {G}_{\delta } \) and a set of measure zero.
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Yes
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Proposition 4.1 Suppose \( f \in {L}^{1}\left( {\mathbb{R}}^{d}\right) \) . Then \( \widehat{f} \) defined by (14) is continuous and bounded on \( {\mathbb{R}}^{d} \) .
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In fact, since \( \left| {f\left( x\right) {e}^{-{2\pi ix} \cdot \xi }}\right| = \left| {f\left( x\right) }\right| \), the integral representing \( \widehat{f} \) converges for each \( \xi \) and \( \mathop{\sup }\limits_{{\xi \in {\mathbb{R}}^{d}}}\left| {\widehat{f}\left( \xi \right) }\right| \leq {\int }_{{\mathbb{R}}^{d}}\left| {f\left( x\right) }\right| {dx} = \parallel f\parallel \) . To verify the continuity, note that for every \( x, f\left( x\right) {e}^{-{2\pi ix} \cdot \xi } \rightarrow f\left( x\right) {e}^{-{2\pi ix} \cdot {\xi }_{0}} \) as \( \xi \rightarrow {\xi }_{0} \) , where \( {\xi }_{0} \) is any point in \( {\mathbb{R}}^{d} \) ; hence \( \widehat{f}\left( \xi \right) \rightarrow \widehat{f}\left( {\xi }_{0}\right) \) by the dominated convergence theorem.
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Yes
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Theorem 4.2 Suppose \( f \in {L}^{1}\left( {\mathbb{R}}^{d}\right) \) and assume also that \( \widehat{f} \in {L}^{1}\left( {\mathbb{R}}^{d}\right) \) . Then the inversion formula (15) holds for almost every \( x \) .
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The proof of the theorem requires only that we adapt the earlier arguments carried out for Schwartz functions in Chapter 5 of Book I to the present context. We begin with the \
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No
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Lemma 4.4 Suppose \( f \) and \( g \) belong to \( {L}^{1}\left( {\mathbb{R}}^{d}\right) \) . Then\n\n\[ \n{\int }_{{\mathbb{R}}^{d}}\widehat{f}\left( \xi \right) g\left( \xi \right) {d\xi } = {\int }_{{\mathbb{R}}^{d}}f\left( y\right) \widehat{g}\left( y\right) {dy}.\n\]
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Note that both integrals converge in view of the proposition above. Consider the function \( F\left( {\xi, y}\right) = g\left( \xi \right) f\left( y\right) {e}^{-{2\pi i\xi } \cdot y} \) defined for \( \left( {\xi, y}\right) \in {\mathbb{R}}^{d} \times \) \( {\mathbb{R}}^{d} = {\mathbb{R}}^{2d} \) . It is measurable as a function on \( {\mathbb{R}}^{2d} \) in view of Corollary 3.7. We now apply Fubini's theorem to observe first that\n\n\[ \n{\int }_{{\mathbb{R}}^{d}}{\int }_{{\mathbb{R}}^{d}}\left| {F\left( {\xi, y}\right) }\right| {d\xi dy} = {\int }_{{\mathbb{R}}^{d}}\left| {g\left( \xi \right) }\right| {d\xi }{\int }_{{\mathbb{R}}^{d}}\left| {f\left( y\right) }\right| {dy} < \infty .\n\]\n\nNext, if we evaluate \( {\int }_{{\mathbb{R}}^{d}}{\int }_{{\mathbb{R}}^{d}}F\left( {\xi, y}\right) \;{d\xi }\;{dy} \) by writing it as \( {\int }_{{\mathbb{R}}^{d}}\left( {{\int }_{{\mathbb{R}}^{d}}F\left( {\xi, y}\right) \;{d\xi }}\right) \;{dy} \) we get the left-hand side of the desired equality. Evaluating the double integral in the reverse order gives as the right-hand side, proving the lemma.
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Yes
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Theorem 1.1 Suppose \( f \) is integrable on \( {\mathbb{R}}^{d} \) . Then:\n\n(i) \( {f}^{ * } \) is measurable.\n\n(ii) \( {f}^{ * }\left( x\right) < \infty \) for a.e. \( x \) .\n\n(iii) \( {f}^{ * } \) satisfies\n\n\[ m\left( \left\{ {x \in {\mathbb{R}}^{d} : {f}^{ * }\left( x\right) > \alpha }\right\} \right) \leq \frac{A}{\alpha }\parallel f{\parallel }_{{L}^{1}\left( {\mathbb{R}}^{d}\right) } \] for all \( \alpha > 0 \), where \( A = {3}^{d} \), and \( \parallel f{\parallel }_{{L}^{1}\left( {\mathbb{R}}^{d}\right) } = {\int }_{{\mathbb{R}}^{d}}\left| {f\left( x\right) }\right| {dx} \).
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Before we come to the proof we want to clarify the nature of the main conclusion (iii). As we shall observe, one has that \( {f}^{ * }\left( x\right) \geq \left| {f\left( x\right) }\right| \) for a.e. \( x \) ; the effect of (iii) is that, broadly speaking, \( {f}^{ * } \) is not much larger than \( \left| f\right| \) . From this point of view, we would have liked to conclude that \( {f}^{ * } \) is integrable, as a result of the assumed integrability of \( f \) . However, this is not the case, and (iii) is the best substitute available (see Exercises 4 and 5).\n\nAn inequality of the type (1) is called a weak-type inequality because it is weaker than the corresponding inequality for the \( {L}^{1} \) -norms. Indeed, this can be seen from the Tchebychev inequality (Exercise 9 in Chapter 2), which states that for an arbitrary integrable function \( g \) ,\n\n\[ m\left( {\{ x : \left| {g\left( x\right) }\right| > \alpha \} }\right) \leq \frac{1}{\alpha }\parallel g{\parallel }_{{L}^{1}\left( {\mathbb{R}}^{d}\right) },\;\text{ for all }\alpha > 0. \]\n\nWe should add that the exact value of \( A \) in the inequality (1) is unimportant for us. What matters is that this constant be independent of \( \alpha \) and \( f \) .\n\nThe only simple assertion in the theorem is that \( {f}^{ * } \) is a measurable function. Indeed, the set \( {E}_{\alpha } = \left\{ {x \in {\mathbb{R}}^{d} : {f}^{ * }\left( x\right) > \alpha }\right\} \) is open, because if \( \bar{x} \in {E}_{\alpha } \), there exists a ball \( B \) such that \( \bar{x} \in B \) and\n\n\[ \frac{1}{m\left( B\right) }{\int }_{B}\left| {f\left( y\right) }\right| {dy} > \alpha \]\n\nNow any point \( x \) close enough to \( \bar{x} \) will also belong to \( B \) ; hence \( x \in {E}_{\alpha } \) as well.\n\nThe two other properties of \( {f}^{ * } \) in the theorem are deeper, with (ii) being a consequence of (iii). This follows at once if we observe that\n\n\[ \left\{ {x : {f}^{ * }\left( x\right) = \infty }\right\} \subset \left\{ {x : {f}^{ * }\left( x\right) > \alpha }\right\} \]\nfor all \( \alpha \) . Taking the limit as \( \alpha \) tends to infinity, the third property yields \( m\left( \left\{ {x : {f}^{ * }\left( x\right) = \infty }\right\} \right) = 0 \).\n\nThe proof of inequality (1) relies on an elementary version of a Vitali covering argument. \( {}^{1} \)
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No
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Lemma 1.2 Suppose \( \mathcal{B} = \left\{ {{B}_{1},{B}_{2},\ldots ,{B}_{N}}\right\} \) is a finite collection of open balls in \( {\mathbb{R}}^{d} \) . Then there exists a disjoint sub-collection \( {B}_{{i}_{1}},{B}_{{i}_{2}},\ldots ,{B}_{{i}_{k}} \) of \( \mathcal{B} \) that satisfies\n\n\[ m\left( {\mathop{\bigcup }\limits_{{\ell = 1}}^{N}{B}_{\ell }}\right) \leq {3}^{d}\mathop{\sum }\limits_{{j = 1}}^{k}m\left( {B}_{{i}_{j}}\right) \]
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Proof. The argument we give is constructive and relies on the following simple observation: Suppose \( B \) and \( {B}^{\prime } \) are a pair of balls that intersect, with the radius of \( {B}^{\prime } \) being not greater than that of \( B \) . Then \( {B}^{\prime } \) is contained in the ball \( \widetilde{B} \) that is concentric with \( B \) but with 3 times its radius.\n\nAs a first step, we pick a ball \( {B}_{{i}_{1}} \) in \( \mathcal{B} \) with maximal (that is, largest) radius, and then delete from \( \mathcal{B} \) the ball \( {B}_{{i}_{1}} \) as well as any balls that intersect \( {B}_{{i}_{1}} \) . Thus all the balls that are deleted are contained in the ball \( {\widetilde{B}}_{{i}_{1}} \) concentric with \( {B}_{{i}_{1}} \), but with 3 times its radius.\n\nThe remaining balls yield a new collection \( {\mathcal{B}}^{\prime } \), for which we repeat the procedure. We pick \( {B}_{{i}_{2}} \) with largest radius in \( {\mathcal{B}}^{\prime } \), and then delete from \( {\mathcal{B}}^{\prime } \) the ball \( {B}_{{i}_{2}} \) and any ball that intersects \( {B}_{{i}_{2}} \) . Continuing this way we find, after at most \( N \) steps, a collection of disjoint balls \( {B}_{{i}_{1}},{B}_{{i}_{2}},\ldots ,{B}_{{i}_{k}} \).\n\nFinally, to prove that this disjoint collection of balls satisfies the inequality in the lemma, we use the observation made at the beginning of the proof. We let \( {\widetilde{B}}_{{i}_{j}} \) denote the ball concentric with \( {B}_{{i}_{j}} \), but with 3 times its radius. Since any ball \( B \) in \( \mathcal{B} \) must intersect a ball \( {B}_{{i}_{j}} \) and have equal or smaller radius than \( {B}_{{i}_{j}} \), we must have \( B \subset {\widetilde{B}}_{{i}_{j}} \), thus\n\n\[ m\left( {\mathop{\bigcup }\limits_{{\ell = 1}}^{N}{B}_{\ell }}\right) \leq m\left( {\mathop{\bigcup }\limits_{{j = 1}}^{k}{\widetilde{B}}_{{i}_{j}}}\right) \leq \mathop{\sum }\limits_{{j = 1}}^{k}m\left( {\widetilde{B}}_{{i}_{j}}\right) = {3}^{d}\mathop{\sum }\limits_{{j = 1}}^{k}m\left( {B}_{{i}_{j}}\right) . \]
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Yes
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Theorem 1.3 If \( f \) is integrable on \( {\mathbb{R}}^{d} \), then\n\n\[ \mathop{\lim }\limits_{\substack{{m\left( B\right) \rightarrow 0} \\ {x \in B} }}\frac{1}{m\left( B\right) }{\int }_{B}f\left( y\right) {dy} = f\left( x\right) \;\text{ for a.e. }x. \]
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Proof. It suffices to show that for each \( \alpha > 0 \) the set\n\n\[ {E}_{\alpha } = \left\{ {x : \mathop{\limsup }\limits_{\substack{{m\left( B\right) \rightarrow 0} \\ {x \in B} }}\left| {\frac{1}{m\left( B\right) }{\int }_{B}f\left( y\right) {dy} - f\left( x\right) }\right| > {2\alpha }}\right\} \]\n\nhas measure zero, because this assertion then guarantees that the set \( E = \mathop{\bigcup }\limits_{{n = 1}}^{\infty }{E}_{1/n} \) has measure zero, and the limit in (4) holds at all points of \( {E}^{c} \) .\n\nWe fix \( \alpha \), and recall Theorem 2.4 in Chapter 2, which states that for each \( \epsilon > 0 \) we may select a continuous function \( g \) of compact support with \( \parallel f - g{\parallel }_{{L}^{1}\left( {\mathbb{R}}^{d}\right) } < \epsilon \) . As we remarked earlier, the continuity of \( g \) implies that\n\n\[ \mathop{\lim }\limits_{\substack{{m\left( B\right) \rightarrow 0} \\ {x \in B} }}\frac{1}{m\left( B\right) }{\int }_{B}g\left( y\right) {dy} = g\left( x\right) ,\;\text{ for all }x. \]\n\nSince we may write the difference \( \frac{1}{m\left( B\right) }{\int }_{B}f\left( y\right) {dy} - f\left( x\right) \) as\n\n\[ \frac{1}{m\left( B\right) }{\int }_{B}\left( {f\left( y\right) - g\left( y\right) }\right) {dy} + \frac{1}{m\left( B\right) }{\int }_{B}g\left( y\right) {dy} - g\left( x\right) + g\left( x\right) - f\left( x\right) \]\nwe find that\n\n\[ \mathop{\limsup }\limits_{\substack{{m\left( B\right) \rightarrow 0} \\ {x \in B} }}\left| {\frac{1}{m\left( B\right) }{\int }_{B}f\left( y\right) {dy} - f\left( x\right) }\right| \leq {\left( f - g\right) }^{ * }\left( x\right) + \left| {g\left( x\right) - f\left( x\right) }\right| \]\n\nwhere the symbol \( * \) indicates the maximal function. Consequently, if\n\n\[ {F}_{\alpha } = \left\{ {x : {\left( f - g\right) }^{ * }\left( x\right) > \alpha }\right\} \;\text{ and }\;{G}_{\alpha } = \{ x : \left| {f\left( x\right) - g\left( x\right) }\right| > \alpha \} \]\n\nthen \( {E}_{\alpha } \subset \left( {{F}_{\alpha } \cup {G}_{\alpha }}\right) \), because if \( {u}_{1} \) and \( {u}_{2} \) are positive, then \( {u}_{1} + {u}_{2} > \) \( {2\alpha } \) only if \( {u}_{i} > \alpha \) for at least one \( {u}_{i} \) . On the one hand, Tchebychev’s inequality yields\n\n\[ m\left( {G}_{\alpha }\right) \leq \frac{1}{\alpha }\parallel f - g{\parallel }_{{L}^{1}\left( {\mathbb{R}}^{d}\right) } \]\n\nand on the other hand, the weak type estimate for the maximal function gives\n\n\[ m\left( {F}_{\alpha }\right) \leq \frac{A}{\alpha }\parallel f - g{\parallel }_{{L}^{1}\left( {\mathbb{R}}^{d}\right) }.\]\n\nThe function \( g \) was selected so that \( \parallel f - g{\parallel }_{{L}^{1}\left( {\mathbb{R}}^{d}\right) } < \epsilon \) . Hence we get\n\n\[ m\left( {E}_{\alpha }\right) \leq \frac{A}{\alpha }\epsilon + \frac{1}{\alpha }\epsilon \]\n\nSince \( \epsilon \) is arbitrary, we must have \( m\left( {E}_{\alpha }\right) = 0 \), and the proof of the theorem is complete.
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Yes
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Theorem 1.4 If \( f \in {L}_{\text{loc }}^{1}\left( {\mathbb{R}}^{d}\right) \), then\n\n\[ \mathop{\lim }\limits_{\substack{{m\left( B\right) \rightarrow 0} \\ {x \in B} }}\frac{1}{m\left( B\right) }{\int }_{B}f\left( y\right) {dy} = f\left( x\right) ,\;\text{ for a.e. }x. \]
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Proof. An application of Theorem 1.4 to the function \( \left| {f\left( y\right) - r}\right| \) shows that for each rational \( r \), there exists a set \( {E}_{r} \) of measure zero, such that\n\n\[ \mathop{\lim }\limits_{\substack{{m\left( B\right) \rightarrow 0} \\ {x \in B} }}\frac{1}{m\left( B\right) }{\int }_{B}\left| {f\left( y\right) - r}\right| {dy} = \left| {f\left( x\right) - r}\right| \;\text{ whenever }x \notin {E}_{r}. \]\n\nIf \( E = \mathop{\bigcup }\limits_{{r \in \mathbb{O}}}{E}_{r} \), then \( m\left( E\right) = 0 \) . Now suppose that \( \bar{x} \notin E \) and \( f\left( \bar{x}\right) \) is finite. Given \( \epsilon > 0 \), there exists a rational \( r \) such that \( \left| {f\left( \bar{x}\right) - r}\right| < \epsilon \) . Since\n\n\[ \frac{1}{m\left( B\right) }{\int }_{B}\left| {f\left( y\right) - f\left( \bar{x}\right) }\right| {dy} \leq \frac{1}{m\left( B\right) }{\int }_{B}\left| {f\left( y\right) - r}\right| {dy} + \left| {f\left( \bar{x}\right) - r}\right| \]\n\nwe must have\n\n\[ \mathop{\limsup }\limits_{\substack{{m\left( B\right) \rightarrow 0} \\ {\bar{x} \in B} }}\frac{1}{m\left( B\right) }{\int }_{B}\left| {f\left( y\right) - f\left( \bar{x}\right) }\right| {dy} \leq {2\epsilon } \]\n\nand thus \( \bar{x} \) is in the Lebesgue set of \( f \) . The corollary is therefore proved.
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Yes
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Corollary 1.6 If \( f \) is locally integrable on \( {\mathbb{R}}^{d} \), then almost every point belongs to the Lebesgue set of \( f \) .
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Proof. An application of Theorem 1.4 to the function \( \left| {f\left( y\right) - r}\right| \) shows that for each rational \( r \), there exists a set \( {E}_{r} \) of measure zero, such that\n\n\[ \mathop{\lim }\limits_{\substack{{m\left( B\right) \rightarrow 0} \\ {x \in B} }}\frac{1}{m\left( B\right) }{\int }_{B}\left| {f\left( y\right) - r}\right| {dy} = \left| {f\left( x\right) - r}\right| \;\text{ whenever }x \notin {E}_{r}. \]\n\nIf \( E = \mathop{\bigcup }\limits_{{r \in \mathbb{O}}}{E}_{r} \), then \( m\left( E\right) = 0 \) . Now suppose that \( \bar{x} \notin E \) and \( f\left( \bar{x}\right) \) is finite. Given \( \epsilon > 0 \), there exists a rational \( r \) such that \( \left| {f\left( \bar{x}\right) - r}\right| < \epsilon \) . Since\n\n\[ \frac{1}{m\left( B\right) }{\int }_{B}\left| {f\left( y\right) - f\left( \bar{x}\right) }\right| {dy} \leq \frac{1}{m\left( B\right) }{\int }_{B}\left| {f\left( y\right) - r}\right| {dy} + \left| {f\left( \bar{x}\right) - r}\right| \]\n\nwe must have\n\n\[ \mathop{\limsup }\limits_{\substack{{m\left( B\right) \rightarrow 0} \\ {\bar{x} \in B} }}\frac{1}{m\left( B\right) }{\int }_{B}\left| {f\left( y\right) - f\left( \bar{x}\right) }\right| {dy} \leq {2\epsilon } \]\n\nand thus \( \bar{x} \) is in the Lebesgue set of \( f \) . The corollary is therefore proved.
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Yes
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Corollary 1.6 If \( f \) is locally integrable on \( {\mathbb{R}}^{d} \), then almost every point belongs to the Lebesgue set of \( f \) .
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Proof. An application of Theorem 1.4 to the function \( \left| {f\left( y\right) - r}\right| \) shows that for each rational \( r \), there exists a set \( {E}_{r} \) of measure zero, such that\n\n\[ \mathop{\lim }\limits_{\substack{{m\left( B\right) \rightarrow 0} \\ {x \in B} }}\frac{1}{m\left( B\right) }{\int }_{B}\left| {f\left( y\right) - r}\right| {dy} = \left| {f\left( x\right) - r}\right| \;\text{ whenever }x \notin {E}_{r}. \]\n\nIf \( E = \mathop{\bigcup }\limits_{{r \in \mathbb{O}}}{E}_{r} \), then \( m\left( E\right) = 0 \) . Now suppose that \( \bar{x} \notin E \) and \( f\left( \bar{x}\right) \) is finite. Given \( \epsilon > 0 \), there exists a rational \( r \) such that \( \left| {f\left( \bar{x}\right) - r}\right| < \epsilon \) . Since\n\n\[ \frac{1}{m\left( B\right) }{\int }_{B}\left| {f\left( y\right) - f\left( \bar{x}\right) }\right| {dy} \leq \frac{1}{m\left( B\right) }{\int }_{B}\left| {f\left( y\right) - r}\right| {dy} + \left| {f\left( \bar{x}\right) - r}\right| \]\n\nwe must have\n\n\[ \mathop{\limsup }\limits_{\substack{{m\left( B\right) \rightarrow 0} \\ {\bar{x} \in B} }}\frac{1}{m\left( B\right) }{\int }_{B}\left| {f\left( y\right) - f\left( \bar{x}\right) }\right| {dy} \leq {2\epsilon } \]\n\nand thus \( \bar{x} \) is in the Lebesgue set of \( f \) . The corollary is therefore proved.
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Yes
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Corollary 1.7 Suppose \( f \) is locally integrable on \( {\mathbb{R}}^{d} \) . If \( \left\{ {U}_{\alpha }\right\} \) shrinks regularly to \( \bar{x} \), then\n\n\[ \mathop{\lim }\limits_{\substack{{m\left( {U}_{\alpha }\right) \rightarrow 0} \\ {x \in {U}_{\alpha }} }}\frac{1}{m\left( {U}_{\alpha }\right) }{\int }_{{U}_{\alpha }}f\left( y\right) {dy} = f\left( \bar{x}\right) \]\n\nfor every point \( \bar{x} \) in the Lebesgue set of \( f \) .
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The proof is immediate once we observe that if \( \bar{x} \in B \) with \( {U}_{\alpha } \subset B \) and \( m\left( {U}_{\alpha }\right) \geq {cm}\left( B\right) \), then\n\n\[ \frac{1}{m\left( {U}_{\alpha }\right) }{\int }_{{U}_{\alpha }}\left| {f\left( y\right) - f\left( \bar{x}\right) }\right| {dy} \leq \frac{1}{{cm}\left( B\right) }{\int }_{B}\left| {f\left( y\right) - f\left( \bar{x}\right) }\right| {dy}. \]
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Yes
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Theorem 2.1 If \( {\left\{ {K}_{\delta }\right\} }_{\delta > 0} \) is an approximation to the identity and \( f \) is integrable on \( {\mathbb{R}}^{d} \), then\n\n\[ \n\left( {f * {K}_{\delta }}\right) \left( x\right) \rightarrow f\left( x\right) \;\text{ as }\delta \rightarrow 0 \n\]\n\nfor every \( x \) in the Lebesgue set of \( f \) . In particular, the limit holds for a.e. \( x \) .
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Since the integral of each kernel \( {K}_{\delta } \) is equal to 1, we may write\n\n\[ \n\left( {f * {K}_{\delta }}\right) \left( x\right) - f\left( x\right) = \int \left\lbrack {f\left( {x - y}\right) - f\left( x\right) }\right\rbrack {K}_{\delta }\left( y\right) {dy}. \n\]\n\nConsequently,\n\n\[ \n\left| {\left( {f * {K}_{\delta }}\right) \left( x\right) - f\left( x\right) }\right| \leq \int \left| {f\left( {x - y}\right) - f\left( x\right) }\right| \left| {{K}_{\delta }\left( y\right) }\right| {dy} \n\]\n\nand it now suffices to prove that the right-hand side tends to 0 as \( \delta \) goes to 0 . The argument we give depends on a simple result that we isolate in the next lemma.
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Yes
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Lemma 2.2 Suppose that \( f \) is integrable on \( {\mathbb{R}}^{d} \), and that \( x \) is a point of the Lebesgue set of \( f \) . Let\n\n\[ \mathcal{A}\left( r\right) = \frac{1}{{r}^{d}}{\int }_{\left| y\right| \leq r}\left| {f\left( {x - y}\right) - f\left( x\right) }\right| {dy},\;\text{ whenever }r > 0. \]\n\nThen \( \mathcal{A}\left( r\right) \) is a continuous function of \( r > 0 \), and\n\n\[ \mathcal{A}\left( r\right) \rightarrow 0\;\text{ as }r \rightarrow 0. \]\n\nMoreover, \( \mathcal{A}\left( r\right) \) is bounded, that is, \( \mathcal{A}\left( r\right) \leq M \) for some \( M > 0 \) and all \( r > 0 \) .
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Proof. The continuity of \( \mathcal{A}\left( r\right) \) follows by invoking the absolute continuity in Proposition 1.12 of Chapter 2.\n\nThe fact that \( \mathcal{A}\left( r\right) \) tends to 0 as \( r \) tends to 0 follows since \( x \) belongs to the Lebesgue set of \( f \), and the measure of a ball of radius \( r \) is \( {v}_{d}{r}^{d} \) . This and the continuity of \( \mathcal{A}\left( r\right) \) for \( 0 < r \leq 1 \) show that \( \mathcal{A}\left( r\right) \) is bounded when \( 0 < r \leq 1 \) . To prove that \( \mathcal{A}\left( r\right) \) is bounded for \( r > 1 \), note that\n\n\[ \mathcal{A}\left( r\right) \leq \frac{1}{{r}^{d}}{\int }_{\left| y\right| \leq r}\left| {f\left( {x - y}\right) }\right| {dy} + \frac{1}{{r}^{d}}{\int }_{\left| y\right| \leq r}\left| {f\left( x\right) }\right| {dy} \]\n\n\[ \leq {r}^{-d}\parallel f{\parallel }_{{L}^{1}\left( {\mathbb{R}}^{d}\right) } + {v}_{d}\left| {f\left( x\right) }\right| \]\n\nand this concludes the proof of the lemma.
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Yes
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Theorem 2.3 Suppose that \( f \) is integrable on \( {\mathbb{R}}^{d} \) and that \( {\left\{ {K}_{\delta }\right\} }_{\delta > 0} \) is an approximation to the identity. Then, for each \( \delta > 0 \), the convolution\n\n\[ \n\left( {f * {K}_{\delta }}\right) \left( x\right) = {\int }_{{\mathbb{R}}^{d}}f\left( {x - y}\right) {K}_{\delta }\left( y\right) {dy} \n\]\n\nis integrable, and\n\n\[ \n{\begin{Vmatrix}\left( f * {K}_{\delta }\right) - f\end{Vmatrix}}_{{L}^{1}\left( {\mathbb{R}}^{d}\right) } \rightarrow 0,\;\text{ as }\delta \rightarrow 0. \n\]
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The proof is merely a repetition in a more general context of the argument in the special case where \( {K}_{\delta }\left( x\right) = {\delta }^{-d/2}{e}^{-\pi {\left| x\right| }^{2}/\delta } \) given in Section \( {4}^{ * } \) , Chapter 2, and so will not be repeated.
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No
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Theorem 3.1 A curve parametrized by \( \left( {x\left( t\right), y\left( t\right) }\right), a \leq t \leq b \), is rectifiable if and only if both \( x\left( t\right) \) and \( y\left( t\right) \) are of bounded variation.
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The proof is immediate once we observe that if \( F\left( t\right) = x\left( t\right) + {iy}\left( t\right) \), then\n\n\[ F\left( {t}_{j}\right) - F\left( {t}_{j - 1}\right) = \left( {x\left( {t}_{j}\right) - x\left( {t}_{j - 1}\right) }\right) + i\left( {y\left( {t}_{j}\right) - y\left( {t}_{j - 1}\right) }\right) ,\]
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Yes
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Lemma 3.2 Suppose \( F \) is real-valued and of bounded variation on \( \left\lbrack {a, b}\right\rbrack \) . Then for all \( a \leq x \leq b \) one has \[ F\left( x\right) - F\left( a\right) = {P}_{F}\left( {a, x}\right) - {N}_{F}\left( {a, x}\right) ,\] and \[ {T}_{F}\left( {a, x}\right) = {P}_{F}\left( {a, x}\right) + {N}_{F}\left( {a, x}\right) .
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Proof. Given \( \epsilon > 0 \) there exists a partition \( a = {t}_{0} < \cdots < {t}_{N} = x \) of \( \left\lbrack {a, x}\right\rbrack \), such that \[ \left| {{P}_{F} - \mathop{\sum }\limits_{\left( +\right) }F\left( {t}_{j}\right) - F\left( {t}_{j - 1}\right) }\right| < \epsilon \text{ and }\left| {{N}_{F} - \mathop{\sum }\limits_{\left( -\right) } - \left\lbrack {F\left( {t}_{j}\right) - F\left( {t}_{j - 1}\right) }\right\rbrack }\right| < \epsilon . \] (To see this, it suffices to use the definition to obtain similar estimates for \( {P}_{F} \) and \( {N}_{F} \) with possibly different partitions, and then to consider a common refinement of these two partitions.) Since we also note that \[ F\left( x\right) - F\left( a\right) = \mathop{\sum }\limits_{\left( +\right) }F\left( {t}_{j}\right) - F\left( {t}_{j - 1}\right) - \mathop{\sum }\limits_{\left( -\right) } - \left\lbrack {F\left( {t}_{j}\right) - F\left( {t}_{j - 1}\right) }\right\rbrack \] we find that \( \left| {F\left( x\right) - F\left( a\right) - \left\lbrack {{P}_{F} - {N}_{F}}\right\rbrack }\right| < {2\epsilon } \), which proves the first identity. For the second identity, we also note that for any partition of \( a = {t}_{0} < \cdots < {t}_{N} = x \) of \( \left\lbrack {a, x}\right\rbrack \) we have \[ \mathop{\sum }\limits_{{j = 1}}^{N}\left| {F\left( {t}_{j}\right) - F\left( {t}_{j - 1}\right) }\right| = \mathop{\sum }\limits_{\left( +\right) }F\left( {t}_{j}\right) - F\left( {t}_{j - 1}\right) + \mathop{\sum }\limits_{\left( -\right) } - \left\lbrack {F\left( {t}_{j}\right) - F\left( {t}_{j - 1}\right) }\right\rbrack \] hence \( {T}_{F} \leq {P}_{F} + {N}_{F} \). Also, the above implies \[ \mathop{\sum }\limits_{\left( +\right) }F\left( {t}_{j}\right) - F\left( {t}_{j - 1}\right) + \mathop{\sum }\limits_{\left( -\right) } - \left\lbrack {F\left( {t}_{j}\right) - F\left( {t}_{j - 1}\right) }\right\rbrack \leq {T}_{F} \] Once again, one can argue using common refinements of partitions in the definitions of \( {P}_{F} \) and \( {N}_{F} \) to deduce the inequality \( {P}_{F} + {N}_{F} \leq {T}_{F} \), and the lemma is proved.
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Yes
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Theorem 3.3 A real-valued function \( F \) on \( \left\lbrack {a, b}\right\rbrack \) is of bounded variation if and only if \( F \) is the difference of two increasing bounded functions.
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Proof. Clearly, if \( F = {F}_{1} - {F}_{2} \), where each \( {F}_{j} \) is bounded and increasing, then \( F \) is of bounded variation.\n\nConversely, suppose \( F \) is of bounded variation. Then, we let \( {F}_{1}\left( x\right) = \) \( {P}_{F}\left( {a, x}\right) + F\left( a\right) \) and \( {F}_{2}\left( x\right) = {N}_{F}\left( {a, x}\right) \) . Clearly, both \( {F}_{1} \) and \( {F}_{2} \) are increasing, of bounded variation, and by the lemma \( F\left( x\right) = {F}_{1}\left( x\right) - {F}_{2}\left( x\right) \) .
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Yes
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Lemma 3.5 Suppose \( G \) is real-valued and continuous on \( \mathbb{R} \) . Let \( E \) be the set of points \( x \) such that\n\n\[ G\left( {x + h}\right) > G\left( x\right) \;\text{ for some }h = {h}_{x} > 0.\]\n\nIf \( E \) is non-empty, then it must be open, and hence can be written as a countable disjoint union of open intervals \( E = \bigcup \left( {{a}_{k},{b}_{k}}\right) \) . If \( \left( {{a}_{k},{b}_{k}}\right) \) is a finite interval in this union, then\n\n\[ G\left( {b}_{k}\right) - G\left( {a}_{k}\right) = 0.\]
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Proof. Since \( G \) is continuous, it is clear that \( E \) is open whenever it is non-empty and can therefore be written as a disjoint union of countably many open intervals (Theorem 1.3 in Chapter 1). If \( \left( {{a}_{k},{b}_{k}}\right) \) denotes a finite interval in this decomposition, then \( {a}_{k} \notin E \) ; therefore we cannot have \( G\left( {b}_{k}\right) > G\left( {a}_{k}\right) \) . We now suppose that \( G\left( {b}_{k}\right) < G\left( {a}_{k}\right) \) . By continuity, there exists \( {a}_{k} < c < {b}_{k} \) so that\n\n\[ G\left( c\right) = \frac{G\left( {a}_{k}\right) + G\left( {b}_{k}\right) }{2} \]\n\nand in fact we may choose \( c \) farthest to the right in the interval \( \left( {{a}_{k},{b}_{k}}\right) \) . Since \( c \in E \), there exists \( d > c \) such that \( G\left( d\right) > G\left( c\right) \) . Since \( {b}_{k} \notin E \), we must have \( G\left( x\right) \leq G\left( {b}_{k}\right) \) for all \( x \geq {b}_{k} \) ; therefore \( d < {b}_{k} \) . Since \( G\left( d\right) > \) \( G\left( c\right) \), there exists (by continuity) \( {c}^{\prime } > d \) with \( {c}^{\prime } < {b}_{k} \) and \( G\left( {c}^{\prime }\right) = G\left( c\right) \) , which contradicts the fact that \( c \) was chosen farthest to the right in \( \left( {{a}_{k},{b}_{k}}\right) \) . This shows that we must have \( G\left( {a}_{k}\right) = G\left( {b}_{k}\right) \), and the lemma is proved.
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Yes
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If \( F \) is increasing and continuous, then \( {F}^{\prime } \) exists almost everywhere. Moreover \( {F}^{\prime } \) is measurable, non-negative, and\n\n\[{\int }_{a}^{b}{F}^{\prime }\left( x\right) {dx} \leq F\left( b\right) - F\left( a\right)\]\n\nIn particular, if \( F \) is bounded on \( \mathbb{R} \), then \( {F}^{\prime } \) is integrable on \( \mathbb{R} \).
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Proof. For \( n \geq 1 \), we consider the quotient\n\n\[{G}_{n}\left( x\right) = \frac{F\left( {x + 1/n}\right) - F\left( x\right) }{1/n}.\n\]\n\nBy the previous theorem, we have that \( {G}_{n}\left( x\right) \rightarrow {F}^{\prime }\left( x\right) \) for a.e. \( x \), which shows in particular that \( {F}^{\prime } \) is measurable and non-negative.\n\nWe now extend \( F \) as a continuous function on all of \( \mathbb{R} \). By Fatou’s lemma (Lemma 1.7 in Chapter 2) we know that\n\n\[{\int }_{a}^{b}{F}^{\prime }\left( x\right) {dx} \leq \mathop{\liminf }\limits_{{n \rightarrow \infty }}{\int }_{a}^{b}{G}_{n}\left( x\right) {dx}.\n\]\n\nTo complete the proof, it suffices to note that\n\n\[{\int }_{a}^{b}{G}_{n}\left( x\right) {dx} = \frac{1}{1/n}{\int }_{a}^{b}F\left( {x + 1/n}\right) {dx} - \frac{1}{1/n}{\int }_{a}^{b}F\left( x\right) {dx}\n\]\n\n\[= \frac{1}{1/n}{\int }_{a + 1/n}^{b + 1/n}F\left( y\right) {dy} - \frac{1}{1/n}{\int }_{a}^{b}F\left( x\right) {dx}\n\]\n\n\[= \frac{1}{1/n}{\int }_{b}^{b + 1/n}F\left( x\right) {dx} - \frac{1}{1/n}{\int }_{a}^{a + 1/n}F\left( x\right) {dx}.\n\]\n\nSince \( F \) is continuous, the first and second terms converge to \( F\left( b\right) \) and \( F\left( a\right) \), respectively, as \( n \) goes to infinity, so the proof of the corollary is complete.
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Yes
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Theorem 3.8 If \( F \) is absolutely continuous on \( \left\lbrack {a, b}\right\rbrack \), then \( {F}^{\prime }\left( x\right) \) exists almost everywhere. Moreover, if \( {F}^{\prime }\left( x\right) = 0 \) for a.e. \( x \), then \( F \) is constant.
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Since an absolutely continuous function is the difference of two continuous monotonic functions, as we have seen above, the existence of \( {F}^{\prime }\left( x\right) \) for a.e. \( x \) follows from what we have already proved. To prove that \( {F}^{\prime }\left( x\right) = 0 \) a.e. implies \( F \) is constant requires a more elaborate version of the covering argument in Lemma 1.2. For the moment we revert to the generality of \( d \) dimensions to describe this.
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Yes
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Corollary 3.10 We can arrange the choice of the balls so that\n\n\[ m\left( {E - \mathop{\bigcup }\limits_{{i = 1}}^{N}{B}_{i}}\right) < {2\delta } \]
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In fact, let \( \mathcal{O} \) be an open set, with \( \mathcal{O} \supset E \) and \( m\left( {\mathcal{O} - E}\right) < \delta \) . Since we are dealing with a Vitali covering of \( E \), we can restrict all of our choices above to balls contained in \( \mathcal{O} \) . If we do this, then \( \left( {E - \mathop{\bigcup }\limits_{{i = 1}}^{N}{B}_{i}}\right) \cup \) \( \mathop{\bigcup }\limits_{{i = 1}}^{N}{B}_{i} \subset \mathcal{O} \), where the union on the left-hand side is a disjoint union. Hence\n\n\[ m\left( {E - \mathop{\bigcup }\limits_{{i = 1}}^{N}{B}_{i}}\right) \leq m\left( \mathcal{O}\right) - m\left( {\mathop{\bigcup }\limits_{{i = 1}}^{N}{B}_{i}}\right) \leq m\left( E\right) + \delta - \left( {m\left( E\right) - \delta }\right) = {2\delta }. \]
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Yes
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Theorem 3.11 Suppose \( F \) is absolutely continuous on \( \left\lbrack {a, b}\right\rbrack \) . Then \( {F}^{\prime } \) exists almost everywhere and is integrable. Moreover,\n\n\[ F\left( x\right) - F\left( a\right) = {\int }_{a}^{x}{F}^{\prime }\left( y\right) {dy},\;\text{ for all }a \leq x \leq b. \]
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Proof. Since we know that a real-valued absolutely continuous function is the difference of two continuous increasing functions, Corollary 3.7 shows that \( {F}^{\prime } \) is integrable on \( \left\lbrack {a, b}\right\rbrack \) . Now let \( G\left( x\right) = {\int }_{a}^{x}{F}^{\prime }\left( y\right) {dy} \) . Then \( G \) is absolutely continuous; hence so is the difference \( G\left( x\right) - F\left( x\right) \) . By the Lebesgue differentiation theorem (Theorem 1.4), we know that \( {G}^{\prime }\left( x\right) = {F}^{\prime }\left( x\right) \) for a.e. \( x \) ; hence the difference \( F - G \) has derivative 0 almost everywhere. By the previous theorem we conclude that \( F - G \) is constant, and evaluating this expression at \( x = a \) gives the desired result.
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Yes
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Lemma 3.12 A bounded increasing function \( F \) on \( \left\lbrack {a, b}\right\rbrack \) has at most countably many discontinuities.
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Proof. If \( F \) is discontinuous at \( x \), we may choose a rational number \( {r}_{x} \) so that \( F\left( {x}^{ - }\right) < {r}_{x} < F\left( {x}^{ + }\right) \) . If \( f \) is discontinuous at \( x \) and \( z \) with \( x < z \), we must have \( F\left( {x}^{ + }\right) \leq F\left( {z}^{ - }\right) \), hence \( {r}_{x} < {r}_{z} \) . Consequently, to each rational number corresponds at most one discontinuity of \( F \), hence \( F \) can have at most a countable number of discontinuities.
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Yes
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Lemma 3.13 If \( F \) is increasing and bounded on \( \left\lbrack {a, b}\right\rbrack \), then:\n\n(i) \( J\left( x\right) \) is discontinuous precisely at the points \( \left\{ {x}_{n}\right\} \) and has a jump at \( {x}_{n} \) equal to that of \( F \) .\n\n(ii) The difference \( F\left( x\right) - J\left( x\right) \) is increasing and continuous.
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Proof. If \( x \neq {x}_{n} \) for all \( n \), each \( {j}_{n} \) is continuous at \( x \), and since the series converges uniformly, \( J \) must be continuous at \( x \) . If \( x = {x}_{N} \) for some \( N \), then we write\n\n\[ J\left( x\right) = \mathop{\sum }\limits_{{n = 1}}^{N}{\alpha }_{n}{j}_{n}\left( x\right) + \mathop{\sum }\limits_{{n = N + 1}}^{\infty }{\alpha }_{n}{j}_{n}\left( x\right) .\n\]\n\nBy the same argument as above, the series on the right-hand side is continuous at \( x \) . Clearly, the finite sum has a jump discontinuity at \( {x}_{N} \) of size \( {\alpha }_{N} \) .\n\nFor (ii), we note that (i) implies at once that \( F - J \) is continuous. Finally, if \( y > x \) we have\n\n\[ J\left( y\right) - J\left( x\right) \leq \mathop{\sum }\limits_{{x < {x}_{n} \leq y}}{\alpha }_{n} \leq F\left( y\right) - F\left( x\right)\n\]\nwhere the last inequality follows since \( F \) is increasing. Hence\n\n\[ F\left( x\right) - J\left( x\right) \leq F\left( y\right) - J\left( y\right)\n\]\n\nand the difference \( F - J \) is increasing, as desired.
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Yes
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Theorem 3.14 If \( J \) is the jump function considered above, then \( {J}^{\prime }\left( x\right) \) exists and vanishes almost everywhere.
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Proof. Given any \( \epsilon > 0 \), we note that the set \( E \) of those \( x \) where\n\n(10)\n\n\[ \mathop{\limsup }\limits_{{h \rightarrow 0}}\frac{J\left( {x + h}\right) - J\left( x\right) }{h} > \epsilon \]\n\nis a measurable set. (The proof of this little fact is outlined in Exercise 14 below.) Suppose \( \delta = m\left( E\right) \) . We need to show that \( \delta = 0 \) . Now observe that since the series \( \sum {\alpha }_{n} \) arising in the definition of \( J \) converges, then for any \( \eta \), to be chosen later, we can find an \( N \) so large that \( \mathop{\sum }\limits_{{n > N}}{\alpha }_{n} < \eta \) . We then write\n\n\[ {J}_{0}\left( x\right) = \mathop{\sum }\limits_{{n > N}}{\alpha }_{n}{j}_{n}\left( x\right) \]\n\nand because of our choice of \( N \) we have\n\n(11)\n\n\[ {J}_{0}\left( b\right) - {J}_{0}\left( a\right) < \eta \]\n\nHowever, \( J - {J}_{0} \) is a finite sum of terms \( {\alpha }_{n}{j}_{n}\left( x\right) \), and therefore the set of points where (10) holds, with \( J \) replaced by \( {J}_{0} \), differs from \( E \) by at most a finite set, the points \( \left\{ {{x}_{1},{x}_{2},\ldots ,{x}_{N}}\right\} \) . Thus we can find a compact set \( K \), with \( m\left( K\right) \geq \delta /2 \), so that \( \mathop{\limsup }\limits_{{h \rightarrow 0}}\frac{{J}_{0}\left( {x + h}\right) - {J}_{0}\left( x\right) }{h} > \epsilon \) for each \( x \in K \) . Hence there are intervals \( \left( {{a}_{x},{b}_{x}}\right) \) containing \( x, x \in K \), so that \( {J}_{0}\left( {b}_{x}\right) - {J}_{0}\left( {a}_{x}\right) > \epsilon \left( {{b}_{x} - {a}_{x}}\right) \) . We can first choose a finite collection of these intervals that covers \( K \), and then apply Lemma 1.2 to select intervals \( {I}_{1},{I}_{2},\ldots ,{I}_{n} \) which are disjoint, and for which \( \mathop{\sum }\limits_{{j = 1}}^{n}m\left( {I}_{j}\right) \geq \) \( m\left( K\right) /3 \) . The intervals \( {I}_{j} = \left( {{a}_{j},{b}_{j}}\right) \) of course satisfy\n\n\[ {J}_{0}\left( {b}_{j}\right) - {J}_{0}\left( {a}_{j}\right) > \epsilon \left( {{b}_{j} - {a}_{j}}\right) \]\n\nNow,\n\n\[ {J}_{0}\left( b\right) - {J}_{0}\left( a\right) \geq \mathop{\sum }\limits_{{j = 1}}^{N}{J}_{0}\left( {b}_{j}\right) - {J}_{0}\left( {a}_{j}\right) > \epsilon \sum \left( {{b}_{j} - {a}_{j}}\right) \geq \frac{\epsilon }{3}m\left( K\right) \geq \frac{\epsilon }{6}\delta . \]\n\nThus by (11), \( {\epsilon \delta }/6 < \eta \), and since we are free to choose \( \eta \), it follows that \( \delta = 0 \) and the theorem is proved.
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No
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Theorem 4.1 Suppose \( \left( {x\left( t\right), y\left( t\right) }\right) \) is a curve defined for \( a \leq t \leq b \) . If both \( x\left( t\right) \) and \( y\left( t\right) \) are absolutely continuous, then the curve is rectifiable, and if \( L \) denotes its length, we have\n\n\[ L = {\int }_{a}^{b}{\left( {x}^{\prime }{\left( t\right) }^{2} + {y}^{\prime }{\left( t\right) }^{2}\right) }^{1/2}{dt}. \]
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Note that if \( F\left( t\right) = x\left( t\right) + {iy}\left( t\right) \) is absolutely continuous then it is automatically of bounded variation, and hence the curve is rectifiable. The identity (12) is an immediate consequence of the proposition below, which can be viewed as a more precise version of Corollary 3.7 for absolutely continuous functions.\n\nProposition 4.2 Suppose \( F \) is complex-valued and absolutely continuous on \( \left\lbrack {a, b}\right\rbrack \) . Then\n\n\[ {T}_{F}\left( {a, b}\right) = {\int }_{a}^{b}\left| {{F}^{\prime }\left( t\right) }\right| {dt} \]
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Yes
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Proposition 4.2 Suppose \( F \) is complex-valued and absolutely continuous on \( \left\lbrack {a, b}\right\rbrack \) . Then\n\n\[ \n{T}_{F}\left( {a, b}\right) = {\int }_{a}^{b}\left| {{F}^{\prime }\left( t\right) }\right| {dt} \n\]
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In fact, because of Theorem 3.11, for any partition \( a = {t}_{0} < {t}_{1} < \cdots < \) \( {t}_{N} = b \) of \( \left\lbrack {a, b}\right\rbrack \), we have\n\n\[ \n\mathop{\sum }\limits_{{j = 1}}^{N}\left| {F\left( {t}_{j}\right) - F\left( {t}_{j - 1}\right) }\right| = \mathop{\sum }\limits_{{j = 1}}^{N}\left| {{\int }_{{t}_{j - 1}}^{{t}_{j}}{F}^{\prime }\left( t\right) {dt}}\right| \n\]\n\n\[ \n\leq \mathop{\sum }\limits_{{j = 1}}^{N}{\int }_{{t}_{j - 1}}^{{t}_{j}}\left| {{F}^{\prime }\left( t\right) }\right| {dt} \n\]\n\n\[ \n= {\int }_{a}^{b}\left| {{F}^{\prime }\left( t\right) }\right| {dt} \n\]\n\nSo this proves\n\n(13)\n\n\[ \n{T}_{F}\left( {a, b}\right) \leq {\int }_{a}^{b}\left| {{F}^{\prime }\left( t\right) }\right| {dt} \n\]\n\nTo prove the reverse inequality, fix \( \epsilon > 0 \), and using Theorem 2.4 in Chapter 2 find a step function \( g \) on \( \left\lbrack {a, b}\right\rbrack \), such that \( {F}^{\prime } = g + h \) with \( {\int }_{a}^{b}\left| {h\left( t\right) }\right| {dt} < \epsilon \) . Set \( G\left( x\right) = {\int }_{a}^{x}g\left( t\right) {dt} \), and \( H\left( x\right) = {\int }_{a}^{x}h\left( t\right) {dt} \) . Then \( F = \) \( G + H \), and as is easily seen\n\n\[ \n{T}_{F}\left( {a, b}\right) \geq {T}_{G}\left( {a, b}\right) - {T}_{H}\left( {a, b}\right) \n\]\n\nHowever, by (13) \( {T}_{H}\left( {a, b}\right) < \epsilon \), so that\n\n\[ \n{T}_{F}\left( {a, b}\right) \geq {T}_{G}\left( {a, b}\right) - \epsilon . \n\]\n\nNow partition the interval \( \left\lbrack {a, b}\right\rbrack \), as \( a = {t}_{0} < \cdots < {t}_{N} = b \), so that the step function \( g \) is constant on each of the intervals \( \left( {{t}_{j - 1},{t}_{j}}\right), j = 1,2,\ldots, N \) . Then\n\n\[ \n{T}_{G}\left( {a, b}\right) \geq \mathop{\sum }\limits_{{j = 1}}^{N}\left| {G\left( {t}_{j}\right) - G\left( {t}_{j - 1}\right) }\right| \n\]\n\n\[ \n= \mathop{\sum }\limits_{{j = 1}}^{N}\left| {{\int }_{{t}_{j} - 1}^{{t}_{j}}g\left( t\right) {dt}}\right| \n\]\n\n\[ \n= \sum {\int }_{{t}_{j - 1}}^{{t}_{j}}\left| {g\left( t\right) }\right| {dt} \n\]\n\n\[ \n= {\int }_{a}^{b}\left| {g\left( t\right) }\right| {dt} \n\]\nSince \( {\int }_{a}^{b}\left| {g\left( t\right) }\right| {dt} \geq {\int }_{a}^{b}\left| {{F}^{\prime }\left( t\right) }\right| {dt} - \epsilon \), we obtain as a consequence that\n\n\[ \n{T}_{F}\left( {a, b}\right) \geq {\int }_{a}^{b}\left| {{F}^{\prime }\left( t\right) }\right| {dt} - {2\epsilon } \n\]\n\nand letting \( \epsilon \rightarrow 0 \) we establish the assertion and also the theorem.
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Yes
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Theorem 4.3 Suppose \( \left( {x\left( t\right), y\left( t\right) }\right), a \leq t \leq b \), is a rectifiable curve that has length \( L \) . Consider the arc-length parametrization \( \widetilde{z}\left( s\right) = \left( {\widetilde{x}\left( s\right) ,\widetilde{y}\left( s\right) }\right) \) described above. Then \( \widetilde{x} \) and \( \widetilde{y} \) are absolutely continuous, \( \left| {{\widetilde{z}}^{\prime }\left( s\right) }\right| = 1 \) for almost every \( s \in \left\lbrack {0, L}\right\rbrack \), and\n\n\[ L = {\int }_{0}^{L}{\left( {\widetilde{x}}^{\prime }{\left( s\right) }^{2} + {\widetilde{y}}^{\prime }{\left( s\right) }^{2}\right) }^{1/2}{ds}. \]
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Proof. We noted that \( \left| {\widetilde{z}\left( {s}_{1}\right) - \widetilde{z}\left( {s}_{2}\right) }\right| \leq \left| {{s}_{1} - {s}_{2}}\right| \), so it follows immediately that \( \widetilde{z}\left( s\right) \) is absolutely continuous, hence differentiable almost everywhere. Moreover, this inequality also proves that \( \left| {{\widetilde{z}}^{\prime }\left( s\right) }\right| \leq 1 \), for almost every \( s \) . By definition the total variation of \( \widetilde{z} \) equals \( L \), and by the previous theorem we must have \( L = {\int }_{0}^{L}\left| {{\widetilde{z}}^{\prime }\left( s\right) }\right| {ds} \) . Finally, we note that this identity is possible only when \( \left| {{\widetilde{z}}^{\prime }\left( s\right) }\right| = 1 \) almost everywhere.
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Yes
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Theorem 4.4 Suppose \( \Gamma = \{ z\left( t\right), a \leq t \leq b\} \) is a quasi-simple curve. The Minkowski content of \( \Gamma \) exists if and only if \( \Gamma \) is rectifiable. When this is the case and \( L \) is the length of the curve, then \( \mathcal{M}\left( \Gamma \right) = L \) .
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To prove the theorem, we also consider for any compact set \( K \)\n\n\[ \n{\mathcal{M}}^{ * }\left( K\right) = \mathop{\limsup }\limits_{{\delta \rightarrow 0}}\frac{m\left( {K}^{\delta }\right) }{2\delta }\;\text{ and }\;{\mathcal{M}}_{ * }\left( K\right) = \mathop{\liminf }\limits_{{\delta \rightarrow 0}}\frac{m\left( {K}^{\delta }\right) }{2\delta }\n\]\n\n(both taken as extended positive numbers). Of course \( {\mathcal{M}}_{ * }\left( K\right) \leq {\mathcal{M}}^{ * }\left( K\right) \) . To say that the Minkowski content exists is the same as saying that \( {\mathcal{M}}^{ * }\left( K\right) < \infty \) and \( {\mathcal{M}}_{ * }\left( K\right) = {\mathcal{M}}^{ * }\left( K\right) \) . Their common value is then \( \mathcal{M}\left( K\right) \) .
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No
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Lemma 4.6 If \( \Gamma = \{ z\left( t\right), a \leq t \leq b\} \) is any curve, and \( \Delta = \left| {z\left( b\right) - z\left( a\right) }\right| \) is the distance between its end-points, then \( m\left( {\Gamma }^{\delta }\right) \geq {2\delta \Delta } \) .
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Proof. Since the distance function and the Lebesgue measure are invariant under translations and rotations (see Section 3 in Chapter 1 and Problem 4 in Chapter 2) we may transform the situation by an appropriate composition of these motions. Therefore we may assume that the end-points of the curve have been placed on the \( x \) -axis, and thus we may suppose that \( z\left( a\right) = \left( {A,0}\right), z\left( b\right) = \left( {B,0}\right) \) with \( A < B \), and \( \Delta = B - A \) (in the case \( A = B \) the conclusion is automatically verified).\n\nBy the continuity of the function \( x\left( t\right) \), there is for each \( x \) in \( \left\lbrack {A, B}\right\rbrack \) a value \( \bar{t} \) in \( \left\lbrack {a, b}\right\rbrack \), such that \( x = x\left( \bar{t}\right) \) . Since \( \bar{Q} = \left( {x\left( \bar{t}\right), y\left( \bar{t}\right) }\right) \in \Gamma \), the set\n\n\( {\Gamma }^{\delta } \) contains a segment parallel to the \( y \) -axis, of length \( {2\delta } \) centered at \( \bar{Q} \) lying above \( x \) (see Figure 10). In other words the slice \( {\left( {\Gamma }^{\delta }\right) }_{x} \) contains the interval \( \left( {y\left( \bar{t}\right) - \delta, y\left( \bar{t}\right) + \delta }\right) \), and hence \( {m}_{1}\left( {\left( {\Gamma }^{\delta }\right) }^{x}\right) \geq {2\delta } \) (where \( {m}_{1} \) is the one-dimensional Lebesgue measure). However by Fubini's theorem\n\n\[ m\left( {\Gamma }^{\delta }\right) = {\int }_{\mathbb{R}}{m}_{1}\left( {\left( {\Gamma }^{\delta }\right) }_{x}\right) {dx} \geq {\int }_{A}^{B}{m}_{1}\left( {\left( {\Gamma }^{\delta }\right) }_{x}\right) {dx} \geq {2\delta }\left( {B - A}\right) = {2\delta \Delta }, \]\n\nand the lemma is proved.
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Yes
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Proposition 4.7 Suppose \( \Gamma = \{ z\left( t\right), a \leq t \leq b\} \) is a rectifiable curve with length \( L \) . Then\n\n\[{\mathcal{M}}^{ * }\left( \Gamma \right) \leq L\]
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The quantities \( {\mathcal{M}}^{ * }\left( \Gamma \right) \) and \( L \) are of course independent of the parametrization used; since the curve is rectifiable, it will be convenient to use the arclength parametrization. Thus we write the curve as \( z\left( s\right) = \left( {x\left( s\right), y\left( s\right) }\right) \) , with \( 0 \leq s \leq L \), and recall that then \( z\left( s\right) \) is absolutely continuous and \( \left| {{z}^{\prime }\left( s\right) }\right| = 1 \) for a.e. \( s \in \left\lbrack {0, L}\right\rbrack \) .\n\nWe first fix any \( 0 < \epsilon < 1 \), and find a measurable set \( {E}_{\epsilon } \subset \mathbb{R} \) and a positive number \( {r}_{\epsilon } \) such that \( m\left( {E}_{\epsilon }\right) < \epsilon \) and\n\n(14)\n\n\[ \mathop{\sup }\limits_{{0 < \left| h\right| < {r}_{\epsilon }}}\left| {\frac{z\left( {s + h}\right) - z\left( s\right) }{h} - {z}^{\prime }\left( s\right) }\right| < \epsilon \;\text{ for all }s \in \left\lbrack {0, L}\right\rbrack - {E}_{\epsilon }. \]\n\nIndeed, for each integer \( n \), let\n\n\[ {F}_{n}\left( s\right) = \mathop{\sup }\limits_{{0 < \left| h\right| < 1/n}}\left| {\frac{z\left( {s + h}\right) - z\left( s\right) }{h} - {z}^{\prime }\left( s\right) }\right| \]\n\n(where \( z\left( s\right) \) has been extended outside \( \left\lbrack {0, L}\right\rbrack \), so that \( z\left( s\right) = z\left( 0\right) \), when \( s < 0 \), and \( z\left( s\right) = z\left( L\right) \) when \( s > L \) ). Because \( z\left( s\right) \) is continuous the supremum of \( h \) in the definition of \( {F}_{n}\left( s\right) \) can be replaced by a supremum of countably many measurable functions, and hence each \( {F}_{n} \) is measurable. However, \( {F}_{n}\left( s\right) \rightarrow 0 \), as \( n \rightarrow \infty \) for a.e \( s \in \left\lbrack {a, b}\right\rbrack \) . Thus by Egorov’s theorem the convergence is uniform outside a set \( {E}_{\epsilon } \) with \( m\left( {E}_{\epsilon }\right) < \epsilon \) , and so we merely need to choose \( {r}_{\epsilon } = 1/n \) for sufficiently large \( n \) to establish (14). It will be convenient in what follows to assume, as we may, that \( {z}^{\prime }\left( s\right) \) exists and \( \left| {{z}^{\prime }\left( s\right) }\right| = 1 \) for every \( s \notin {E}_{\epsilon } \) .\n\nNow for any \( 0 < \rho < {r}_{\epsilon } \) (with \( \rho < 1 \) ), we partition the interval \( \left\lbrack {0, L}\right\rbrack \) into consecutive closed intervals, each of length \( \rho \) ,(except that the last interval may have length \( \leq \rho \) ). Then there is a total of \( N \leq L/\rho + 1 \) such intervals that arise. We call these intervals \( {I}_{1},{I}_{2},\ldots ,{I}_{N} \), and divide them into two classes. The first class, those intervals \( {I}_{j} \) we call \
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Yes
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Proposition 1.1 The space \( {L}^{2}\left( {\mathbb{R}}^{d}\right) \) has the following properties:\n\n(i) \( {L}^{2}\left( {\mathbb{R}}^{d}\right) \) is a vector space.\n\n(ii) \( f\left( x\right) \overline{g\left( x\right) } \) is integrable whenever \( f, g \in {L}^{2}\left( {\mathbb{R}}^{d}\right) \), and the Cauchy-Schwarz inequality holds: \( \left| \left( {f, g}\right) \right| \leq \parallel f\parallel \parallel g\parallel \) .\n\n(iii) If \( g \in {L}^{2}\left( {\mathbb{R}}^{d}\right) \) is fixed, the map \( f \mapsto \left( {f, g}\right) \) is linear in \( f \), and also \( \left( {f, g}\right) = \overline{\left( g, f\right) }.\n\n(iv) The triangle inequality holds: \( \parallel f + g\parallel \leq \parallel f\parallel + \parallel g\parallel \) .
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Proof. If \( f, g \in {L}^{2}\left( {\mathbb{R}}^{d}\right) \), then since \( \left| {f\left( x\right) + g\left( x\right) }\right| \leq 2\max \left( {\left| {f\left( x\right) }\right| ,\left| {g\left( x\right) }\right| }\right) \), we have\n\n\[{\left| f\left( x\right) + g\left( x\right) \right| }^{2} \leq 4\left( {{\left| f\left( x\right) \right| }^{2} + {\left| g\left( x\right) \right| }^{2}}\right) ,\]\n\ntherefore\n\n\[ \int {\left| f + g\right| }^{2} \leq 4\int {\left| f\right| }^{2} + 4\int {\left| g\right| }^{2} < \infty \]\n\nhence \( f + g \in {L}^{2}\left( {\mathbb{R}}^{d}\right) \) . Also, if \( \lambda \in \mathbb{C} \) we clearly have \( {\lambda f} \in {L}^{2}\left( {\mathbb{R}}^{d}\right) \), and part (i) is proved.\n\nTo see why \( f\bar{g} \) is integrable whenever \( f \) and \( g \) are in \( {L}^{2}\left( {\mathbb{R}}^{d}\right) \), it suffices to recall that for all \( A, B \geq 0 \), one has \( {2AB} \leq {A}^{2} + {B}^{2} \), so that\n\n(1)\n\n\[ \int \left| {f\bar{g}}\right| \leq \frac{1}{2}\left\lbrack {\parallel f{\parallel }^{2} + \parallel g{\parallel }^{2}}\right\rbrack \]\n\nTo prove the Cauchy-Schwarz inequality, we first observe that if either \( \parallel f\parallel = 0 \) or \( \parallel g\parallel = 0 \), then \( {fg} = 0 \) is zero almost everywhere, hence \( \left( {f, g}\right) = \) 0 and the inequality is obvious. Next, if we assume that \( \parallel f\parallel = \parallel g\parallel = 1 \) , then we get the desired inequality \( \left| \left( {f, g}\right) \right| \leq 1 \) . This follows from the fact that \( \left| \left( {f, g}\right) \right| \leq \int \left| {f\bar{g}}\right| \), and inequality (1). Finally, in the case when both \( \parallel f\parallel \) and \( \parallel g\parallel \) are non-zero, we normalize \( f \) and \( g \) by setting\n\n\[ \widetilde{f} = f/\parallel f\parallel \;\text{ and }\;\widetilde{g} = g/\parallel g\parallel \]\n\nso that \( \parallel \widetilde{f}\parallel = \parallel \widetilde{g}\parallel = 1 \) . By our previous observation we then find\n\n\[ \left| \left( {\widetilde{f},\widetilde{g}}\right) \right| \leq 1 \]\n\nMultiplying both sides of the above by \( \parallel f\parallel \parallel g\parallel \) yields the Cauchy-Schwarz inequality.\n\nPart (iii) follows from the linearity of the integral.\n\nFinally, to prove the triangle inequality, we use the Cauchy-Schwarz inequality as follows:\n\n\[ \parallel f + g{\parallel }^{2} = \left( {f + g, f + g}\right) \]\n\n\[ = \parallel f{\parallel }^{2} + \left( {f, g}\right) + \left( {g, f}\right) + \parallel g{\parallel }^{2} \]\n\n\[ \leq \parallel f{\parallel }^{2} + 2\left| \left( {f, g}\right) \right| + \parallel g{\parallel }^{2} \]\n\n\[ \leq {\begin{Vmatrix}f\end{Vmatrix}}^{2} + 2\begin{Vmatrix}f\end{Vmatrix}\begin{Vmatrix}g\end{Vmatrix} + {\begin{Vmatrix}g\end{Vmatrix}}^{2} \]\n\n\[ = {\left( \parallel f\parallel + \parallel g\parallel \right) }^{2}, \]\n\nand taking square roots completes the argument.
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Yes
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Theorem 1.3 The space \( {L}^{2}\left( {\mathbb{R}}^{d}\right) \) is separable, in the sense that there exists a countable collection \( \left\{ {f}_{k}\right\} \) of elements in \( {L}^{2}\left( {\mathbb{R}}^{d}\right) \) such that their linear combinations are dense in \( {L}^{2}\left( {\mathbb{R}}^{d}\right) \) .
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Proof. Consider the family of functions of the form \( r{\chi }_{R}\left( x\right) \), where \( r \) is a complex number with rational real and imaginary parts, and \( R \) is a rectangle in \( {\mathbb{R}}^{d} \) with rational coordinates. We claim that finite linear combinations of these type of functions are dense in \( {L}^{2}\left( {\mathbb{R}}^{d}\right) \).\n\nSuppose \( f \in {L}^{2}\left( {\mathbb{R}}^{d}\right) \) and let \( \epsilon > 0 \). Consider for each \( n \geq 1 \) the function \( {g}_{n} \) defined by\n\n\[ \n{g}_{n}\left( x\right) = \left\{ \begin{matrix} f\left( x\right) & \text{ if }\left| x\right| \leq n\text{ and }\left| {f\left( x\right) }\right| \leq n, \\ 0 & \text{ otherwise. } \end{matrix}\right.\n\]\n\nThen \( {\left| f - {g}_{n}\right| }^{2} \leq 4{\left| f\right| }^{2} \) and \( {g}_{n}\left( x\right) \rightarrow f\left( x\right) \) almost everywhere. \( {}^{1} \) The dominated convergence theorem implies that \( {\begin{Vmatrix}f - {g}_{n}\end{Vmatrix}}_{{L}^{2}\left( {\mathbb{R}}^{d}\right) }^{2} \rightarrow 0 \) as \( n \) tends to infinity; therefore we have\n\n\[ \n{\begin{Vmatrix}f - {g}_{N}\end{Vmatrix}}_{{L}^{2}\left( {\mathbb{R}}^{d}\right) } < \epsilon /2\;\text{ for some }N.\n\]\n\nLet \( g = {g}_{N} \), and note that \( g \) is a bounded function supported on a bounded set; thus \( g \in {L}^{1}\left( {\mathbb{R}}^{d}\right) \). We may now find a step function \( \varphi \) so that \( \left| \varphi \right| \leq N \) and \( \int \left| {g - \varphi }\right| < {\epsilon }^{2}/{16N} \) (Theorem 2.4, Chapter 2). By replacing the coefficients and rectangles that appear in the canonical form of \( \varphi \) by complex numbers with rational real and imaginary parts, and rectangles with rational coordinates, we may find a \( \psi \) with \( \left| \psi \right| \leq N \) and \( \int \left| {g - \psi }\right| < {\epsilon }^{2}/{8N} \). Finally, we note that\n\n\[ \n\int {\left| g - \psi \right| }^{2} \leq {2N}\int \left| {g - \psi }\right| < {\epsilon }^{2}/4\n\]\n\nConsequently \( \parallel g - \psi \parallel < \epsilon /2 \), therefore \( \parallel f - \psi \parallel < \epsilon \), and the proof is complete.
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Yes
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Proposition 2.1 If \( f \bot g \), then \( \parallel f + g{\parallel }^{2} = \parallel f{\parallel }^{2} + \parallel g{\parallel }^{2} \) .
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Proof. It suffices to note that \( \left( {f, g}\right) = 0 \) implies \( \left( {g, f}\right) = 0 \), and therefore\n\n\[ \n\parallel f + g{\parallel }^{2} = \left( {f + g, f + g}\right) = \parallel f{\parallel }^{2} + \left( {f, g}\right) + \left( {g, f}\right) + \parallel g{\parallel }^{2}\n\]\n\n\[ \n= \parallel f{\parallel }^{2} + \parallel g{\parallel }^{2}.\n\]
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Yes
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Proposition 2.2 If \( {\left\{ {e}_{k}\right\} }_{k = 1}^{\infty } \) is orthonormal, and \( f = \sum {a}_{k}{e}_{k} \in \mathcal{H} \) where the sum is finite, then\n\n\[ \parallel f{\parallel }^{2} = \sum {\left| {a}_{k}\right| }^{2} \]
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The proof is a simple application of the Pythagorean theorem.
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No
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Corollary 2.5 Any two infinite-dimensional Hilbert spaces are unitarily equivalent.
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Proof. If \( \mathcal{H} \) and \( {\mathcal{H}}^{\prime } \) are two infinite-dimensional Hilbert spaces, we may select for each an orthonormal basis, say\n\n\[ \left\{ {{e}_{1},{e}_{2},\ldots }\right\} \subset \mathcal{H}\;\text{ and }\;\left\{ {{e}_{1}^{\prime },{e}_{2}^{\prime },\ldots }\right\} \subset {\mathcal{H}}^{\prime }.\]\n\nThen, consider the mapping defined as follows: if \( f = \mathop{\sum }\limits_{{k = 1}}^{\infty }{a}_{k}{e}_{k} \), then\n\n\[ U\left( f\right) = g,\;\text{ where }\;g = \mathop{\sum }\limits_{{k = 1}}^{\infty }{a}_{k}{e}_{k}^{\prime }.\]\n\nClearly, the mapping \( U \) is both linear and invertible. Moreover, by Parseval's identity, we must have\n\n\[ \parallel {Uf}{\parallel }_{{\mathcal{H}}^{\prime }}^{2} = \parallel g{\parallel }_{{\mathcal{H}}^{\prime }}^{2} = \mathop{\sum }\limits_{{k = 1}}^{\infty }{\left| {a}_{k}\right| }^{2} = \parallel f{\parallel }_{\mathcal{H}}^{2} \]\n\nand the corollary is proved.
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Yes
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Corollary 2.6 Any two finite-dimensional Hilbert spaces are unitarily equivalent if and only if they have the same dimension.
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Thus every finite-dimensional Hilbert space over \( \mathbb{C} \) (or over \( \mathbb{R} \) ) is equivalent with \( {\mathbb{C}}^{d} \) (or \( {\mathbb{R}}^{d} \) ), for some \( d \) .
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No
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Proposition 2.7 Suppose we are given a pre-Hilbert space \( {\mathcal{H}}_{0} \) with inner product \( {\left( \cdot , \cdot \right) }_{0} \) . Then we can find a Hilbert space \( \mathcal{H} \) with inner product \( \left( {\cdot , \cdot }\right) \) such that\n\n(i) \( {\mathcal{H}}_{0} \subset \mathcal{H} \) .\n\n(ii) \( {\left( f, g\right) }_{0} = \left( {f, g}\right) \) whenever \( f, g \in {\mathcal{H}}_{0} \) .\n\n(iii) \( {\mathcal{H}}_{0} \) is dense in \( \mathcal{H} \) .
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A Hilbert space satisfying properties like \( \mathcal{H} \) in the above proposition is called a completion of \( {\mathcal{H}}_{0} \) . We shall only sketch the construction of \( \mathcal{H} \), since it follows closely Cantor’s familiar method of obtaining the real numbers as the completion of the rationals in terms of Cauchy sequences of rationals.\n\nIndeed, consider the collection of all Cauchy sequences \( \left\{ {f}_{n}\right\} \) with \( {f}_{n} \in \) \( {\mathcal{H}}_{0},1 \leq n < \infty \) . One defines an equivalence relation in this collection by saying that \( \left\{ {f}_{n}\right\} \) is equivalent to \( \left\{ {f}_{n}^{\prime }\right\} \) if \( {f}_{n} - {f}_{n}^{\prime } \) converges to 0 as \( n \rightarrow \infty \) . The collection of equivalence classes is then taken to be \( \mathcal{H} \) . One then easily verifies that \( \mathcal{H} \) inherits the structure of a vector space, with an inner product \( \left( {f, g}\right) \) defined as \( \mathop{\lim }\limits_{{n \rightarrow \infty }}\left( {{f}_{n},{g}_{n}}\right) \), where \( \left\{ {f}_{n}\right\} \) and \( \left\{ {g}_{n}\right\} \) are Cauchy sequences in \( {\mathcal{H}}_{0} \), representing, respectively, the elements \( f \) and \( g \) in \( \mathcal{H} \) . Next, if \( f \in {\mathcal{H}}_{0} \) we take the sequence \( \left\{ {f}_{n}\right\} \), with \( {f}_{n} = f \) for all \( n \), to represent \( f \) as an element of \( \mathcal{H} \), giving \( {\mathcal{H}}_{0} \subset \mathcal{H} \) . To see that \( \mathcal{H} \) is complete, let \( {\left\{ {F}^{k}\right\} }_{k = 1}^{\infty } \) be a Cauchy sequence in \( \mathcal{H} \), with each \( {F}^{k} \) represented by \( {\left\{ {f}_{n}^{k}\right\} }_{n = 1}^{\infty },{f}_{n}^{k} \in {\mathcal{H}}_{0} \) . If we define \( F \in \mathcal{H} \) as represented by the sequence \( \left\{ {f}_{n}\right\} \) with \( {f}_{n} = {f}_{N\left( n\right) }^{n} \), where \( N\left( n\right) \) is so that \( \left| {{f}_{N\left( n\right) }^{n} - {f}_{j}^{n}}\right| \leq \) \( 1/n \) for \( j \geq N\left( n\right) \), then we note that \( {F}^{k} \rightarrow F \) in \( \mathcal{H} \) .
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Yes
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Theorem 3.1 Suppose \( f \) is integrable on \( \left\lbrack {-\pi ,\pi }\right\rbrack \) .\n\n(i) If \( {a}_{n} = 0 \) for all \( n \), then \( f\left( x\right) = 0 \) for a.e. \( x \) .\n\n(ii) \( \mathop{\sum }\limits_{{n = - \infty }}^{\infty }{a}_{n}{r}^{\left| n\right| }{e}^{inx} \) tends to \( f\left( x\right) \) for a.e. \( x \), as \( r \rightarrow 1, r < 1 \) .
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Proof. The first conclusion is an immediate consequence of the second. To prove the latter we recall the identity\n\n\[ \mathop{\sum }\limits_{{n = - \infty }}^{\infty }{r}^{\left| n\right| }{e}^{iny} = {P}_{r}\left( y\right) = \frac{1 - {r}^{2}}{1 - {2r}\cos y + {r}^{2}} \]\n\nfor the Poisson kernel; see Book I, Chapter 2. Starting with our given \( f \in {L}^{1}\left( \left\lbrack {-\pi ,\pi }\right\rbrack \right) \) we extend it as a function on \( \mathbb{R} \) by making it periodic of period \( {2\pi }.{}^{3} \) We then claim that for every \( x \)\n\n(3)\n\n\[ \mathop{\sum }\limits_{{n = - \infty }}^{\infty }{a}_{n}{r}^{\left| n\right| }{e}^{inx} = \frac{1}{2\pi }{\int }_{-\pi }^{\pi }f\left( {x - y}\right) {P}_{r}\left( y\right) {dy}. \]\n\nIndeed, by the dominated convergence theorem the right-hand side equals\n\n\[ \sum {r}^{\left| n\right| }\frac{1}{2\pi }{\int }_{-\pi }^{\pi }f\left( {x - y}\right) {e}^{iny}{dy} \]\n\nMoreover, for each \( x \) and \( n \)\n\n\[ {\int }_{-\pi }^{\pi }f\left( {x - y}\right) {e}^{iny}{dy} = {\int }_{-\pi + x}^{\pi + x}f\left( y\right) {e}^{{in}\left( {x - y}\right) }{dy} \]\n\n\[ = {e}^{inx}{\int }_{-\pi }^{\pi }f\left( y\right) {e}^{-{iny}}{dy} = {e}^{inx}{2\pi }{a}_{n}. \]\n\n\( {}^{3} \) Note that we may without loss of generality assume that \( f\left( \pi \right) = f\left( {-\pi }\right) \) so as to make the periodic extension unambiguous.\n\nThe first equality follows by translation invariance (see Section 3, Chapter 2), and the second since \( {\int }_{-\pi }^{\pi }F\left( y\right) {dy} = {\int }_{I}F\left( y\right) {dy} \) whenever \( F \) is periodic of period \( {2\pi } \) and \( I \) is an interval of length \( {2\pi } \) (Exercise 3, Chapter 2). With these observations, the identity (3) is established. We can now invoke the facts about approximations to the identity (Theorem 2.1 and Example 4, Chapter 3) to conclude that the left-hand side of (3) tends to \( f\left( x\right) \) at every point of the Lebesgue set of \( f \), hence almost everywhere. (To be correct, the hypotheses of the theorem require that \( f \) be integrable on all of \( \mathbb{R} \) . We can achieve this for our periodic function by setting \( f \) equal to zero outside \( \left\lbrack {-{2\pi },{2\pi }}\right\rbrack \), and then (3) still holds for this modified \( f \), whenever \( x \in \left\lbrack {-\pi ,\pi }\right\rbrack \) .)
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Yes
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Theorem 3.2 Suppose \( f \in {L}^{2}\left( \left\lbrack {-\pi ,\pi }\right\rbrack \right) \) . Then:\n\n(i) We have Parseval's relation\n\n\[ \mathop{\sum }\limits_{{n = - \infty }}^{\infty }{\left| {a}_{n}\right| }^{2} = \frac{1}{2\pi }{\int }_{-\pi }^{\pi }{\left| f\left( x\right) \right| }^{2}{dx} \]
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To apply the previous results, we let \( \mathcal{H} = {L}^{2}\left( \left\lbrack {-\pi ,\pi }\right\rbrack \right) \) with inner product \( \left( {f, g}\right) = \frac{1}{2\pi }{\int }_{-\pi }^{\pi }f\left( x\right) \overline{g\left( x\right) }{dx} \), and take the orthonormal set \( {\left\{ {e}_{k}\right\} }_{k = 1}^{\infty } \) to be the exponentials \( {\left\{ {e}^{inx}\right\} }_{n = - \infty }^{\infty } \), with \( k = 1 \) when \( n = 0, k = {2n} \) for \( n > 0 \), and \( k = 2\left| n\right| - 1 \) for \( n < 0 \) .\n\nBy the previous result, assertion (ii) of Theorem 2.3 holds and thus all the other conclusions hold. We therefore have Parseval's relation, and from (iv) we conclude that \( {\begin{Vmatrix}f - {S}_{N}\left( f\right) \end{Vmatrix}}^{2} = \mathop{\sum }\limits_{{\left| n\right| > N}}{\left| {a}_{n}\right| }^{2} \rightarrow 0 \) as \( N \rightarrow \infty \) .
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Yes
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Theorem 3.3 A bounded holomorphic function \( F\left( {r{e}^{i\theta }}\right) \) on the unit disc has radial limits at almost every \( \theta \) .
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Proof. We know that \( F\left( z\right) \) has a power series expansion \( \mathop{\sum }\limits_{{n = 0}}^{\infty }{a}_{n}{z}^{n} \) in \( \mathbb{D} \) that converges absolutely and uniformly whenever \( z = r{e}^{i\theta } \) and \( r < 1 \) . In fact, for \( r < 1 \) the series \( \mathop{\sum }\limits_{{n = 0}}^{\infty }{a}_{n}{r}^{n}{e}^{in\theta } \) is the Fourier series of the function \( F\left( {r{e}^{i\theta }}\right) \), that is,\n\n\[ \n{a}_{n}{r}^{n} = \frac{1}{2\pi }{\int }_{-\pi }^{\pi }F\left( {r{e}^{i\theta }}\right) {e}^{-{in\theta }}{d\theta }\;\text{ when }n \geq 0, \n\]\n\nand the integral vanishes when \( n < 0 \) . (See also Chapter 3, Section 7 in Book II).\n\nWe pick \( M \) so that \( \left| {F\left( z\right) }\right| \leq M \), for all \( z \in \mathbb{D} \) . By Parseval’s identity\n\n\[ \n\mathop{\sum }\limits_{{n = 0}}^{\infty }{\left| {a}_{n}\right| }^{2}{r}^{2n} = \frac{1}{2\pi }{\int }_{-\pi }^{\pi }{\left| F\left( r{e}^{i\theta }\right) \right| }^{2}{d\theta }\;\text{ for each }0 \leq r < 1. \n\]\n\nLetting \( r \rightarrow 1 \) one sees that \( \sum {\left| {a}_{n}\right| }^{2} \) converges (and is \( \leq {M}^{2} \) ). We now let \( F\left( {e}^{i\theta }\right) \) be the \( {L}^{2} \) -function whose Fourier coefficients are \( {a}_{n} \) when \( n \geq 0 \) , and 0 when \( n < 0 \) . Hence by conclusion (ii) in Theorem 3.1\n\n\[ \n\mathop{\sum }\limits_{{n = 0}}^{\infty }{a}_{n}{r}^{n}{e}^{in\theta } \rightarrow F\left( {e}^{i\theta }\right) ,\;\text{ for a.e }\theta , \n\]\n\nconcluding the proof of the theorem.
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Yes
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Proposition 4.2 If \( \mathcal{S} \) is a closed subspace of a Hilbert space \( \mathcal{H} \), then\n\n\[ \mathcal{H} = \mathcal{S} \oplus {\mathcal{S}}^{ \bot } \]\n\nThe notation in the proposition means that every \( f \in \mathcal{H} \) can be written uniquely as \( f = g + h \), where \( g \in \mathcal{S} \) and \( h \in {\mathcal{S}}^{ \bot } \) ; we say that \( \mathcal{H} \) is the direct sum of \( S \) and \( {S}^{ \bot } \) . This is equivalent to saying that any \( f \) in \( \mathcal{H} \) is the sum of two elements, one in \( \mathcal{S} \), the other in \( {\mathcal{S}}^{ \bot } \), and that \( \mathcal{S} \cap {\mathcal{S}}^{ \bot } \) contains only 0 .
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The proof of the proposition relies on the previous lemma giving the closest element of \( f \) in \( \mathcal{S} \) . In fact, for any \( f \in \mathcal{H} \), we choose \( {g}_{0} \) as in the lemma and write\n\n\[ f = {g}_{0} + \left( {f - {g}_{0}}\right) \]\n\nBy construction \( {g}_{0} \in \mathcal{S} \), and the lemma implies \( f - {g}_{0} \in {S}^{ \bot } \), and this shows that \( f \) is the sum of an element in \( \mathcal{S} \) and one in \( {\mathcal{S}}^{ \bot } \) . To prove that this decomposition is unique, suppose that\n\n\[ f = g + h = \widetilde{g} + \widetilde{h}\;\text{ where }g,\widetilde{g} \in \mathcal{S}\text{ and }h,\widetilde{h} \in {\mathcal{S}}^{ \bot }.\]\n\nThen, we must have \( g - \widetilde{g} = \widetilde{h} - h \) . Since the left-hand side belongs to \( \mathcal{S} \) while the right-hand side belongs to \( {\mathcal{S}}^{ \bot } \) the fact that \( \mathcal{S} \cap {\mathcal{S}}^{ \bot } = \{ 0\} \) implies \( g - \widetilde{g} = 0 \) and \( \widetilde{h} - h = 0 \) . Therefore \( g = \widetilde{g} \) and \( h = \widetilde{h} \) and the uniqueness is established.
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Yes
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Lemma 5.1 \( \parallel T\parallel = \sup \{ \left| \left( {{Tf}, g}\right) \right| : \parallel f\parallel \leq 1,\parallel g\parallel \leq 1\} \)
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Proof. If \( \parallel T\parallel \leq M \), the Cauchy-Schwarz inequality gives\n\n\[ \left| \left( {{Tf}, g}\right) \right| \leq M\;\text{ whenever }\parallel f\parallel \leq 1\text{ and }\parallel g\parallel \leq 1 \]\n\nthus \( \sup \{ \left| \left( {{Tf}, g}\right) \right| : \parallel f\parallel \leq 1,\parallel g\parallel \leq 1\} \leq \parallel T\parallel \) .\n\nConversely, if \( \sup \{ \left| \left( {{Tf}, g}\right) \right| : \parallel f\parallel \leq 1,\parallel g\parallel \leq 1\} \leq M \), we claim that \( \parallel {Tf}\parallel \leq M\parallel f\parallel \) for all \( f \) . If \( f \) or \( {Tf} \) is zero, there is nothing to prove. Otherwise, \( {f}^{\prime } = f/\parallel f\parallel \) and \( {g}^{\prime } = {Tf}/\parallel {Tf}\parallel \) have norm 1, so by assumption\n\n\[ \left| \left( {T{f}^{\prime },{g}^{\prime }}\right) \right| \leq M \]\n\nBut since \( \left| \left( {T{f}^{\prime },{g}^{\prime }}\right) \right| = \parallel {Tf}\parallel /\parallel f\parallel \) this gives \( \parallel {Tf}\parallel \leq M\parallel f\parallel \), and the lemma is proved.
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Yes
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Proposition 5.2 A linear operator \( T : {\mathcal{H}}_{1} \rightarrow {\mathcal{H}}_{2} \) is bounded if and only if it is continuous.
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Proof. If \( T \) is bounded, then \( {\begin{Vmatrix}T\left( f\right) - T\left( {f}_{n}\right) \end{Vmatrix}}_{{\mathcal{H}}_{2}} \leq M{\begin{Vmatrix}f - {f}_{n}\end{Vmatrix}}_{{\mathcal{H}}_{1}} \) , hence \( T \) is continuous. Conversely, suppose that \( T \) is continuous but not bounded. Then for each \( n \) there exists \( {f}_{n} \neq 0 \) such that \( \begin{Vmatrix}{T\left( {f}_{n}\right) }\end{Vmatrix} \geq \) \( n\begin{Vmatrix}{f}_{n}\end{Vmatrix} \) . The element \( {g}_{n} = {f}_{n}/\left( {n\begin{Vmatrix}{f}_{n}\end{Vmatrix}}\right) \) has norm \( 1/n \), hence \( {g}_{n} \rightarrow 0 \) . Since \( T \) is continuous at 0, we must have \( T\left( {g}_{n}\right) \rightarrow 0 \), which contradicts the fact that \( \begin{Vmatrix}{T\left( {g}_{n}\right) }\end{Vmatrix} \geq 1 \) . This proves the proposition.
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Yes
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Proposition 5.4 Let \( T : \mathcal{H} \rightarrow \mathcal{H} \) be a bounded linear transformation. There exists a unique bounded linear transformation \( {T}^{ * } \) on \( \mathcal{H} \) so that:\n\n(i) \( \left( {{Tf}, g}\right) = \left( {f,{T}^{ * }g}\right) \) ,\n\n(ii) \( \parallel T\parallel = \begin{Vmatrix}{T}^{ * }\end{Vmatrix} \) ,\n\n(iii) \( {\left( {T}^{ * }\right) }^{ * } = T \) .
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The linear operator \( {T}^{ * } : \mathcal{H} \rightarrow \mathcal{H} \) satisfying the above conditions is called the adjoint of \( T \).\n\nTo prove the existence of an operator satisfying (i) above, we observe that for each fixed \( g \in \mathcal{H} \), the linear functional \( \ell = {\ell }_{g} \), defined by\n\n\[ \ell \left( f\right) = \left( {{Tf}, g}\right) \]\n\nis bounded. Indeed, since \( T \) is bounded one has \( \parallel {Tf}\parallel \leq M\parallel f\parallel \) ; hence the Cauchy-Schwarz inequality implies that\n\n\[ \left| {\ell \left( f\right) }\right| \leq \parallel {Tf}\parallel \parallel g\parallel \leq B\parallel f\parallel \]\n\nwhere \( B = M\parallel g\parallel \) . Consequently, the Riesz representation theorem guarantees the existence of a unique \( h \in \mathcal{H}, h = {h}_{g} \), such that\n\n\[ \ell \left( f\right) = \left( {f, h}\right) . \]\n\nThen we define \( {T}^{ * }g = h \), and note that the association \( {T}^{ * } : g \mapsto h \) is linear and satisfies (i).\n\nThe fact that \( \parallel T\parallel = \begin{Vmatrix}{T}^{ * }\end{Vmatrix} \) follows at once from (i) and Lemma 5.1:\n\n\[ \parallel T\parallel = \sup \left\{ {\left| \left( {{Tf}, g}\right) \right| : \parallel f\parallel \leq 1,\parallel g\parallel \leq 1}\right\} \]\n\n\[ = \sup \left\{ {\left| \left( {f,{T}^{ * }g}\right) \right| : \parallel f\parallel \leq 1,\parallel g\parallel \leq 1}\right\} \]\n\n\[ = \begin{Vmatrix}{T}^{ * }\end{Vmatrix}\text{.} \]\n\nTo prove (iii), note that \( \left( {{Tf}, g}\right) = \left( {f,{T}^{ * }g}\right) \) for all \( f \) and \( g \) if and only if \( \left( {{T}^{ * }f, g}\right) = \left( {f,{Tg}}\right) \) for all \( f \) and \( g \), as one can see by taking complex conjugates and reversing the roles of \( f \) and \( g \) .
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Yes
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Proposition 5.5 Let \( T \) be a Hilbert-Schmidt operator on \( {L}^{2}\left( {\mathbb{R}}^{d}\right) \) with kernel \( K \).\n\n(i) If \( f \in {L}^{2}\left( {\mathbb{R}}^{d}\right) \), then for almost every \( x \) the function \( y \mapsto K\left( {x, y}\right) f\left( y\right) \) is integrable.\n\n(ii) The operator \( T \) is bounded from \( {L}^{2}\left( {\mathbb{R}}^{d}\right) \) to itself, and\n\n\[ \parallel T\parallel \leq \parallel K{\parallel }_{{L}^{2}\left( {{\mathbb{R}}^{d} \times {\mathbb{R}}^{d}}\right) } \]\n\nwhere \( \parallel K{\parallel }_{{L}^{2}\left( {{\mathbb{R}}^{d} \times {\mathbb{R}}^{d}}\right) } \) is the \( {L}^{2} \) -norm of \( K \) on \( {\mathbb{R}}^{d} \times {\mathbb{R}}^{d} = {\mathbb{R}}^{2d} \).\n\n(iii) The adjoint \( {T}^{ * } \) has kernel \( \overline{K\left( {y, x}\right) } \).
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Proof. By Fubini’s theorem we know that for almost every \( x \), the function \( y \mapsto {\left| K\left( x, y\right) \right| }^{2} \) is integrable. Then, part (i) follows directly from an application of the Cauchy-Schwarz inequality.\n\nFor (ii), we make use again of the Cauchy-Schwarz inequality as follows\n\n\[ \left| {\int K\left( {x, y}\right) f\left( y\right) {dy}}\right| \leq \int \left| {K\left( {x, y}\right) }\right| \left| {f\left( y\right) }\right| {dy} \]\n\n\[ \leq {\left( \int {\left| K\left( x, y\right) \right| }^{2}dy\right) }^{1/2}{\left( \int {\left| f\left( y\right) \right| }^{2}dy\right) }^{1/2}. \]\n\nTherefore, squaring this and integrating in \( x \) yields\n\n\[ \parallel {Tf}{\parallel }_{{L}^{2}\left( {\mathbb{R}}^{d}\right) }^{2} \leq \int \left( {\int {\left| K\left( x, y\right) \right| }^{2}{dy}\int {\left| f\left( y\right) \right| }^{2}{dy}}\right) {dx} \]\n\n\[ = \parallel K{\parallel }_{{L}^{2}\left( {{\mathbb{R}}^{d} \times {\mathbb{R}}^{d}}\right) }^{2}\parallel f{\parallel }_{{L}^{2}\left( {\mathbb{R}}^{d}\right) }^{2}. \]\n\nFinally, part (iii) follows by writing out \( \left( {{Tf}, g}\right) \) in terms of a double integral, and then interchanging the order of integration, as is permissible by Fubini's theorem.
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Yes
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Proposition 6.1 Suppose \( T \) is a bounded linear operator on \( \mathcal{H} \). (i) If \( S \) is compact on \( \mathcal{H} \), then \( {ST} \) and \( {TS} \) are also compact.
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Proof. Part (i) is immediate.
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No
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Lemma 6.3 Suppose \( T \) is a bounded symmetric linear operator on a Hilbert space \( \mathcal{H} \). (i) If \( \lambda \) is an eigenvalue of \( T \), then \( \lambda \) is real. (ii) If \( {f}_{1} \) and \( {f}_{2} \) are eigenvectors corresponding to two distinct eigenvalues, then \( {f}_{1} \) and \( {f}_{2} \) are orthogonal.
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Proof. To prove (i), we first choose a non-zero eigenvector \( f \) such that \( T\left( f\right) = {\lambda f} \). Since \( T \) is symmetric (that is, \( T = {T}^{ * } \)), we find that \[ \lambda \left( {f, f}\right) = \left( {{Tf}, f}\right) = \left( {f,{Tf}}\right) = \left( {f,{\lambda f}}\right) = \bar{\lambda }\left( {f, f}\right) , \] where we have used in the last equality the fact that the inner product is conjugate linear in the second variable. Since \( f \neq 0 \), we must have \( \lambda = \bar{\lambda } \) and hence \( \lambda \in \mathbb{R} \). For (ii), suppose \( {f}_{1} \) and \( {f}_{2} \) have eigenvalues \( {\lambda }_{1} \) and \( {\lambda }_{2} \), respectively. By the previous argument both \( {\lambda }_{1} \) and \( {\lambda }_{2} \) are real, and we note that \[ {\lambda }_{1}\left( {{f}_{1},{f}_{2}}\right) = \left( {{\lambda }_{1}{f}_{1},{f}_{2}}\right) = \left( {T{f}_{1},{f}_{2}}\right) = \left( {{f}_{1}, T{f}_{2}}\right) = \left( {{f}_{1},{\lambda }_{2}{f}_{2}}\right) = {\lambda }_{2}\left( {{f}_{1},{f}_{2}}\right) \] Since by assumption \( {\lambda }_{1} \neq {\lambda }_{2} \) we must have \( \left( {{f}_{1},{f}_{2}}\right) = 0 \) as desired.
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Yes
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Lemma 6.5 Suppose \( T \neq 0 \) is compact and symmetric. Then either \( \parallel T\parallel \) or \( - \parallel T\parallel \) is an eigenvalue of \( T \) .
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Proof. By the observation (7) made earlier, either\n\n\[ \parallel T\parallel = \sup \{ \left( {{Tf}, f}\right) : \parallel f\parallel = 1\} \;\text{ or }\; - \parallel T\parallel = \inf \{ \left( {{Tf}, f}\right) : \parallel f\parallel = 1\} .\n\]\n\nWe assume the first case, that is,\n\n\[ \lambda = \parallel T\parallel = \sup \{ \left( {{Tf}, f}\right) : \parallel f\parallel = 1\}\n\]\n\nand prove that \( \lambda \) is an eigenvalue of \( T \) . (The proof of the other case is similar.)\n\nWe pick a sequence \( \left\{ {f}_{n}\right\} \subset \mathcal{H} \) such that \( \begin{Vmatrix}{f}_{n}\end{Vmatrix} = 1 \) and \( \left( {T{f}_{n},{f}_{n}}\right) \rightarrow \lambda \) . Since \( T \) is compact, we may assume also (by passing to a subsequence of \( \left\{ {f}_{n}\right\} \) if necessary) that \( \left\{ {T{f}_{n}}\right\} \) converges to a limit \( g \in \mathcal{H} \) . We claim that \( g \) is an eigenvector of \( T \) with eigenvalue \( \lambda \) . To see this, we first observe that \( T{f}_{n} - \lambda {f}_{n} \rightarrow 0 \) because\n\n\[ {\begin{Vmatrix}T{f}_{n} - \lambda {f}_{n}\end{Vmatrix}}^{2} = {\begin{Vmatrix}T{f}_{n}\end{Vmatrix}}^{2} - {2\lambda }\left( {T{f}_{n},{f}_{n}}\right) + {\lambda }^{2}{\begin{Vmatrix}{f}_{n}\end{Vmatrix}}^{2}\n\]\n\n\[ \leq \parallel T{\parallel }^{2}{\begin{Vmatrix}{f}_{n}\end{Vmatrix}}^{2} - {2\lambda }\left( {T{f}_{n},{f}_{n}}\right) + {\lambda }^{2}{\begin{Vmatrix}{f}_{n}\end{Vmatrix}}^{2}\n\]\n\n\[ \leq 2{\lambda }^{2} - {2\lambda }\left( {T{f}_{n},{f}_{n}}\right) \rightarrow 0\n\]\n\nSince \( T{f}_{n} \rightarrow g \), we must have \( \lambda {f}_{n} \rightarrow g \), and since \( T \) is continuous, this implies that \( {\lambda T}{f}_{n} \rightarrow {Tg} \) . This proves that \( {\lambda g} = {Tg} \) . Finally, we must have \( g \neq 0 \), for otherwise \( \begin{Vmatrix}{{T}_{n}{f}_{n}}\end{Vmatrix} \rightarrow 0 \), hence \( \left( {T{f}_{n},{f}_{n}}\right) \rightarrow 0 \), and \( \lambda = \) \( \parallel T\parallel = 0 \), which is a contradiction.
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Yes
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Theorem 1.1 The Fourier transform \( {\mathcal{F}}_{0} \), initially defined on \( \mathcal{S}\left( {\mathbb{R}}^{d}\right) \), has a (unique) extension \( \mathcal{F} \) to a unitary mapping of \( {L}^{2}\left( {\mathbb{R}}^{d}\right) \) to itself. In particular,\n\n\[ \parallel \mathcal{F}\left( f\right) {\parallel }_{{L}^{2}\left( {\mathbb{R}}^{d}\right) } = \parallel f{\parallel }_{{L}^{2}\left( {\mathbb{R}}^{d}\right) } \]\n\nfor all \( f \in {L}^{2}\left( {\mathbb{R}}^{d}\right) \) .
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The extension \( \mathcal{F} \) will be given by a limiting process: if \( \left\{ {f}_{n}\right\} \) is a sequence in the Schwartz space that converges to \( f \) in \( {L}^{2}\left( {\mathbb{R}}^{d}\right) \), then \( \left\{ {{\mathcal{F}}_{0}\left( {f}_{n}\right) }\right\} \) will converge to an element in \( {L}^{2}\left( {\mathbb{R}}^{d}\right) \) which we will define as the Fourier transform of \( f \) . To implement this approach we have to see that every \( {L}^{2} \) function can be approximated by elements in the Schwartz space.
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Yes
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Lemma 1.2 The space \( \mathcal{S}\left( {\mathbb{R}}^{d}\right) \) is dense in \( {L}^{2}\left( {\mathbb{R}}^{d}\right) \) . In other words, given any \( f \in {L}^{2}\left( {\mathbb{R}}^{d}\right) \), there exists a sequence \( \left\{ {f}_{n}\right\} \subset \mathcal{S}\left( {\mathbb{R}}^{d}\right) \) such that\n\n\[ \n{\begin{Vmatrix}f - {f}_{n}\end{Vmatrix}}_{{L}^{2}\left( {\mathbb{R}}^{d}\right) } \rightarrow 0\;\text{ as }n \rightarrow \infty .\n\]
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For the proof of the lemma, we fix \( f \in {L}^{2}\left( {\mathbb{R}}^{d}\right) \) and \( \epsilon > 0 \) . Then, for each \( M > 0 \), we define\n\n\[ \n{g}_{M}\left( x\right) = \left\{ \begin{matrix} f\left( x\right) & \text{ if }\left| x\right| \leq M\text{ and }\left| {f\left( x\right) }\right| \leq M, \\ 0 & \text{ otherwise. } \end{matrix}\right.\n\]\n\nThen, \( \left| {f\left( x\right) - {g}_{M}\left( x\right) }\right| \leq 2\left| {f\left( x\right) }\right| \), hence \( {\left| f\left( x\right) - {g}_{M}\left( x\right) \right| }^{2} \leq 4{\left| f\left( x\right) \right| }^{2} \), and since \( {g}_{M}\left( x\right) \rightarrow f\left( x\right) \) as \( M \rightarrow \infty \) for almost every \( x \), the dominated convergence theorem guarantees that for some \( M \), we have\n\n\[ \n{\begin{Vmatrix}f - {g}_{M}\end{Vmatrix}}_{{L}^{2}\left( {\mathbb{R}}^{d}\right) } < \epsilon .\n\]\n\nWe write \( g = {g}_{M} \), note that this function is bounded and supported on a bounded set, and observe that it now suffices to approximate \( g \) by functions in the Schwartz space. To achieve this goal, we use a method called regularization, which consists of \
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No
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Lemma 1.3 Let \( {\mathcal{H}}_{1} \) and \( {\mathcal{H}}_{2} \) denote Hilbert spaces with norms \( \parallel \cdot {\parallel }_{1} \) and \( \parallel \cdot {\parallel }_{2} \), respectively. Suppose \( \mathcal{S} \) is a dense subspace of \( {\mathcal{H}}_{1} \) and \( {T}_{0} : \mathcal{S} \rightarrow {\mathcal{H}}_{2} \) a linear transformation that satisfies \( {\begin{Vmatrix}{T}_{0}\left( f\right) \end{Vmatrix}}_{2} \leq c\parallel f{\parallel }_{1} \) whenever \( f \in \mathcal{S} \) . Then \( {T}_{0} \) extends to a (unique) linear transformation \( T : {\mathcal{H}}_{1} \rightarrow {\mathcal{H}}_{2} \) that satisfies \( \parallel T\left( f\right) {\parallel }_{2} \leq c\parallel f{\parallel }_{1} \) for all \( f \in {\mathcal{H}}_{1} \) .
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Proof. Given \( f \in {\mathcal{H}}_{1} \), let \( \left\{ {f}_{n}\right\} \) be a sequence in \( \mathcal{S} \) that converges to \( f \) , and define \[ T\left( f\right) = \mathop{\lim }\limits_{{n \rightarrow \infty }}{T}_{0}\left( {f}_{n}\right) \] where the limit is taken in \( {\mathcal{H}}_{2} \) . To see that \( T \) is well-defined we must verify that the limit exists, and that it is independent of the sequence \( \left\{ {f}_{n}\right\} \) used to approximate \( f \) . Indeed, for the first point, we note that \( \left\{ {T\left( {f}_{n}\right) }\right\} \) is a Cauchy sequence in \( {\mathcal{H}}_{2} \) because by construction \( \left\{ {f}_{n}\right\} \) is Cauchy in \( {\mathcal{H}}_{1} \), and the inequality verified by \( {T}_{0} \) yields \[ {\begin{Vmatrix}{T}_{0}\left( {f}_{n}\right) - {T}_{0}\left( {f}_{m}\right) \end{Vmatrix}}_{2} \leq c{\begin{Vmatrix}{f}_{n} - {f}_{m}\end{Vmatrix}}_{1} \rightarrow 0\;\text{ as }n, m \rightarrow \infty ; \] thus \( \left\{ {{T}_{0}\left( {f}_{n}\right) }\right\} \) is Cauchy, hence converges in \( {\mathcal{H}}_{2} \) . Second, to justify that the limit is independent of the approximating sequence, let \( \left\{ {g}_{n}\right\} \) be another sequence in \( \mathcal{S} \) that converges to \( f \) in \( {\mathcal{H}}_{1} \) . Then \[ {\begin{Vmatrix}{T}_{0}\left( {f}_{n}\right) - {T}_{0}\left( {g}_{n}\right) \end{Vmatrix}}_{2} \leq c{\begin{Vmatrix}{f}_{n} - {g}_{n}\end{Vmatrix}}_{1} \] and since \( {\begin{Vmatrix}{f}_{n} - {g}_{n}\end{Vmatrix}}_{1} \leq {\begin{Vmatrix}{f}_{n} - f\end{Vmatrix}}_{1} + {\begin{Vmatrix}f - {g}_{n}\end{Vmatrix}}_{1} \), we conclude that \( \left\{ {{T}_{0}\left( {g}_{n}\right) }\right\} \) converges to a limit in \( {\mathcal{H}}_{2} \) that equals the limit of \( \left\{ {{T}_{0}\left( {f}_{n}\right) }\right\} \) . Finally, we recall that if \( {f}_{n} \rightarrow f \) and \( {T}_{0}\left( {f}_{n}\right) \rightarrow T\left( f\right) \), then \( {\begin{Vmatrix}{f}_{n}\end{Vmatrix}}_{1} \rightarrow \) \( \parallel f{\parallel }_{1} \) and \( {\begin{Vmatrix}{T}_{0}\left( {f}_{n}\right) \end{Vmatrix}}_{2} \rightarrow \parallel T\left( f\right) {\parallel }_{2} \), so in the limit as \( n \rightarrow \infty \), the inequality \( \parallel T\left( f\right) {\parallel }_{2} \leq c\parallel f{\parallel }_{1} \) holds for all \( f \in {\mathcal{H}}_{1} \) .
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Yes
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Theorem 2.1 The elements \( F \) in \( {H}^{2}\left( {\mathbb{R}}_{ + }^{2}\right) \) are exactly the functions given by (6), with \( {\widehat{F}}_{0} \in {L}^{2}\left( {0,\infty }\right) \) . Moreover \[ \parallel F{\parallel }_{{H}^{2}\left( {\mathbb{R}}_{ + }^{2}\right) } = {\begin{Vmatrix}{\widehat{F}}_{0}\end{Vmatrix}}_{{L}^{2}\left( {0,\infty }\right) }.\]
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This shows incidentally that \( {H}^{2}\left( {\mathbb{R}}_{ + }^{2}\right) \) is a Hilbert space that is isomorphic to \( {L}^{2}\left( {0,\infty }\right) \) via the correspondence (6). The crucial point in the proof of the theorem is the following fact. For any fixed strictly positive \( y \), we let \( {\widehat{F}}_{y}\left( \xi \right) \) denote the Fourier transform of the \( {L}^{2} \) function \( F\left( {x + {iy}}\right), x \in \mathbb{R} \) . Then for any pair of choices of \( y \) , \( {y}_{1} \) and \( {y}_{2} \), we have that (7) \[ {\widehat{F}}_{{y}_{1}}\left( \xi \right) {e}^{{2\pi }{y}_{1}\xi } = {\widehat{F}}_{{y}_{2}}\left( \xi \right) {e}^{{2\pi }{y}_{2}\xi }\;\text{ for a.e. }\xi . \] To establish this assertion we rely on a useful technical observation.
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No
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Lemma 2.2 If \( F \) belongs to \( {H}^{2}\left( {\mathbb{R}}_{ + }^{2}\right) \), then \( F \) is bounded in any proper half-plane \( \{ z = x + {iy}, y \geq \delta \} \), where \( \delta > 0 \) .
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To prove this we exploit the mean-value property of holomorphic functions. This property may be stated in two alternative ways. First, in terms of averages over circles,\n\n(8)\n\n\[ F\left( \zeta \right) = \frac{1}{2\pi }{\int }_{0}^{2\pi }F\left( {\zeta + r{e}^{i\theta }}\right) {d\theta }\;\text{ if }0 < r \leq \delta .\n\]\n\n(Note that if \( \zeta \) lies in the upper half-plane, \( \operatorname{Im}\left( \zeta \right) > \delta \), then the disc centered at \( \zeta \) of radius \( r \) belongs to \( {\mathbb{R}}_{ + }^{2} \).) Alternatively, integrating over \( r \), we have the mean-value property in terms of discs,\n\n(9)\n\n\[ F\left( \zeta \right) = \frac{1}{\pi {\delta }^{2}}{\int }_{\left| z\right| < \delta }F\left( {\zeta + z}\right) {dxdy},\;z = x + {iy}.\n\]\n\nThese assertions actually hold for harmonic functions in \( {\mathbb{R}}^{2} \) (see Corollary 7.2, Chapter 3 in Book II for the result about holomorphic functions, and Lemma 2.8, Chapter 5 in Book I for the case of harmonic functions); later in this chapter we in fact prove the extension of (9) to \( {\mathbb{R}}^{d} \).\n\nFrom (9) we see from the Cauchy-Schwarz inequality that\n\n\[ {\left| F\left( \zeta \right) \right| }^{2} \leq \frac{1}{\pi {\delta }^{2}}{\int }_{\left| z\right| < \delta }{\left| F\left( \zeta + z\right) \right| }^{2}{dxdy}.\n\]\n\nWriting \( z = x + {iy} \) and \( \zeta = \xi + {i\eta } \), with \( \eta > \delta \), we see that the disc \( {B}_{\delta }\left( \zeta \right) \) of center \( \zeta \) and radius \( \delta \) is contained in the strip \( \{ z + \zeta : z = \) \( x + {iy}, - \delta < y < \delta \} \), and moreover this strip lies in the half-plane \( {\mathbb{R}}_{ + }^{2} \). See Figure 1.\n\nThis gives the following majorization:\n\n\[ {\int }_{\left| z\right| < \delta }{\left| F\left( \zeta + z\right) \right| }^{2}{dxdy} \leq {\int }_{\left| y\right| < \delta }{\int }_{\mathbb{R}}{\left| F\left( \zeta + x + iy\right) \right| }^{2}{dxdy}\n\]\n\n\[ \leq {2\delta }\mathop{\sup }\limits_{{-\delta < y < \delta }}{\int }_{\mathbb{R}}{\left| F\left( x + i\left( \eta + y\right) \right) \right| }^{2}{dx}.\n\]\n\nRecalling that \( \eta > \delta \), we see that the last expression is in fact majorized by\n\n\[ {2\delta }\mathop{\sup }\limits_{{y > 0}}{\int }_{\mathbb{R}}{\left| F\left( x + iy\right) \right| }^{2}{dx} = {2\delta }\parallel F{\parallel }_{{H}^{2}\left( {\mathbb{R}}_{ + }^{2}\right) }^{2}.\n\]\n\nIn all \( {\left| F\left( \zeta \right) \right| }^{2} \leq \frac{2}{\pi \delta }\parallel F{\parallel }_{{H}^{2}}^{2} \) in the half-plane \( \operatorname{Im}\left( \zeta \right) > 0 \), which proves the lemma.
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Yes
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Theorem 2.3 Suppose \( F \) belongs to \( {H}^{2}\left( {\mathbb{R}}_{ + }^{2}\right) \) . Then \( \mathop{\lim }\limits_{{y \rightarrow 0}}F\left( {x + {iy}}\right) = \) \( {F}_{0}\left( x\right) \) exists in the following two senses:\n\n(i) As a limit in the \( {L}^{2}\left( \mathbb{R}\right) \) -norm.\n\n(ii) As a limit for almost every \( x \) .
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Thus \( F \) has boundary values (denoted by \( {F}_{0} \) ) in either of the two senses above. The function \( {F}_{0} \) is sometimes referred to as the boundary-value function of \( f \) . The proof of (i) is immediate from what we already know. Indeed, if \( {F}_{0} \) is the \( {L}^{2} \) function whose Fourier transform is \( {\widehat{F}}_{0} \), then\n\n\[ {\begin{Vmatrix}F\left( x + iy\right) - {F}_{0}\left( x\right) \end{Vmatrix}}_{{L}^{2}\left( \mathbb{R}\right) }^{2} = {\int }_{0}^{\infty }{\left| {\widehat{F}}_{0}\left( \xi \right) \right| }^{2}{\left| {e}^{-{2\pi \xi y}} - 1\right| }^{2}{dy} \]\n\nand this tends to zero as \( y \rightarrow 0 \) by the dominated convergence theorem.
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Yes
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Lemma 3.1 The space \( {C}_{0}^{\infty }\left( \Omega \right) \) is dense in \( {L}^{2}\left( \Omega \right) \) in the norm \( \parallel \cdot {\parallel }_{{L}^{2}\left( \Omega \right) } \) .
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The proof is essentially a repetition of that of Lemma 1.2. We take the precaution of modifying the definition of \( {g}_{M} \) given there to be: \( {g}_{M}\left( x\right) = \) \( f\left( x\right) \) if \( \left| x\right| \leq M, d\left( {x,{\Omega }^{c}}\right) \geq 1/M \) and \( \left| {f\left( x\right) }\right| \leq M \), and \( {g}_{M}\left( x\right) = 0 \) otherwise. Also, when we regularize \( {g}_{M} \), we replace it with \( {g}_{M} * {\varphi }_{\delta } \), with \( \delta < 1/{2M} \) . Then the support of \( {g}_{M} * {\varphi }_{\delta } \) is still compact and at a distance \( \geq 1/{2M} \) from \( {\Omega }^{c} \) .
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Yes
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