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Lemma 12.2.2 Let \( S \) be a nonempty subset of a metric space, \( \left( {X, d}\right) \) . Define\n\n\[ f\left( x\right) \equiv \operatorname{dist}\left( {x, S}\right) \equiv \inf \{ d\left( {x, y}\right) : y \in S\} .\n\]\n\nThen \( f \) is continuous. | Proof: Consider \( \left| {f\left( x\right) - f\left( {x}_{1}\right) }\right| \) and suppose without loss of generality that \( f\left( {x}_{1}\right) \geq f\left( x\right) \) . Then choose \( y \in S \) such that \( f\left( x\right) + \varepsilon > d\left( {x, y}\right) \) . Then\n\n\[ \left| {f\left( {x}_{1}\right) -... | Yes |
Theorem 12.2.3 (Urysohn's lemma for metric space) Let \( H \) be a closed subset of an open set, \( U \) in a metric space, \( \left( {X, d}\right) \) . Then there exists a continuous function, \( g : X \rightarrow \left\lbrack {0,1}\right\rbrack \) such that \( g\left( x\right) = 1 \) for all \( x \in H \) and \( g\le... | Proof: If \( x \notin C \), a closed set, then \( \operatorname{dist}\left( {x, C}\right) > 0 \) because if not, there would exist a sequence of points of \( C \) converging to \( x \) and it would follow that \( x \in C \) . Therefore, \( \operatorname{dist}\left( {x, H}\right) + \operatorname{dist}\left( {x,{U}^{C}}\... | Yes |
Lemma 12.2.6 If \( \left( {X,\tau }\right) \) is a locally compact Hausdorff space, then \( \left( {\widetilde{X},\widetilde{\tau }}\right) \) is a compact Hausdorff space. Also if \( U \) is an open set of \( \widetilde{\tau } \), then \( U \smallsetminus \{ \infty \} \) is an open set of \( \tau \) . | Proof: Since \( \left( {X,\tau }\right) \) is a locally compact Hausdorff space, it follows \( \left( {\widetilde{X},\widetilde{\tau }}\right) \) is a Hausdorff topological space. The only case which needs checking is the one of \( p \in X \) and \( \infty \) . Since \( \left( {X,\tau }\right) \) is locally compact, th... | Yes |
Theorem 12.2.11 (Partition of unity) Let \( K \) be a compact subset of a locally compact Hausdorff topological space satisfying Theorem 12.2.7 or 12.2.8 and suppose\n\n\[ K \subseteq V = { \cup }_{i = 1}^{n}{V}_{i},{V}_{i}\text{open.} \]\n\nThen there exist \( {\psi }_{i} \prec {V}_{i} \) with\n\n\[ \mathop{\sum }\lim... | Proof: Let \( {K}_{1} = K \smallsetminus { \cup }_{i = 2}^{n}{V}_{i} \) . Thus \( {K}_{1} \) is compact and \( {K}_{1} \subseteq {V}_{1} \) . Let \( {K}_{1} \subseteq \) \( {W}_{1} \subseteq {\bar{W}}_{1} \subseteq {V}_{1} \) with \( {\bar{W}}_{1} \) compact. To obtain \( {W}_{1} \), use Theorem 12.2.7 or 12.2.8 to get... | Yes |
Corollary 12.2.12 If \( H \) is a compact subset of \( {V}_{i} \), there exists a partition of unity such that \( {\psi }_{i}\left( x\right) = 1 \) for all \( x \in H \) in addition to the conclusion of Theorem 12.2.11. | Proof: Keep \( {V}_{i} \) the same but replace \( {V}_{j} \) with \( \widetilde{{V}_{j}} \equiv {V}_{j} \smallsetminus H \) . Now in the proof above, applied to this modified collection of open sets, if \( j \neq i,{\phi }_{j}\left( x\right) = 0 \) whenever \( x \in H \) . Therefore, \( {\psi }_{i}\left( x\right) = 1 \... | Yes |
Theorem 12.3.2 (Riesz representation theorem) Let \( \\left( {\\Omega ,\\tau }\\right) \) be a locally compact Hausdorff space and let \( L \) be a positive linear functional on \( {C}_{c}\\left( \\Omega \\right) \) . Then there exists a \( \\sigma \) algebra \( \\mathcal{S} \) containing the Borel sets and a unique me... | The plan is to define an outer measure and then to show that it, together with the \( \\sigma \) algebra of sets measurable in the sense of Caratheodory, satisfies the conclusions of the theorem. Always, \( K \) will be a compact set and \( V \) will be an open set.\n\nDefinition 12.3.3 \( \\mu \\left( V\\right) \\equi... | Yes |
Lemma 12.3.4 \( \mu \) is a well-defined outer measure. | Proof: First it is necessary to verify \( \mu \) is well defined because there are two descriptions of it on open sets. Suppose then that \( {\mu }_{1}\left( V\right) \equiv \inf \{ \mu \left( U\right) : U \supseteq V \) and \( U \) is open \( \} \) . It is required to verify that \( {\mu }_{1}\left( V\right) = \mu \le... | Yes |
Lemma 12.3.5 Let \( K \) be compact, \( g \geq 0, g \in {C}_{c}\left( \Omega \right) \), and \( g = 1 \) on \( K \) . Then \( \mu \left( K\right) \leq {Lg} \) . Also \( \mu \left( K\right) < \infty \) whenever \( K \) is compact. | Proof: Let \( \alpha \in \left( {0,1}\right) \) and \( {V}_{\alpha } = \{ x : g\left( x\right) > \alpha \} \) so \( {V}_{\alpha } \supseteq K \) and let \( h \prec {V}_{\alpha } \) . Then \( h \leq 1 \) on \( {V}_{\alpha } \) while \( g{\alpha }^{-1} \geq 1 \) on \( {V}_{\alpha } \) and so \( g{\alpha }^{-1} \geq h \) ... | Yes |
Lemma 12.3.6 If \( A \) and \( B \) are disjoint compact subsets of \( \Omega \), then \( \mu \left( {A \cup B}\right) = \) \( \mu \left( A\right) + \mu \left( B\right) \) . | Proof: By Theorem 12.2.7 or 12.2.8, there exists \( h \in {C}_{c}\left( \Omega \right) \) such that \( A \prec h \prec \) \( {B}^{C} \) . Let \( {U}_{1} = {h}^{-1}\left( \left( {\frac{1}{2},1}\right\rbrack \right) ,{V}_{1} = {h}^{-1}\left( {\lbrack 0,\frac{1}{2}}\right) ) \) . Then \( A \subseteq {U}_{1}, B \subseteq {... | Yes |
Lemma 12.3.7 Let \( f \in {C}_{c}\left( \Omega \right), f\left( \Omega \right) \subseteq \left\lbrack {0,1}\right\rbrack \) . Then \( \mu \left( {\operatorname{spt}\left( f\right) }\right) \geq {Lf} \) . Also, every open set, \( V \) satisfies \[ \mu \left( V\right) = \sup \{ \mu \left( K\right) : K \subseteq V\} . \] | Proof: Let \( V \supseteq \operatorname{spt}\left( f\right) \) and let \( \operatorname{spt}\left( f\right) \prec g \prec V \) . Then \( {Lf} \leq {Lg} \leq \mu \left( V\right) \) because \( f \leq g \) . Since this holds for all \( V \supseteq \operatorname{spt}\left( f\right) ,{Lf} \leq \mu \left( {\operatorname{spt}... | Yes |
Theorem 12.3.10 Let \( \left( {\Omega ,\tau }\right) \) be a metric space in which the closures of the balls are compact and let \( L \) be a positive linear functional defined on \( {C}_{c}\left( \Omega \right) \) . Then there exists a measure representing the positive linear functional which satisfies all the conclus... | Proof: Let \( \mu \) and \( \mathcal{S} \) be as described in Theorem 12.3.2. The outer regularity comes automatically as a conclusion of Theorem 12.3.2. It remains to verify inner regularity. Let \( F \in \mathcal{S} \) and let \( l < k < \mu \left( F\right) \) . Now let \( z \in \Omega \) and \( {\Omega }_{n} = \over... | Yes |
Corollary 12.3.13 Let \( \left( {\Omega ,\tau }\right) \) be a locally compact Hausdorff space which is also \( \sigma \) compact meaning\n\n\[ \Omega = { \cup }_{n = 1}^{\infty }{\Omega }_{n},{\Omega }_{n}\text{is compact,}\]\n\nand let \( L \) be a positive linear functional defined on \( {C}_{c}\left( \Omega \right)... | Proof: Suppose \( \left( {{\mu }_{1},{\mathcal{S}}_{1}}\right) \) and \( \left( {{\mu }_{2},{\mathcal{S}}_{2}}\right) \) both work. It will be shown the two measures are equal on every compact set. Let \( K \) be compact and let \( V \) be an open set containing \( K \) . Then let \( K \prec f \prec V \) . Then\n\n\[ {... | Yes |
Lemma 12.3.14 Let \( \left( {\Omega ,\mathcal{F},\mu }\right) \) be a measure space where \( \Omega \) is a topological space. Suppose \( \mu \) is a Radon measure and \( f \) is measurable with respect to \( \mathcal{F} \) . Then there exists a Borel measurable function, \( g \), such that \( g = f \) a.e. | Proof: Assume without loss of generality that \( f \geq 0 \) . Then let \( {s}_{n} \uparrow f \) pointwise. Say\n\n\[ \n{s}_{n}\left( \omega \right) = \mathop{\sum }\limits_{{k = 1}}^{{P}_{n}}{c}_{k}^{n}{\mathcal{X}}_{{E}_{k}^{n}}\left( \omega \right) \n\] \n\nwhere \( {E}_{k}^{n} \in \mathcal{F} \) . By the outer regu... | Yes |
Lemma 12.6.2 If \( \left\{ {f}_{n}\right\} \) is an increasing sequence of functions converging pointwise to \( f \) then\n\n\[ \mu \left( \left\lbrack {f > t}\right\rbrack \right) = \mathop{\lim }\limits_{{n \rightarrow \infty }}\mu \left( \left\lbrack {{f}_{n} > t}\right\rbrack \right) \] | Proof: The sets, \( \left\lbrack {{f}_{n} > t}\right\rbrack \) are increasing and their union is \( \left\lbrack {f > t}\right\rbrack \) because if \( f\left( \omega \right) > t \), then for all \( n \) large enough, \( {f}_{n}\left( \omega \right) > t \) also. Therefore, the desired conclusion follows from properties ... | Yes |
Lemma 12.6.3 Suppose \( s \geq 0 \) is a measurable simple function,\n\n\[ s\left( \omega \right) \equiv \mathop{\sum }\limits_{{k = 1}}^{n}{a}_{k}{\mathcal{X}}_{{E}_{k}}\left( \omega \right) \]\n\nwhere the \( {a}_{k} \) are the distinct nonzero values of \( s,0 < {a}_{1} < {a}_{2} < \cdots < {a}_{n} \) . Suppose \( \... | Proof: First note that if \( \mu \left( {E}_{k}\right) = \infty \) for any \( k \) then both sides equal \( \infty \) and so without loss of generality, assume \( \mu \left( {E}_{k}\right) < \infty \) for all \( k \) . Letting \( {a}_{0} \equiv 0 \), the left side equals\n\n\[ \mathop{\sum }\limits_{{k = 1}}^{n}{\int }... | Yes |
Theorem 12.6.4 Let \( f \geq 0 \) be measurable and let \( \phi \) be a \( {C}^{1} \) function defined on \( \lbrack 0,\infty ) \) which satisfies \( {\phi }^{\prime }\left( t\right) > 0 \) for all \( t > 0 \) and \( \phi \left( 0\right) = 0 \) . Then\n\n\[ \int \phi \left( f\right) {d\mu } = {\int }_{0}^{\infty }{\phi... | Proof: By Theorem 11.3.9 on Page 257 there exists an increasing sequence of nonnegative simple functions, \( \left\{ {s}_{n}\right\} \) which converges pointwise to \( f \) . By the monotone convergence theorem and Lemma 12.6.2,\n\n\[ \int \phi \left( f\right) {d\mu } = \mathop{\lim }\limits_{{n \rightarrow \infty }}\i... | Yes |
Lemma 12.6.5 Suppose \( s \geq 0 \) is a measurable simple function,\n\n\[ s\left( \omega \right) \equiv \mathop{\sum }\limits_{{k = 1}}^{n}{a}_{k}{\mathcal{X}}_{{E}_{k}}\left( \omega \right) \]\n\nwhere the \( {a}_{k} \) are the distinct nonzero values of \( s,{a}_{1} < {a}_{2} < \cdots < {a}_{n} \) . Suppose \( F \) ... | Proof: This follows from the following computation and Proposition 12.5.1. Since \( F \) is continuous at 0 and the values \( {a}_{k} \),\n\n\[ {\int }_{0}^{\infty }\mu \left( \left\lbrack {s > t}\right\rbrack \right) {d\nu }\left( t\right) = \mathop{\sum }\limits_{{k = 1}}^{n}{\int }_{\left( {a}_{k - 1},{a}_{k}\right\... | Yes |
Theorem 12.6.6 Let \( f \geq 0 \) be measurable with respect to \( \mathcal{F} \) where \( \left( {\Omega ,\mathcal{F},\mu }\right) \) a measure space, and let \( F \) be an increasing continuous function defined on \( \lbrack 0,\infty ) \) and \( F\left( 0\right) = 0 \) . Then\n\n\[{\int }_{\Omega }F\left( f\right) {d... | Proof: By Theorem 11.3.9 on Page 257 there exists an increasing sequence of nonnegative simple functions, \( \left\{ {s}_{n}\right\} \) which converges pointwise to \( f \) . By the monotone convergence theorem and Lemma 12.6.5,\n\n\[{\int }_{\Omega }F\left( f\right) {d\mu } = \mathop{\lim }\limits_{{n \rightarrow \inf... | Yes |
Theorem 12.7.1 Let \( \\left( {\\Omega ,\\mathcal{F},\\mu }\\right) \) be a finite measure space and let \( F \) be a continuous increasing function defined on \( \\lbrack 0,\\infty ) \) such that \( F\\left( 0\\right) = 0 \) . Suppose also that for all \( \\alpha > 1 \), there exists a constant \( {C}_{\\alpha } \) su... | Proof: Let \( \\beta > 1 \) be as given above. First suppose \( f \) is bounded.\n\n\[ {\\int }_{\\Omega }F\\left( f\\right) {d\\mu } = {\\int }_{\\Omega }F\\left( {\\beta \\frac{f}{\\beta }}\\right) {d\\mu } \\leq {C}_{\\beta }{\\int }_{\\Omega }F\\left( \\frac{f}{\\beta }}\\right) {d\\mu }\n\n= {C}_{\\beta }{\\int }_... | Yes |
Lemma 12.8.1 If \( T \) satisfies 12.8.18, then \( f \circ T \) is measurable whenever \( f \) is measurable. | Proof: Let \( U \) be an open set. Then\n\n\[
{\left( f \circ T\right) }^{-1}\left( U\right) = {T}^{-1}\left( {{f}^{-1}\left( U\right) }\right) \in \mathcal{F}
\]\n\nby 12.8.18. | Yes |
Lemma 12.8.2 If \( T \) satisfies 12.8.18 and 12.8.19 then whenever \( f \) is nonnegative and measurable,\n\n\[ \n{\int }_{\Omega }f\left( \omega \right) {d\mu } = {\int }_{\Omega }f\left( {T\omega }\right) {d\mu } \n\]\n\n(12.8.20)\n\nAlso 12.8.20 holds whenever \( f \in {L}^{1}\left( \Omega \right) \) . | Proof: Let \( f \geq 0 \) and \( f \) is measurable. Let \( A \in \mathcal{F} \) . Then from 12.8.19,\n\n\[ \n{\int }_{\Omega }{\mathcal{X}}_{A}\left( \omega \right) {d\mu } = \mu \left( A\right) = \mu \left( {{T}^{-1}\left( A\right) }\right) = {\int }_{\Omega }{\mathcal{X}}_{{T}^{-1}\left( A\right) }\left( \omega \rig... | Yes |
Lemma 12.8.4 Let \( f \in {L}^{1}\left( \mu \right) \) where \( f \) has real values. Then \( {\int }_{\left\lbrack {M}_{\infty }f > 0\right\rbrack }{fd\mu } \geq 0 \) . | Proof: First note that \( {M}_{n}f\left( \omega \right) \geq 0 \) for all \( n \) and \( \omega \) . This follows easily from the observation that by definition, \( {S}_{0}f\left( \omega \right) = 0 \) and so \( {M}_{n}f\left( \omega \right) \) is at least as large. There is certainly something to show here because the... | Yes |
Corollary 12.8.7 In the situation of Theorem 12.8.5, if \( T \) is ergodic, then\n\n\[ g\left( \omega \right) = \int f\left( \omega \right) {d\mu } \] \n\nfor a.e. \( \omega \) . | Proof: Let \( g \) be the function of Theorem 12.8.5 and let \( {R}_{1} \) be a rectangle in \( {\mathbb{R}}^{2} = \mathbb{C} \) of the form \( \left\lbrack {-a, a}\right\rbrack \times \left\lbrack {-a, a}\right\rbrack \) such that \( {g}^{-1}\left( {R}_{1}\right) \) has measure greater than 0 . This set is invariant b... | Yes |
Lemma 12.9.2 Given \( C \times D \) and \( {\left\{ {A}_{i} \times {B}_{i}\right\} }_{i = 1}^{n} \), there exist finitely many disjoint rectangles, \( {\left\{ {C}_{i}^{\prime } \times {D}_{i}^{\prime }\right\} }_{i = 1}^{p} \) such that none of these sets intersect any of the \( {A}_{i} \times {B}_{i} \) , each set is... | Proof: From the above picture, you see that \[ \left( {C \times D}\right) \smallsetminus \left( {{A}_{1} \times {B}_{1}}\right) = C \times \left( {D \smallsetminus {B}_{1}}\right) \cup \left( {C \smallsetminus {A}_{1}}\right) \times \left( {D \cap {B}_{1}}\right) \] and these last two sets are disjoint, have empty inte... | Yes |
Lemma 12.9.3 If \( Q = { \cup }_{i = 1}^{\infty }{A}_{i} \times {B}_{i} \in \mathcal{R} \), then there exist disjoint sets, of the form \( {A}_{i}^{\prime } \times {B}_{i}^{\prime } \) such that \( Q = { \cup }_{i = 1}^{\infty }{A}_{i}^{\prime } \times {B}_{i}^{\prime } \), each \( {A}_{i}^{\prime } \times {B}_{i}^{\pr... | \[ \rho \left( Q\right) = \mathop{\sum }\limits_{i}\mu \left( {A}_{i}^{\prime }\right) \nu \left( {B}_{i}^{\prime }\right) = \mathop{\sum }\limits_{i}\rho \left( {{A}_{i}^{\prime } \times {B}_{i}^{\prime }}\right) . \] Proof: Let \( Q \) be given as above. Let \( {A}_{1}^{\prime } \times {B}_{1}^{\prime } = {A}_{1} \ti... | Yes |
Lemma 12.9.4 Suppose \( {\left\{ {R}_{i}\right\} }_{i = 1}^{\infty } \) is a sequence of sets of \( \mathcal{R} \) then\n\n\[ \rho \left( {{ \cup }_{i = 1}^{\infty }{R}_{i}}\right) \leq \mathop{\sum }\limits_{{i = 1}}^{\infty }\rho \left( {R}_{i}\right) \] | Proof: Let \( {R}_{i} = { \cup }_{j = 1}^{\infty }{A}_{j}^{i} \times {B}_{j}^{i} \) . Using Lemma 12.9.3, let \( {\left\{ {A}_{m}^{\prime } \times {B}_{m}^{\prime }\right\} }_{m = 1}^{\infty } \) be a sequence of disjoint rectangles each of which is contained in some \( {A}_{j}^{i} \times {B}_{j}^{i} \) for some \( i, ... | Yes |
Proposition 12.9.6 \( \left( \overline{\mu \times \nu }\right) \left( S\right) = \inf \left\{ {\mathop{\sum }\limits_{{i = 1}}^{\infty }\mu \left( {A}_{i}\right) \nu \left( {B}_{i}\right) : S \subseteq { \cup }_{i = 1}^{\infty }{A}_{i} \times {B}_{i}}\right\} \) | Proof: Let \( \lambda \left( S\right) \equiv \inf \left\{ {\mathop{\sum }\limits_{{i = 1}}^{\infty }\mu \left( {A}_{i}\right) \nu \left( {B}_{i}\right) : S \subseteq { \cup }_{i = 1}^{\infty }{A}_{i} \times {B}_{i}}\right\} \) . Suppose \( S \subseteq \) \( { \cup }_{i = 1}^{\infty }{A}_{i} \times {B}_{i} \equiv Q \in ... | Yes |
Lemma 12.9.7 \( \overline{\mu \times \nu } \) is an outer measure on \( X \times Y \) and for \( R \in \mathcal{R} \n\[ \left( \overline{\mu \times \nu }\right) \left( R\right) = \rho \left( R\right) \]\n(12.9.36) | Proof: First consider 12.9.36. Since \( R \supseteq R \), it follows \( \rho \left( R\right) \geq \left( \overline{\mu \times \nu }\right) \left( R\right) \) . On the other hand, if \( Q \in \mathcal{R} \) and \( Q \supseteq R \), then \( \rho \left( Q\right) \geq \rho \left( R\right) \) and so, taking the infimum on t... | Yes |
Corollary 12.9.12 (Fubini) Let \( \left( {X,\mathcal{S},\mu }\right) \) and \( \left( {Y,\mathcal{T},\nu }\right) \) be complete measure spaces such that \( \left( {X,\mathcal{S},\mu }\right) \) and \( \left( {Y,\mathcal{T},\nu }\right) \) are both \( \sigma \) finite and let\n\n\[ \left( \overline{\mu \times \nu }\rig... | Proof: Let \( { \cup }_{n = 1}^{\infty }{X}_{n} = X \) and \( { \cup }_{n = 1}^{\infty }{Y}_{n} = Y \) where \( {X}_{n} \in \mathcal{S},{Y}_{n} \in \mathcal{T},{X}_{n} \subseteq \) \( {X}_{n + 1},{Y}_{n} \subseteq {Y}_{n + 1} \) for all \( n \) and \( \mu \left( {X}_{n}\right) < \infty ,\nu \left( {Y}_{n}\right) < \inf... | Yes |
Corollary 12.9.13 If \( f \in {L}^{1}\left( {X \times Y}\right) \), then\n\n\[ \int {fd}\left( \overline{\mu \times \nu }\right) = \iint f\left( {x, y}\right) {d\mu d\nu } = \iint f\left( {x, y}\right) {d\nu d\mu }.\n\]\n\nIf \( \mu \) and \( \nu \) are \( \sigma \) finite, then if \( f \) is \( \overline{\mu \times \n... | Proof: Without loss of generality, it can be assumed that \( f \) has real values.\n\nThen\n\[ f = \frac{\left| f\right| + f - \left( {\left| f\right| - f}\right) }{2} \]\n\nand both \( {f}^{ + } \equiv \frac{\left| f\right| + f}{2} \) and \( {f}^{ - } \equiv \frac{\left| f\right| - f}{2} \) are nonnegative and are les... | Yes |
Lemma 12.10.2 Suppose \( \mathcal{R} \) and \( \mathcal{E} \) are subsets of \( \mathcal{P}{\left( Z\right) }^{3} \) such that \( \mathcal{E} \) is defined as the set of all finite disjoint unions of sets of \( \mathcal{R} \) . Suppose also\n\n\[ \varnothing, Z \in \mathcal{R} \]\n\n\[ A \cap B \in \mathcal{R}\text{whe... | Proof: Note first that if \( A \in \mathcal{R} \), then \( {A}^{C} \in \mathcal{E} \) because \( {A}^{C} = Z \smallsetminus A \) .\n\nNow suppose that \( {E}_{1} \) and \( {E}_{2} \) are in \( \mathcal{E} \) ,\n\n\[ {E}_{1} = { \cup }_{i = 1}^{m}{R}_{i},\;{E}_{2} = { \cup }_{j = 1}^{n}{R}_{j} \]\n\nwhere the \( {R}_{i}... | Yes |
Corollary 12.10.3 Let \( \left( {{Z}_{1},{\mathcal{R}}_{1},{\mathcal{E}}_{1}}\right) \) and \( \left( {{Z}_{2},{\mathcal{R}}_{2},{\mathcal{E}}_{2}}\right) \) be as described in Lemma 14.1.2. Then \( \left( {{Z}_{1} \times {Z}_{2},\mathcal{R},\mathcal{E}}\right) \) also satisfies the conditions of Lemma 14.1.2 if \( \ma... | Proof: It is clear \( \varnothing ,{Z}_{1} \times {Z}_{2} \in \mathcal{R} \) . Let \( A \times B \) and \( C \times D \) be two elements of \( \mathcal{R} \) .\n\n\[ A \times B \cap C \times D = A \cap C \times B \cap D \in \mathcal{R} \]\n\nby assumption.\n\n\[ A \times B \smallsetminus \left( {C \times D}\right) = \]... | Yes |
Theorem 12.10.9 If \( E \in \mathcal{S} \times \mathcal{F} \), then \( {E}_{x} \in \mathcal{F} \) and \( {E}^{y} \in \mathcal{S} \) for all \( x \in X \) and \( y \in Y \) . | Proof: Let\n\n\[ \mathcal{M} = \left\{ {E \subseteq \mathcal{S} \times \mathcal{F}}\right. \text{such that for all}x \in X,\;{E}_{x} \in \mathcal{F}\text{,}\]\n\n\[ \text{and for all}\left. {y \in Y,{E}^{y} \in \mathcal{S}\text{.}}\right\}\]\n\nThen \( \mathcal{M} \) contains all measurable rectangles. If \( {E}_{i} \i... | Yes |
Theorem 12.10.10 If \( \left( {X,\mathcal{S},\mu }\right) \) and \( \left( {Y,\mathcal{F},\lambda }\right) \) are both finite measure spaces \( (\mu \left( X\right) \) , \( \lambda \left( Y\right) < \infty ) \), then for every \( E \in \mathcal{S} \times \mathcal{F} \) ,\na.) \( x \rightarrow \lambda \left( {E}_{x}\rig... | Proof: Let\n\n\[ \mathcal{M} = \{ E \in \mathcal{S} \times \mathcal{F}\text{such that both}a\text{.) and}b\text{.) hold}\} \text{.} \]\n\nSince \( \mu \) and \( \lambda \) are both finite, the monotone convergence and dominated convergence theorems imply \( \mathcal{M} \) is a monotone class.\n\nNext I will argue \( \m... | Yes |
Theorem 12.10.11 If \( \left( {X,\mathcal{S},\mu }\right) \) and \( \left( {Y,\mathcal{F},\lambda }\right) \) are both \( \sigma \) finite measure spaces, then for every \( E \in \mathcal{S} \times \mathcal{F} \) , a.) \( x \rightarrow \lambda \left( {E}_{x}\right) \) is \( \mu \) measurable, \( y \rightarrow \mu \left... | Proof: Let \( X = { \cup }_{n = 1}^{\infty }{X}_{n}, Y = { \cup }_{n = 1}^{\infty }{Y}_{n} \) where, \[ {X}_{n} \subseteq {X}_{n + 1},{Y}_{n} \subseteq {Y}_{n + 1},\mu \left( {X}_{n}\right) < \infty ,\lambda \left( {Y}_{n}\right) < \infty . \] Let \[ {\mathcal{S}}_{n} = \left\{ {A \cap {X}_{n} : A \in \mathcal{S}}\righ... | Yes |
Theorem 12.10.13 If \( A \in \mathcal{S}, B \in \mathcal{F} \), then \( \left( {\mu \times \lambda }\right) \left( {A \times B}\right) = \mu \left( A\right) \lambda \left( B\right) \), and \( \mu \times \lambda \) is a measure on \( \mathcal{S} \times \mathcal{F} \) called product measure. | Proof: The first assertion about the measure of a measurable rectangle was established above. Now suppose \( {\left\{ {E}_{i}\right\} }_{i = 1}^{\infty } \) is a disjoint collection of sets of \( \mathcal{S} \times \mathcal{F} \) . Then using the monotone convergence theorem along with the observation that \( {\left( {... | Yes |
Theorem 12.10.14 Let \( f : X \times Y \rightarrow \left\lbrack {0,\infty }\right\rbrack \) be measurable with respect to \( \mathcal{S} \times \mathcal{F} \) and suppose \( \mu \) and \( \lambda \) are \( \sigma \) finite. Then\n\n\[{\int }_{X \times Y}{fd}\left( {\mu \times \lambda }\right) = {\int }_{X}{\int }_{Y}f\... | Proof: For \( E \in \mathcal{S} \times \mathcal{F} \) ,\n\n\[{\int }_{Y}{\mathcal{X}}_{E}\left( {x, y}\right) {d\lambda } = \lambda \left( {E}_{x}\right) ,{\int }_{X}{\mathcal{X}}_{E}\left( {x, y}\right) {d\mu } = \mu \left( {E}^{y}\right) .\]\n\nThus from Definition 12.10.12,12.10.50 holds if \( f = {\mathcal{X}}_{E} ... | Yes |
Corollary 12.10.15 Let \( f : X \times Y \rightarrow \mathbb{C} \) be \( \mathcal{S} \times \mathcal{F} \) measurable. Suppose either \( {\int }_{X}{\int }_{Y}\left| f\right| {d\lambda d\mu } \) or \( {\int }_{Y}{\int }_{X}\left| f\right| {d\mu d\lambda } < \infty \) . Then \( f \in {L}^{1}\left( {X \times Y,\mu \times... | Proof: Suppose first that \( f \) is real valued. Apply Theorem 12.10.14 to \( {f}^{ + } \) and \( {f}^{ - } \) . 12.10.51 follows from observing that \( f = {f}^{ + } - {f}^{ - } \) ; and that all integrals are finite. If \( f \) is complex valued, consider real and imaginary parts. This proves the corollary. | Yes |
Theorem 12.11.2 Let \( \left( {\Omega ,\mathcal{F},\mu }\right) \) be a complete measure space and let \( f \leq g \leq h \) be functions having values in \( \left\lbrack {0,\infty }\right\rbrack \) . Suppose also that \( f\left( \omega \right) = h\left( \omega \right) \) a.e. \( \omega \) and that \( f \) and \( h \) ... | Proof: Let \( \alpha \in \mathbb{R} \) . \[ \left\lbrack {f > \alpha }\right\rbrack \subseteq \left\lbrack {g > \alpha }\right\rbrack \subseteq \left\lbrack {h > \alpha }\right\rbrack \] Thus \[ \left\lbrack {g > \alpha }\right\rbrack = \left\lbrack {f > \alpha }\right\rbrack \cup \left( {\left\lbrack {g > \alpha }\rig... | Yes |
Theorem 12.12.4 Let \( f : X \times Y \rightarrow \left\lbrack {0,\infty }\right\rbrack \) be measurable with respect to the \( \sigma \) algebra, \( \sigma \left( \mathcal{K}\right) \) just defined and let \( \mu \times \nu \) be the product measure of 12.12.57 where \( \mu \) and \( \nu \) are finite measures on \( \... | Proof: Let \( \left\{ {s}_{n}\right\} \) be an increasing sequence of \( \sigma \left( \mathcal{K}\right) \) measurable simple functions which converges pointwise to \( f \) . The above equation holds for \( {s}_{n} \) in place of \( f \) from what was shown above. The final result follows from passing to the limit and... | Yes |
Theorem 12.12.5 Let \( f : X \times Y \rightarrow \left\lbrack {0,\infty }\right\rbrack \) be measurable with respect to the \( \sigma \) algebra, \( \sigma \left( \mathcal{K}\right) \) just defined and let \( \mu \times \nu \) be the product measure of 12.12.57 where \( \mu \) and \( \nu \) are \( \sigma \) finite mea... | Proof: Since the measures are \( \sigma \) finite, there exist increasing sequences of sets, \( \left\{ {X}_{n}\right\} \) and \( \left\{ {Y}_{n}\right\} \) such that \( \mu \left( {X}_{n}\right) < \infty \) and \( \nu \left( {Y}_{n}\right) < \infty \) . Then \( \mu \) and \( \nu \) restricted to \( {X}_{n} \) and \( {... | Yes |
Theorem 12.12.7 Let \( {\left\{ \left( {X}_{i},{\mathcal{F}}_{i},{\mu }_{i}\right) \right\} }_{i = 1}^{n} \) be \( \sigma \) finite measure spaces and let \( \mathop{\prod }\limits_{{i = 1}}^{n}{\mathcal{F}}_{i} \) denote the smallest \( \sigma \) algebra which contains the measurable boxes of the form \( \mathop{\prod... | Proof: This follows immediately from Theorem 12.12.6 and Theorem 12.11.2. By the second theorem, there exists a function \( {f}_{1} \geq f \) such that \( {f}_{1} = f \) for all \( \left( {{x}_{1},\cdots ,{x}_{n}}\right) \notin N \), a set of \( \mathop{\prod }\limits_{{i = 1}}^{n}{\mathcal{F}}_{i} \) having measure ze... | Yes |
Corollary 12.12.8 Suppose \( f \in {L}^{1}\left( {\mathop{\prod }\limits_{{i = 1}}^{n}{X}_{i},\overline{\mathop{\prod }\limits_{{i = 1}}^{n}{\mathcal{F}}_{i}},\overline{{\mu }_{1} \times \cdots \times {\mu }_{n}}}\right) \) where each \( {X}_{i} \) is a \( \sigma \) finite measure space. Then if \( \left( {{i}_{1},\cdo... | Proof: Just apply Theorem 12.12.7 to the positive and negative parts of the real and imaginary parts of \( f \) . This proves the theorem. | No |
Corollary 12.12.9 Suppose in the situation of Corollary 12.12.8, \( f = {f}_{1} \) off \( N \), a set of \( \mathop{\prod }\limits_{{i = 1}}^{n}{\mathcal{F}}_{i} \) having \( {\mu }_{1} \times \cdots \times {\mu }_{n} \) measure zero and that \( {f}_{1} \) is a complex valued function measurable with respect to \( \mat... | Proof: Since \( \left| {f}_{1}\right| \) is \( \mathop{\prod }\limits_{{i = 1}}^{n}{\mathcal{F}}_{i} \) measurable, it follows from Theorem 12.12.6 that\n\n\[ \n\infty > {\int }_{{X}_{{j}_{n}}}\cdots {\int }_{{X}_{{j}_{1}}}\left| {f}_{1}\right| d{\mu }_{{j}_{1}}\cdots d{\mu }_{{j}_{n}}\n\]\n\n\[ \n= \int \left| {f}_{1}... | Yes |
Lemma 12.12.10 Let \( \left( {X,\mathcal{F},\mu }\right) \) and \( \left( {Y,\mathcal{S},\nu }\right) \) be \( \sigma \) finite complete measure spaces and suppose \( f \geq 0 \) is \( \overline{\mathcal{F} \times \mathcal{S}} \) measurable. Then for a.e. \( x \) ,\n\n\[ \ny \rightarrow f\left( {x, y}\right)\n\]\n\nis ... | Proof: By Theorem 12.11.2, there exist \( \mathcal{F} \times \mathcal{S} \) measurable functions, \( g \) and \( h \) and a set, \( N \in \mathcal{F} \times \mathcal{S} \) of \( \mu \times \lambda \) measure zero such that \( g \leq f \leq h \) and for \( \left( {x, y}\right) \notin N \) , it follows that \( g\left( {x... | Yes |
Let \( \left\{ {a}_{n}\right\} \) be an increasing sequence of numbers in \( \left( {0,1}\right) \) which converges to 1. Let \( {g}_{n} \in {C}_{c}\left( {{a}_{n},{a}_{n + 1}}\right) \) such that \( \int {g}_{n}{dx} = 1 \) . Now for \( \left( {x, y}\right) \in \) \( \lbrack 0,1) \times \lbrack 0,1) \) define \[ f\left... | Note this is actually a finite sum for each such \( \left( {x, y}\right) \) . Therefore, this is a continuous function on \( \lbrack 0,1) \times \lbrack 0,1) \) . Now for a fixed \( y \) , \[ {\int }_{0}^{1}f\left( {x, y}\right) {dx} = \mathop{\sum }\limits_{{k = 1}}^{\infty }{g}_{n}\left( y\right) {\int }_{0}^{1}\left... | Yes |
This time let \( \mu = m \), Lebesgue measure on \( \left\lbrack {0,1}\right\rbrack \) and let \( \nu \) be counting measure on \( \left\lbrack {0,1}\right\rbrack \), in this case, the \( \sigma \) algebra is \( \mathcal{P}\left( \left\lbrack {0,1}\right\rbrack \right) \) . Let \( l \) denote the line segment in \( \le... | Let \( \mathcal{B} \equiv \left\{ {k \in \mathbb{N} : \nu \left( {B}_{k}\right) = \infty }\right\} \) . If \( m\left( {{ \cup }_{k \in \mathcal{B}}{A}_{k}}\right) \) has measure zero, then there are uncountably many points of \( \left\lbrack {0,1}\right\rbrack \) outside of \( { \cup }_{k \in \mathcal{B}}{A}_{k} \) . F... | Yes |
Let \( X \) be an uncountable set. It follows from the well ordering theorem which says every set can be well ordered which is presented in the appendix that \( X \) can be well ordered. Let \( \omega \in X \) be the first element of \( X \) which is preceded by uncountably many points of \( X \) . Let \( \Omega \) den... | \[ {\int }_{0}^{1}f\left( {x, y}\right) {dy} = 1,{\int }_{0}^{1}f\left( {x, y}\right) {dx} = 0 \] In each case, the integrals make sense. In the first, for fixed \( x, f\left( {x, y}\right) = 1 \) for all but countably many \( y \) so the function of \( y \) is Borel measurable. In the second where \( y \) is fixed, \(... | Yes |
Every open set in \( {\mathbb{R}}^{n} \) is the countable disjoint union of half open boxes of the form\n\n\[ \mathop{\prod }\limits_{{i = 1}}^{n}\left( {{a}_{i},{a}_{i} + {2}^{-k}}\right\rbrack \]\n\nwhere \( {a}_{i} = l{2}^{-k} \) for some integers, \( l, k \) . The sides of these boxes are of equal length. One could... | Proof: Let\n\n\[ {\mathcal{C}}_{k} = \left\{ \right. \text{All half open boxes}\mathop{\prod }\limits_{{i = 1}}^{n}\left( {{a}_{i},{a}_{i} + {2}^{-k}}\right\rbrack \text{where}\n\n\[ \left. {{a}_{i} = l{2}^{-k}\text{ for some integer }l.}\right\} \]\n\nThus \( {\mathcal{C}}_{k} \) consists of a countable disjoint colle... | Yes |
Lemma 13.1.3 Let \( R = \mathop{\prod }\limits_{{i = 1}}^{n}\left\lbrack {{a}_{i},{b}_{i}}\right\rbrack ,{R}_{0} = \mathop{\prod }\limits_{{i = 1}}^{n}\left( {{a}_{i},{b}_{i}}\right) \) . Then \[ {m}_{n}\left( {R}_{0}\right) = {m}_{n}\left( R\right) = \mathop{\prod }\limits_{{i = 1}}^{n}\left( {{b}_{i} - {a}_{i}}\right... | Proof: Let \( k \) be large enough that \[ {a}_{i} + 1/k < {b}_{i} - 1/k \] for \( i = 1,\cdots, n \) and consider functions \( {g}_{i}^{k} \) and \( {f}_{i}^{k} \) having the following graphs.  ![3f4063cc-9f64-45dc-... | Yes |
Lemma 13.1.4 Let \( U \) be an open or closed set. Then \( {m}_{n}\left( U\right) = {m}_{n}\left( {\mathbf{x} + U}\right) \) . | Proof: By Lemma 13.1.2 there is a sequence of disjoint half open rectangles, \( \left\{ {R}_{i}\right\} \) such that \( { \cup }_{i}{R}_{i} = U \) . Therefore, \( \mathbf{x} + U = { \cup }_{i}\left( {\mathbf{x} + {R}_{i}}\right) \) and the \( \mathbf{x} + {R}_{i} \) are also disjoint rectangles which are identical to t... | Yes |
Theorem 13.1.5 Lebesgue measure is translation invariant. That is\n\n\[ \n{m}_{n}\left( E\right) = {m}_{n}\left( {\mathbf{x} + E}\right) \n\]\n\nfor all \( E \) Lebesgue measurable. | Proof: Suppose \( {m}_{n}\left( E\right) < \infty \) . By regularity of the measure, there exist sets \( G, H \) such that \( G \) is a countable intersection of open sets, \( H \) is a countable union of compact sets, \( {m}_{n}\left( {G \smallsetminus H}\right) = 0 \), and \( G \supseteq E \supseteq H \) . Now \( {m}... | Yes |
Let \( D \) be an \( n \times n \) diagonal matrix and let \( U \) be an open set. Then \[ {m}_{n}\left( {DU}\right) = \left| {\det \left( D\right) }\right| {m}_{n}\left( U\right) . \] | Proof: If any of the diagonal entries of \( D \) equals 0 there is nothing to prove because then both sides equal zero. Therefore, it can be assumed none are equal to zero. Suppose these diagonal entries are \( {k}_{1},\cdots ,{k}_{n} \) . From Lemma 13.1.2 there exist half open boxes, \( \left\{ {R}_{i}\right\} \) hav... | Yes |
Corollary 13.1.7 Let \( M > 0 \) . Then \( {m}_{n}\left( {B\left( {\mathbf{a},{Mr}}\right) }\right) = {M}^{n}{m}_{n}\left( {B\left( {\mathbf{0}, r}\right) }\right) \) . | Proof: By Lemma 13.1.4 there is no loss of generality in taking \( \mathbf{a} = \mathbf{0} \) . Let \( D \) be the diagonal matrix which has \( M \) in every entry of the main diagonal so \( \left| {\det \left( D\right) }\right| = \) \( {M}^{n} \) . Note that \( {DB}\left( {\mathbf{0}, r}\right) = B\left( {\mathbf{0},{... | Yes |
Lemma 13.2.2 Let \( \parallel \cdot \parallel \) be a norm on \( {\mathbb{R}}^{n} \) and let \( \mathcal{F} \) be a collection of balls determined by this norm. Suppose\n\n\[ \infty > M \equiv \sup \{ r : B\left( {\mathbf{p}, r}\right) \in \mathcal{F}\} > 0 \]\n\nand \( k \in \left( {0,\infty }\right) \) . Then there e... | Proof: Let \( \mathcal{H} = \{ \mathcal{B} \subseteq \mathcal{F} \) such that 13.2.1 and 13.2.2 hold \( \} \) . If there are no balls with radius larger than \( k \) then \( \mathcal{H} = \varnothing \) and you let \( \mathcal{G} = \varnothing \) . In the other case, \( \mathcal{H} \neq \varnothing \) because there exi... | Yes |
Theorem 13.2.3 (Vitali) Let \( \mathcal{F} \) be a collection of balls and let\n\n\[ A \equiv \cup \{ B : B \in \mathcal{F}\}\]\n\nSuppose\n\n\[ \infty > M \equiv \sup \{ r : B\left( {\mathbf{p}, r}\right) \in \mathcal{F}\} > 0.\]\n\nThen there exists \( \mathcal{G} \subseteq \mathcal{F} \) such that \( \mathcal{G} \) ... | Proof: Using Lemma 13.2.2, there exists \( {\mathcal{G}}_{1} \subseteq \mathcal{F} \equiv {\mathcal{F}}_{0} \) which satisfies\n\n\[ B\left( {\mathbf{p}, r}\right) \in {\mathcal{G}}_{1}\text{ implies }r > \frac{M}{2},\]\n\n\( \left( {13.2.3}\right) \)\n\n\[ {B}_{1},{B}_{2} \in {\mathcal{G}}_{1}\text{ implies }{B}_{1} \... | Yes |
Lemma 13.3.1 Let \( \mathcal{F} \) be a countable collection of balls satisfying\n\n\[ \infty > M \equiv \sup \{ r : B\left( {\mathbf{p}, r}\right) \in \mathcal{F}\} > 0 \]\n\nand let \( k \in \left( {0,\infty }\right) \) . Then there exists \( \mathcal{G} \subseteq \mathcal{F} \) such that\n\n\[ \text{If}B\left( {\mat... | Proof: If no ball of \( \mathcal{F} \) has radius larger than \( k \), let \( \mathcal{G} = \varnothing \) . Assume therefore, that some balls have radius larger than \( k \) . Let \( \mathcal{F} \equiv {\left\{ {B}_{i}\right\} }_{i = 1}^{\infty } \) . Now let \( {B}_{{n}_{1}} \) be the first ball in the list which has... | Yes |
Theorem 13.3.4 (Vitali) Let \( \\mathcal{F} \) be a collection of balls, and let\n\n\[ A \\equiv \\cup \\{ B : B \\in \\mathcal{F}\\} .\n\nSuppose\n\n\[ \\infty > M \\equiv \\sup \\{ r : B\\left( {\\mathbf{p}, r}\\right) \\in \\mathcal{F}\\} > 0.\n\nThen there exists \( \\mathcal{G} \\subseteq \\mathcal{F} \) such that... | Proof: For \( B \) one of these balls, say \( \\overline{B\\left( {\\mathbf{x}, r}\\right) } \\supseteq B \\supseteq B\\left( {\\mathbf{x}, r}\\right) \), denote by \( {B}_{1} \), the ball \( B\\left( {\\mathbf{x},\\frac{5r}{4}}\\right) \) . Let \( {\\mathcal{F}}_{1} \\equiv \\left\\{ {{B}_{1} : B \\in \\mathcal{F}}\\r... | Yes |
Corollary 13.4.3 Let \( E \subseteq {\mathbb{R}}^{n} \) and suppose \( \overline{{m}_{n}}\left( E\right) < \infty \) where \( \overline{{m}_{n}} \) is the outer measure determined by \( {m}_{n}, n \) dimensional Lebesgue measure, and let \( \mathcal{F} \), be a collection of closed balls of bounded radii such that \( \... | Proof: If \( 0 = \overline{{m}_{n}}\left( E\right) \) you simply pick any ball from \( \mathcal{F} \) for your collection of disjoint balls. | No |
Corollary 13.4.4 Let \( E \subseteq {\mathbb{R}}^{n} \) and let \( \mathcal{F} \), be a collection of closed balls of bounded radii such that \( \mathcal{F} \) covers \( E \) in the sense of Vitali. Then there exists a countable collection of disjoint balls from \( \mathcal{F},{\left\{ {B}_{j}\right\} }_{j = 1}^{\infty... | Proof: Let \( {R}_{m} \equiv {\left( -m, m\right) }^{n} \) be the open rectangle having sides of length \( {2m} \) which is centered at \( \mathbf{0} \) and let \( {R}_{0} = \varnothing \) . Let \( {H}_{m} \equiv \overline{{R}_{m}} \smallsetminus {R}_{m} \) . Since both \( \overline{{R}_{m}} \) and \( {R}_{m} \) have t... | Yes |
Corollary 13.4.5 Let \( E \subseteq {\mathbb{R}}^{n} \) and let \( \mathcal{F} \), be a collection of open balls of bounded radii such that \( \mathcal{F} \) covers \( E \) in the sense of Vitali. Then there exists a countable collection of disjoint balls from \( \mathcal{F},{\left\{ {B}_{j}\right\} }_{j = 1}^{\infty }... | Proof: Let \( \overline{\mathcal{F}} \) be the collection of closures of balls in \( \mathcal{F} \) . Then \( \overline{\mathcal{F}} \) covers \( E \) in the sense of Vitali and so from Corollary 13.4.4 there exists a sequence of disjoint closed balls from \( \overline{\mathcal{F}} \) satisfying \( \overline{{m}_{n}}\l... | Yes |
Lemma 13.5.1 Let \( \mathbf{h} \) satisfy 13.5.11. If \( T \subseteq U \) and \( {m}_{n}\left( T\right) = 0 \), then \( {m}_{n}\left( {\mathbf{h}\left( T\right) }\right) = 0 \). | Proof: Let\n\n\[ {T}_{k} \equiv \{ \mathbf{x} \in T : \parallel D\mathbf{h}\left( \mathbf{x}\right) \parallel < k\} \]\n\nand let \( \varepsilon > 0 \) be given. Now by outer regularity, there exists an open set, \( V \) , containing \( {T}_{k} \) which is contained in \( U \) such that \( {m}_{n}\left( V\right) < \var... | Yes |
Lemma 13.5.2 Let \( \mathbf{h} \) satisfy 13.5.11. If \( S \) is a Lebesgue measurable subset of \( U \) , then \( \mathbf{h}\left( S\right) \) is Lebesgue measurable. | Proof: Let \( {S}_{k} = S \cap B\left( {\mathbf{0}, k}\right), k \in \mathbb{N} \) . By inner regularity of Lebesgue measure, there exists a set, \( F \), which is the countable union of compact sets and a set \( T \) with \( {m}_{n}\left( T\right) = 0 \) such that\n\n\[ F \cup T = {S}_{k} \]\n\nThen \( \mathbf{h}\left... | Yes |
Lemma 13.5.4 Let \( R \) be unitary \( \left( {{R}^{ * }R = R{R}^{ * } = I}\right) \) and let \( V \) be a an open or closed set. Then \( {m}_{n}\left( {RV}\right) = {m}_{n}\left( V\right) \) . | Proof: First assume \( V \) is a bounded open set. By Corollary 13.4.6 there is a disjoint sequence of closed balls, \( \left\{ {B}_{i}\right\} \) such that \( V = { \cup }_{i = 1}^{\infty }{B}_{i} \cup N \) where \( {m}_{n}\left( N\right) = 0 \) . Denote by \( {\mathbf{x}}_{i} \) the center of \( {B}_{i} \) and let \(... | Yes |
Lemma 13.5.5 Let \( E \) be Lebesgue measurable set in \( {\mathbb{R}}^{n} \) and let \( R \) be unitary. Then \( {m}_{n}\left( {RE}\right) = {m}_{n}\left( E\right) . | Proof: First suppose \( E \) is bounded. Then there exist sets, \( G \) and \( H \) such that \( H \subseteq E \subseteq G \) and \( H \) is the countable union of closed sets while \( G \) is the countable intersection of open sets such that \( {m}_{n}\left( {G \smallsetminus H}\right) = 0 \) . By Lemma 13.5.4 applied... | Yes |
Lemma 13.5.6 Let \( V \) be an open or closed set in \( {\mathbb{R}}^{n} \) and let \( A \) be an \( n \times n \) matrix. Then \( {m}_{n}\left( {AV}\right) = \left| {\det \left( A\right) }\right| {m}_{n}\left( V\right) \) . | Proof: Let \( {RU} \) be the right polar decomposition (Theorem 5.9.6 on Page 97) of \( A \) and let \( V \) be an open set. Then from Lemma 13.5.5,\n\n\[ \n{m}_{n}\left( {AV}\right) = {m}_{n}\left( {RUV}\right) = {m}_{n}\left( {UV}\right) . \n\] \n\nNow \( U = {Q}^{ * }{DQ} \) where \( D \) is a diagonal matrix such t... | Yes |
Theorem 13.5.7 Let \( E \) be Lebesgue measurable set in \( {\mathbb{R}}^{n} \) and let \( A \) be an \( n \times n \) matrix. Then \( {m}_{n}\left( {AE}\right) = \left| {\det \left( A\right) }\right| {m}_{n}\left( E\right) \) . | Proof: First suppose \( E \) is bounded. Then there exist sets, \( G \) and \( H \) such that \( H \subseteq E \subseteq G \) and \( H \) is the countable union of closed sets while \( G \) is the countable intersection of open sets such that \( {m}_{n}\left( {G \smallsetminus H}\right) = 0 \) . By Lemma 13.5.6 applied... | Yes |
Corollary 13.6.2 Let \( U \) and \( V \) be bounded open sets in \( {\mathbb{R}}^{n} \) and let \( \mathbf{h},{\mathbf{h}}^{-1} \) be \( {C}^{1} \) functions such that \( \mathbf{h}\left( U\right) = V \) and \( \left| {\det \left( {D\mathbf{h}\left( \mathbf{x}\right) }\right) }\right| \) is bounded. Also let \( E \subs... | Proof: By regularity, there exist compact sets, \( {K}_{k} \) and open sets \( {G}_{k} \) such that\n\n\[ \n{K}_{k} \subseteq E \subseteq {G}_{k} \n\]\n\nand \( {m}_{n}\left( {{G}_{k} \smallsetminus {K}_{k}}\right) < {2}^{-k} \) . By Theorem 12.2.7, there exist \( {f}_{k} \) such that \( {K}_{k} \prec {f}_{k} \prec \) ... | Yes |
Corollary 13.6.3 Let \( U \) and \( V \) be open sets in \( {\mathbb{R}}^{n} \) and let \( \mathbf{h},{\mathbf{h}}^{-1} \) be \( {C}^{1} \) functions such that \( \mathbf{h}\left( U\right) = V \) . Also let \( E \subseteq V \) be measurable. Then \[ {\int }_{V}{\mathcal{X}}_{E}\left( \mathbf{y}\right) d{m}_{n} = {\int ... | Proof: For each \( \mathbf{x} \in U \), there exists \( {r}_{\mathbf{x}} \) such that \( \overline{B\left( {\mathbf{x},{r}_{\mathbf{x}}}\right) } \subseteq U \) and \( {r}_{\mathbf{x}} < 1 \) . Then by the mean value inequality Theorem 6.13.4, it follows \( \mathbf{h}\left( {B\left( {\mathbf{x},{r}_{\mathbf{x}}}\right)... | Yes |
Theorem 13.6.4 Let \( U \) and \( V \) be open sets in \( {\mathbb{R}}^{n} \) and let \( \mathbf{h},{\mathbf{h}}^{-1} \) be \( {C}^{1} \) functions such that \( \mathbf{h}\left( U\right) = V \) . Then if \( g \) is a nonnegative Lebesgue measurable function,\n\n\[{\int }_{V}g\left( \mathbf{y}\right) {dy} = {\int }_{U}g... | Proof: From Corollary 13.6.3, 13.6.16 holds for any nonnegative simple function in place of \( g \) . In general, let \( \left\{ {s}_{k}\right\} \) be an increasing sequence of simple functions which converges to \( g \) pointwise. Then from the monotone convergence theorem\n\n\[{\int }_{V}g\left( \mathbf{y}\right) {dy... | Yes |
Theorem 13.6.6 Let \( U \) be an open set and let \( \mathbf{h} \) be a \( 1 - 1,{C}^{1} \) function with values in \( {\mathbb{R}}^{n} \) . Then if \( g \) is a nonnegative Lebesgue measurable function,\n\n\[{\int }_{\mathbf{h}\left( U\right) }g\left( \mathbf{y}\right) {dy} = {\int }_{U}g\left( {\mathbf{h}\left( \math... | Proof: Let \( Z = \{ \mathbf{x} : \det \left( {D\mathbf{h}\left( \mathbf{x}\right) }\right) = 0\} \) . Then by the inverse function theorem, \( {\mathbf{h}}^{-1} \) is \( {C}^{1} \) on \( \mathbf{h}\left( {U \smallsetminus Z}\right) \) and \( \mathbf{h}\left( {U \smallsetminus Z}\right) \) is an open set. Therefore, fr... | Yes |
Lemma 13.7.1 Let \( F \subseteq \mathbf{h}\left( U\right) \) be measurable. Then\n\n\[ \n{\int }_{\mathbf{h}\left( U\right) }n\left( \mathbf{y}\right) {\mathcal{X}}_{F}\left( \mathbf{y}\right) {dy} = {\int }_{U}{\mathcal{X}}_{F}\left( {\mathbf{h}\left( \mathbf{x}\right) }\right) \left| {\det D\mathbf{h}\left( \mathbf{x... | Proof: Using Lemma 13.6.5 and the Monotone convergence Theorem or Fubini's Theorem,\n\n\[ \n{\int }_{\mathbf{h}\left( U\right) }n\left( \mathbf{y}\right) {\mathcal{X}}_{F}\left( \mathbf{y}\right) {dy} = {\int }_{\mathbf{h}\left( U\right) }\left( {\mathop{\sum }\limits_{{i = 1}}^{\infty }{\mathcal{X}}_{\mathbf{h}\left( ... | Yes |
Theorem 13.7.3 Let \( g \geq 0, g \) measurable, and let \( \mathbf{h} \) be \( {C}^{1}\left( U\right) \) . Then\n\n\[{\int }_{\mathbf{h}\left( U\right) }\# \left( \mathbf{y}\right) g\left( \mathbf{y}\right) {dy} = {\int }_{U}g\left( {\mathbf{h}\left( \mathbf{x}\right) }\right) \left| {\det D\mathbf{h}\left( \mathbf{x}... | Proof: From 13.7.18 and Lemma 13.7.1,13.7.19 holds for all \( g \), a nonnegative simple function. Approximating an arbitrary measurable nonnegative function, \( g \) , with an increasing pointwise convergent sequence of simple functions and using the monotone convergence theorem, yields 13.7.19 for an arbitrary nonneg... | Yes |
Theorem 13.8.1 Let \( f \geq 0 \) and suppose \( f \) is a Lebesgue measurable function defined on \( {\mathbb{R}}^{n} \) and \( {\int }_{{\mathbb{R}}^{n}}{fd}{m}_{n} < \infty \) . Then\n\n\[ \n{\int }_{{\mathbb{R}}^{n}}{fd}{m}_{n} = {\int }_{{\mathbb{R}}^{k}}{\int }_{{\mathbb{R}}^{n - k}}{fd}{m}_{n - k}d{m}_{k}.\n\] | This will be accomplished by Fubini's theorem, Theorem 12.9.11 and the following lemma. | No |
Lemma 13.8.2 \( \overline{{m}_{k} \times {m}_{n - k}} = {m}_{n} \) on the \( {m}_{n} \) measurable sets. | Proof: First of all, let \( R = \mathop{\prod }\limits_{{i = 1}}^{n}\left( {{a}_{i},{b}_{i}}\right\rbrack \) be a measurable rectangle and let \( {R}_{k} = \mathop{\prod }\limits_{{i = 1}}^{k}\left( {{a}_{i},{b}_{i}}\right\rbrack ,{R}_{n - k} = \mathop{\prod }\limits_{{i = k + 1}}^{n}\left( {{a}_{i},{b}_{i}}\right\rbra... | Yes |
Corollary 13.8.3 Let \( f \) be a nonnegative real valued measurable function. Then\n\n\[ \n{\int }_{{\mathbb{R}}^{n}}{fd}{m}_{n} = {\int }_{{\mathbb{R}}^{k}}{\int }_{{\mathbb{R}}^{n - k}}{fd}{m}_{n - k}d{m}_{k}.\n\] | Proof: Let \( {S}_{p} \equiv \left\{ {\mathbf{x} \in {\mathbb{R}}^{n} : 0 \leq f\left( \mathbf{x}\right) \leq p}\right\} \cap B\left( {\mathbf{0}, p}\right) \) . Then \( {\int }_{{\mathbb{R}}^{n}}f{\mathcal{X}}_{{S}_{p}}d{m}_{n} < \infty \) . Therefore, from Theorem 13.8.1,\n\n\[ \n{\int }_{{\mathbb{R}}^{n}}f{\mathcal{... | Yes |
Corollary 13.8.4 Let \( f \in {L}^{1}\left( {\mathbb{R}}^{n}\right) \) where the measure is \( {m}_{n} \) . Then\n\n\[{\int }_{{\mathbb{R}}^{n}}{fd}{m}_{n} = {\int }_{{\mathbb{R}}^{k}}{\int }_{{\mathbb{R}}^{n - k}}{fd}{m}_{n - k}d{m}_{k}.\] | Proof: Apply Corollary 13.8.3 to the postive and negative parts of the real and imaginary parts of \( f \) . | No |
Theorem 13.9.1 Let \( \mathbf{y} = {\mathbf{h}}_{p}\left( {\overrightarrow{\phi },\theta ,\rho }\right) \) be the spherical coordinate transformations in \( {\mathbb{R}}^{p} \) . Then letting \( A = \mathop{\prod }\limits_{{i = 1}}^{{p - 2}}\left( {0,\pi }\right) \times \left( {0,{2\pi }}\right) \), it follows \( \math... | Proof: Formula 13.9.20 is obvious from the definition of the spherical coordinates because in the matrix of the derivative, there will be a \( \rho \) in \( p - 1 \) columns. The first claim is also clear from the definition and math induction or from the geometry of the above description. It remains to verify 13.9.21 ... | Yes |
For what values of \( s \) is the integral \( {\int }_{B\left( {\mathbf{0}, R}\right) }{\left( 1 + {\left| \mathbf{x}\right| }^{2}\right) }^{s}{dy} \) bounded independent of \( R \) ? Here \( B\left( {\mathbf{0}, R}\right) \) is the ball, \( \left\{ {\mathbf{x} \in {\mathbb{R}}^{p} : \left| \mathbf{x}\right| \leq R}\ri... | I think you can see immediately that \( s \) must be negative but exactly how negative? It turns out it depends on \( p \) and using polar coordinates, you can find just exactly what is needed. From the polar coordinates formula above,\n\n\[ \n{\int }_{B\left( {\mathbf{0}, R}\right) }{\left( 1 + {\left| \mathbf{x}\righ... | Yes |
Theorem 13.10.3 Let \( {B}_{r} \) be the above closed ball and let \( \mathbf{f} : {B}_{r} \rightarrow {B}_{r} \) be continuous. Then there exists \( \mathbf{x} \in {B}_{r} \) such that \( \mathbf{f}\left( \mathbf{x}\right) = \mathbf{x} \) . | Proof: Let \( {\mathbf{f}}_{k}\left( \mathbf{x}\right) \equiv \frac{\mathbf{f}\left( \mathbf{x}\right) }{1 + {k}^{-1}} \) . Thus\n\n\[ \begin{Vmatrix}{{\mathbf{f}}_{k} - \mathbf{f}}\end{Vmatrix} = \mathop{\max }\limits_{{\mathbf{x} \in {B}_{r}}}\left\{ \left| {\frac{\mathbf{f}\left( \mathbf{x}\right) }{1 + \left( {1/k}... | Yes |
Corollary 13.10.4 Let \( \mathbf{f} : \overline{B\left( {\mathbf{a}, r}\right) } \rightarrow \overline{B\left( {\mathbf{a}, r}\right) } \) be continuous. Then there exists \( \mathbf{x} \in \overline{B\left( {\mathbf{a}, r}\right) } \) such that \( \mathbf{f}\left( \mathbf{x}\right) = \mathbf{x} \) . | Proof: Let \( \mathbf{g} : {B}_{r} \rightarrow {B}_{r} \) be defined by \( \mathbf{g}\left( \mathbf{y}\right) \equiv \mathbf{f}\left( {\mathbf{y} + \mathbf{a}}\right) - \mathbf{a} \) . Then \( \mathbf{g} \) is a continuous map from \( {B}_{r} \) to \( {B}_{r} \) . Therefore, there exists \( \mathbf{y} \in {B}_{r} \) su... | Yes |
Proposition 13.10.6 Let \( A \) be a retract of \( B \) and suppose \( B \) has the fixed point property. Then so does \( A \) . | Proof: Suppose \( \mathbf{f} : A \rightarrow A \) . Let \( \mathbf{h} \) be the retract of \( B \) onto \( A \) . Then \( \mathbf{f} \circ \mathbf{h} \) : \( B \rightarrow B \) is continuous. Thus, it has a fixed point \( \mathbf{x} \in B \) so \( \mathbf{f}\left( {\mathbf{h}\left( \mathbf{x}\right) }\right) = \mathbf{... | Yes |
Lemma 13.11.3 There does not exist \( \mathbf{h} \in {C}^{2}\left( \overline{B\left( {\mathbf{0}, R}\right) }\right) \) such that \( \mathbf{h} : \overline{B\left( {\mathbf{0}, R}\right) } \rightarrow \) \( \partial B\left( {\mathbf{0}, R}\right) \) which also has the property that \( \mathbf{h}\left( \mathbf{x}\right)... | Proof: Here and below, let \( {B}_{R} \) denote \( \overline{B\left( {\mathbf{0}, R}\right) } \) . Suppose such an \( \mathbf{h} \) exists. Let \( \lambda \in \left\lbrack {0,1}\right\rbrack \) and let \( {\mathbf{p}}_{\lambda }\left( \mathbf{x}\right) \equiv \mathbf{x} + \lambda \left( {\mathbf{h}\left( \mathbf{x}\rig... | Yes |
Lemma 13.11.4 If \( \mathbf{h} \in {C}^{2}\left( \overline{B\left( {\mathbf{0}, R}\right) }\right) \) and \( \mathbf{h} : \overline{B\left( {\mathbf{0}, R}\right) } \rightarrow \overline{B\left( {\mathbf{0}, R}\right) } \), then \( \mathbf{h} \) has a fixed point, \( \mathbf{x} \) such that \( \mathbf{h}\left( \mathbf{... | Proof: Suppose the lemma is not true. Then for all \( \mathbf{x},\left| {\mathbf{x} - \mathbf{h}\left( \mathbf{x}\right) }\right| \neq 0 \) . Then define\n\n\[ \mathbf{g}\left( \mathbf{x}\right) = \mathbf{h}\left( \mathbf{x}\right) + \frac{\mathbf{x} - \mathbf{h}\left( \mathbf{x}\right) }{\left| \mathbf{x} - \mathbf{h}... | Yes |
Theorem 13.11.5 Let \( {B}_{R} \) be the above closed ball and let \( \mathbf{f} : {B}_{R} \rightarrow {B}_{R} \) be continuous. Then there exists \( \mathbf{x} \in {B}_{R} \) such that \( \mathbf{f}\left( \mathbf{x}\right) = \mathbf{x} \) . | Proof: Let \( {\mathbf{f}}_{k}\left( \mathbf{x}\right) \equiv \frac{\mathbf{f}\left( \mathbf{x}\right) }{1 + {k}^{-1}} \) . Thus\n\n\[ \begin{Vmatrix}{{\mathbf{f}}_{k} - \mathbf{f}}\end{Vmatrix} = \mathop{\max }\limits_{{\mathbf{x} \in {B}_{R}}}\left\{ \left| {\frac{\mathbf{f}\left( \mathbf{x}\right) }{1 + \left( {1/k}... | Yes |
Corollary 13.11.6 Let \( \mathbf{f} : \overline{B\left( {\mathbf{a}, R}\right) } \rightarrow \overline{B\left( {\mathbf{a}, R}\right) } \) be continuous. Then there exists \( \mathbf{x} \in \overline{B\left( {\mathbf{a}, R}\right) } \) such that \( \mathbf{f}\left( \mathbf{x}\right) = \mathbf{x} \) . | Proof: Let \( \mathbf{g} : {B}_{R} \rightarrow {B}_{R} \) be defined by \( \mathbf{g}\left( \mathbf{y}\right) \equiv \mathbf{f}\left( {\mathbf{y} + \mathbf{a}}\right) - \mathbf{a} \) . Then \( \mathbf{g} \) is a continuous map from \( {B}_{R} \) to \( {B}_{R} \) . Therefore, there exists \( \mathbf{y} \in {B}_{R} \) su... | Yes |
Proposition 13.11.8 Let \( A \) be a retract of \( B \) and suppose \( B \) has the fixed point property. Then so does \( A \) . | Proof: Suppose \( \mathbf{f} : A \rightarrow A \) . Let \( \mathbf{h} \) be the retract of \( B \) onto \( A \) . Then \( \mathbf{f} \circ \mathbf{h} \) : \( B \rightarrow B \) is continuous. Thus, it has a fixed point \( \mathbf{x} \in B \) so \( \mathbf{f}\left( {\mathbf{h}\left( \mathbf{x}\right) }\right) = \mathbf{... | Yes |
Lemma 13.12.1 Let \( B \) be a closed ball in \( {\mathbb{R}}^{n} \) centered at \( \mathbf{a} \) which has radius \( r \) . Let \( \mathbf{f} : B \rightarrow {\mathbb{R}}^{n} \) . Then \( \mathbf{f}\left( \mathbf{a}\right) \) is an interior point of \( \mathbf{f}\left( B\right) \) . | Proof: Since \( \mathbf{f}\left( B\right) \) is compact and \( \mathbf{f} \) is one to one, \( {\mathbf{f}}^{-1} \) is continuous on \( \mathbf{f}\left( B\right) \) . Use Tietze extension theorem on components of \( {\mathbf{f}}^{-1} \) or some such thing to obtain \( \mathbf{g} : {\mathbb{R}}^{n} \rightarrow {\mathbb{... | Yes |
Theorem 13.12.3 Let \( U \) be an open set in \( {\mathbb{R}}^{n} \) and let \( \mathbf{f} : U \rightarrow \mathbf{f}\left( U\right) \subseteq {\mathbb{R}}^{n} \) . Then \( \mathbf{f}\left( U\right) \) is also an open set in \( {\mathbb{R}}^{n} \) . | Proof: For \( \mathbf{a} \in U \), let \( \mathbf{a} \in {B}_{\mathbf{a}} \subseteq U \), where \( {B}_{\mathbf{a}} \) is a closed ball centered at \( \mathbf{a} \) . Then from Lemma 13.12.2, \( \mathbf{f}\left( \mathbf{a}\right) \in {V}_{\mathbf{f}\left( \mathbf{a}\right) } \) an open subset of \( \mathbf{f}\left( {B}... | Yes |
Lemma 13.13.1 Let \( \mathcal{F} \) be a nonempty set of nonempty balls in \( X \) with\n\n\[ \sup \{ \operatorname{diam}\left( B\right) : B \in \mathcal{F}\} = D < \infty \]\n\nand let \( A \) denote the set of centers of these balls. Suppose \( A \) is bounded. Define a sequence of balls from \( \mathcal{F},{\left\{ ... | Proof: First note that \( {B}_{m + 1} \) can be chosen as in 13.13.33. This is because the \( {A}_{m} \) are decreasing and so\n\n\[ \frac{3}{4}\sup \left\{ {r : B\left( {\mathbf{a}, r}\right) \in \mathcal{F},\mathbf{a} \in {A}_{m}}\right\} \]\n\n\[ \leq \frac{3}{4}\sup \left\{ {r : B\left( {\mathbf{a}, r}\right) \in \... | Yes |
Lemma 13.13.3 Let \( \Gamma > 1 \) and \( B\left( {\mathbf{a},{\Gamma r}}\right) \) be a ball and suppose \( {\left\{ B\left( {\mathbf{x}}_{i},{r}_{i}\right) \right\} }_{i = 1}^{m} \) are balls contained in \( B\left( {\mathbf{a},{\Gamma r}}\right) \) such that \( r \leq {r}_{i} \) and none of these balls contains the ... | Proof: Let \( {\mathbf{z}}_{i} = {\mathbf{x}}_{i} - \mathbf{a} \) . Then \( B\left( {{\mathbf{z}}_{i},{r}_{i}}\right) \) are balls contained in \( B\left( {\mathbf{0},{\Gamma r}}\right) \) with no ball containing a center of another. Then \( B\left( {\frac{{\mathbf{z}}_{i}}{\Gamma r},\frac{{r}_{i}}{\Gamma r}}\right) \)... | Yes |
Lemma 13.13.4 Let \( B \) be a ball having radius \( r \) and suppose \( B \) has nonempty intersection with the balls \( {B}_{1},\cdots ,{B}_{m} \) having radii \( {r}_{1},\cdots ,{r}_{m} \) respectively, and as before, no \( {B}_{i} \) contains the center of any other and the centers of the \( {B}_{i} \) are not cont... | Proof: Let \( B = B\left( {\mathbf{a}, r}\right) \) . Then each \( {B}_{i} \) is contained in \( B\left( {\mathbf{a},{2r} + {\alpha r} + {\alpha r}}\right) \) . This is because if \( \mathbf{y} \in {B}_{i} \equiv B\left( {{\mathbf{x}}_{i},{r}_{i}}\right) \), \[ \parallel \mathbf{y} - \mathbf{a}\parallel \leq \begin{Vma... | Yes |
Corollary 13.14.3 Let \( \mu \) be a Radon measure on \( {\mathbb{R}}^{p} \). Letting \( \bar{\mu } \) be the outer measure determined by \( \mu \), suppose \( \mathcal{F} \) is a collection of closed balls which cover \( E \) in the sense of Vitali. Then there exists a sequence of disjoint balls, \( \left\{ {B}_{i}\ri... | Proof: Since \( \mu \) is a Radon measure it is finite on compact sets. Therefore, there are at most countably many numbers, \( {\left\{ {b}_{i}\right\} }_{i = 1}^{\infty } \) such that \( \mu \left( {\partial B\left( {\mathbf{0},{b}_{i}}\right) }\right) > 0 \). It follows there exists an increasing sequence of positiv... | Yes |
Corollary 13.14.4 Let \( \mu \) be a Radon measure on \( {\mathbb{R}}^{p} \). Letting \( \bar{\mu } \) be the outer measure determined by \( \mu \), suppose \( \mathcal{F} \) is a collection of balls which cover \( E \) in the sense that for all \( \varepsilon > 0 \) there are uncountably many balls of \( \mathcal{F} \... | Proof: Let \( \mathbf{x} \in E \). Thus \( \mathbf{x} \) is the center of arbitrarily small balls from \( \mathcal{F} \). Since \( \mu \) is a Radon measure, at most countably many radii, \( r \) of these balls can have the property that \( \mu \left( {\partial B\left( {\mathbf{0}, r}\right) }\right) = 0 \). Let \( {\m... | Yes |
Lemma 14.1.2 Suppose \( \mathcal{R} \) and \( \mathcal{E} \) are subsets of \( \mathcal{P}{\left( Z\right) }^{1} \) such that \( \mathcal{E} \) is defined as the set of all finite disjoint unions of sets of \( \mathcal{R} \) . Suppose also\n\n\[ \varnothing, Z \in \mathcal{R} \]\n\n\[ A \cap B \in \mathcal{R}\text{when... | Proof: Note first that if \( A \in \mathcal{R} \), then \( {A}^{C} \in \mathcal{E} \) because \( {A}^{C} = Z \smallsetminus A \) .\n\nNow suppose that \( {E}_{1} \) and \( {E}_{2} \) are in \( \mathcal{E} \),\n\n\[ {E}_{1} = { \cup }_{i = 1}^{m}{R}_{i},\;{E}_{2} = { \cup }_{j = 1}^{n}{R}_{j} \]\n\nwhere the \( {R}_{i} ... | Yes |
Corollary 14.1.3 Let \( \\left( {{Z}_{1},{\\mathcal{R}}_{1},{\\mathcal{E}}_{1}}\\right) \) and \( \\left( {{Z}_{2},{\\mathcal{R}}_{2},{\\mathcal{E}}_{2}}\\right) \) be as described in Lemma 14.1.2. Then \( \\left( {{Z}_{1} \\times {Z}_{2},\\mathcal{R},\\mathcal{E}}\\right) \) also satisfies the conditions of Lemma 14.1... | Proof: It is clear \( \\varnothing ,{Z}_{1} \\times {Z}_{2} \\in \\mathcal{R} \) . Let \( A \\times B \) and \( C \\times D \) be two elements of \( \\mathcal{R} \) .\n\n\[ \nA \\times B \\cap C \\times D = A \\cap C \\times B \\cap D \\in \\mathcal{R} \n\]\n\nby assumption.\n\n\[ \nA \\times B \\smallsetminus \\left( ... | Yes |
Lemma 14.2.3 Let \( M \) be a metric space with the closed balls compact and suppose \( \mu \) is a measure defined on the Borel sets of \( M \) which is finite on compact sets. Then there exists a unique Radon measure, \( \bar{\mu } \) which equals \( \mu \) on the Borel sets. In particular \( \mu \) must be both inne... | Proof: Define a positive linear functional, \( \Lambda \left( f\right) = \int {fd\mu } \) . Let \( \bar{\mu } \) be the Radon measure which comes from the Riesz representation theorem for positive linear functionals. Thus for all \( f \in {C}_{0}\left( M\right) \) ,\n\n\[ \int {fd\mu } = \int {fd}\bar{\mu } \]\n\nIf \(... | Yes |
Proposition 14.3.5 The product topology is the smallest topology \( \tau \) for \( X \equiv \mathop{\prod }\limits_{{i \in I}}{X}_{i} \) such that each \( {\pi }_{i} \) is continuous. Here \( {\pi }_{i} \) is defined in the following manner. For \( \mathbf{x} \in X,{\pi }_{i}\left( \mathbf{x}\right) \equiv {x}_{i} \) .... | Proof: If each \( {\pi }_{i} \) is continuous, then for \( A \in {\tau }_{i},{\pi }_{i}^{-1}\left( A\right) \) must be in \( \tau \) . However, \( {\pi }_{i}^{-1}\left( A\right) = {P}_{j}\left( A\right) \) having \( A \) in the \( {i}^{th} \) slot and \( {X}_{j} \) in every other. Therefore, \( \tau \) must contain the... | Yes |
Theorem 14.3.6 If \( \left( {{X}_{i},{\tau }_{i}}\right) \) is compact, then so is \( \left( {\mathop{\prod }\limits_{{i \in I}}{X}_{i},\tau }\right) \) where \( \tau \) is the product topology. | Proof: By the Alexander subbasis theorem, the theorem will be proved if every subbasic open cover admits a finite subcover. Therefore, let \( \mathcal{O} \) be a subbasic open cover of \( X \equiv \mathop{\prod }\limits_{{i \in I}}{X}_{i} \) . Let\n\n\[ \n{\mathcal{O}}_{j} = \left\{ {Q \in \mathcal{O} : {\pi }_{i}Q = {... | Yes |
Lemma 14.4.2 The sets, \( \mathcal{E},{\mathcal{E}}_{J} \) defined above form an algebra of sets of \( \mathop{\prod }\limits_{{t \in I}}{M}_{t} \) . | Proof: First consider \( {\mathcal{R}}_{J} \) . If \( \mathbf{A},\mathbf{B} \in {\mathcal{R}}_{J} \), then \( \mathbf{A} \cap \mathbf{B} \in {\mathcal{R}}_{J} \) also. Is \( \mathbf{A} \smallsetminus \mathbf{B} \) a finite disjoint union of sets of \( {\mathcal{R}}_{J} \) ? It suffices to verify that \( {\pi }_{J}\left... | Yes |
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