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Corollary 1.2.11. Every open set \( G \subseteq {\mathbb{R}}^{n} \) can be written as a countably infinite union of nonoverlapping closed intervals \( G = \mathop{\bigcup }\limits_{{k = 1}}^{\infty }{I}_{k} \) with\n\n\[ m\left( G\right) = \mathop{\sum }\limits_{{k = 1}}^{\infty }v\left( {I}_{k}\right) \] | Proof. By Lemma 1.2.10,\n\n\[ G = \mathop{\bigcup }\limits_{{k = 1}}^{\infty }{I}_{k} \]\n\nwhere \( \left\{ {I}_{k}\right\} \) is a countable collection of nonoverlapping closed intervals. By Proposition 1.1.9 and Proposition 1.1.11,\n\n\[ m\left( G\right) \leq \mathop{\sum }\limits_{{k = 1}}^{\infty }v\left( {I}_{k}\... | Yes |
Lemma 1.2.16. If \( d\left( {{E}_{1},{E}_{2}}\right) > 0 \), then\n\n\[ \n{m}^{ * }\left( {{E}_{1} \cup {E}_{2}}\right) = {m}^{ * }\left( {E}_{1}\right) + {m}^{ * }\left( {E}_{2}\right) .\n\] | Proof. If one of \( {m}^{ * }\left( {E}_{1}\right) \) or \( {m}^{ * }\left( {E}_{2}\right) \) is infinite, then \( {m}^{ * }\left( {{E}_{1} \cup {E}_{2}}\right) \) will also be infinite by Proposition 1.1.8, and the result is true. So we will assume both of these quantities are finite.\n\nBy Proposition 1.1.9 we know t... | Yes |
Theorem 1.2.17. Every closed subset of \( {\mathbb{R}}^{n} \) is Lebesgue measurable. | Proof. Let \( F \subseteq {\mathbb{R}}^{n} \) be a closed set. We will consider two cases.\n\n(i) First assume that \( F \) is a bounded set. Hence, \( F \) is a compact set and \( {m}^{ * }\left( F\right) \) is finite. Let \( \epsilon > 0 \) . By Theorem 1.1.13, there is an open set \( G \) containing \( F \) with\n\n... | Yes |
Theorem 1.2.18. Let \( E \subseteq {\mathbb{R}}^{n} \) . If \( E \) is Lebesgue measurable, then\n\n\[ \n{E}^{c} = \left\{ {x = \left( {{x}_{1},{x}_{2},\ldots ,{x}_{n}}\right) \in {\mathbb{R}}^{n} \mid x \notin E}\right\} \n\]\n\nis measurable. | Proof. Assume \( E \subseteq {\mathbb{R}}^{n} \) is measurable. Then for every positive integer \( k \) there exists an open set \( {G}_{k} \) containing \( E \) such that\n\n\[ \n{m}^{ * }\left( {{G}_{k} \smallsetminus E}\right) < \frac{1}{k}. \n\]\n\nFor every \( k \) ,\n\n\[ \n{G}_{k}^{c} \subseteq {E}^{c} \n\]\n\nh... | Yes |
Proposition 1.2.19. Let \( \left\{ {A}_{j}\right\} \) be a countable collection of Lebesgue measurable subsets of \( {\mathbb{R}}^{n} \) . Then the set\n\n\[ A = \bigcap {A}_{j} \]\n\nis Lebesgue measurable. | ## Proof. This is Exercise 11. | No |
The half open interval \( (0,1\rbrack \) in \( {\mathbb{R}}^{1} \) is of type \( {G}_{\delta } \) since | \[ (0,1\rbrack = \mathop{\bigcap }\limits_{{n = 1}}^{\infty }\left( {0,1 + \frac{1}{n}}\right) \] | Yes |
The set\n\n\[ A = \\left\\{ {\\left( {x, y}\\right) \\in {\\mathbb{R}}^{2} \\mid 1 \\leq x < 2\\text{ and }3 < y \\leq 5}\\right\\} \]\nin \( {\\mathbb{R}}^{2} \) is of type \( {F}_{\\sigma } \) | since\n\n\[ A = \\mathop{\\bigcup }\\limits_{{n = 1}}^{\\infty }\\left\\{ {\\left( {x, y}\\right) \\in {\\mathbb{R}}^{2} \\mid 1 \\leq x \\leq 2 - \\frac{1}{n}\\text{ and }3 + \\frac{1}{n} \\leq y \\leq 5}\\right\\} . \] | Yes |
Proposition 1.2.23. Let \( E \subseteq {\mathbb{R}}^{n} \) be a set. \( E \) is Lebesgue measurable if and only if for every \( \epsilon > 0 \) there is a closed set \( F \) with \( F \subseteq E \) and \( {m}^{ * }\left( {E \smallsetminus F}\right) < \epsilon \) | ## Proof. This is Exercise 15 | No |
Let \( E = \{ - 1\} \cup \lbrack 2,3) \cup (4,6\rbrack \) . Then \( m\left( E\right) = 3 \) and | \[ {D}_{E} = \left\lbrack {-7, - 5) \cup \left\lbrack {-4,4}\right\rbrack \cup (5,7}\right\rbrack \] which contains an interval centered at 0 . | No |
Although the Cantor set \( C \) has measure 0, as we will show, the corresponding set of arithmetic differences is\n\n\[ \n{D}_{C} = \left\lbrack {-1,1}\right\rbrack \n\]\n\nSince \( C \subseteq \left\lbrack {0,1}\right\rbrack \), it must be the case that \( {D}_{C} \subseteq \left\lbrack {-1,1}\right\rbrack \) . We wi... | Let \( \alpha \in \left\lbrack {-1,1}\right\rbrack \) . Then \( \frac{1}{2}\left( {\alpha + 1}\right) \in \left\lbrack {0,1}\right\rbrack \) has a ternary expansion,\n\nsay\n\n\[ \n\frac{1}{2}\left( {\alpha + 1}\right) = {.}_{\left( 3\right) }{c}_{1}{c}_{2}{c}_{3}\ldots ,\;\text{where}{c}_{i} = 0,1\text{, or }2\text{.}... | Yes |
Example 1.3.6. By Exercise 25, there is a nonmeasurable subset \( A \) of \( \left\lbrack {0,1}\right\rbrack \) . If \( {m}^{ * }\left( A\right) = 0 \), then \( A \) would be a measurable set by Example 1.2.4. Therefore | \[ 0 < {m}^{ * }\left( A\right) \leq 1 \] Let \( \delta = {m}^{ * }\left( A\right) \) . The set of rational numbers in the interval \( \left\lbrack {0,1}\right\rbrack \) is a countable set, say \( \mathbb{Q} \cap \left\lbrack {0,1}\right\rbrack = \left\{ {r}_{k}\right\} \) . Hence \( \left\{ {A + {r}_{k}}\right\} \) is... | Yes |
Theorem 2.1.4. Let \( f \) be defined on the interval \( I \) . The following four statements are equivalent:\n\n(i) \( f \) is a Lebesgue measurable function.\n\n(ii) For every \( s \in \mathbb{R} \), the set \( \{ x \in I \mid f\left( x\right) \leq s\} \) is a Lebesgue measurable set.\n\n(iii) For every \( s \in \mat... | Proof. We will show that\n\n\[ \left( \mathrm{i}\right) \Rightarrow \left( \mathrm{{ii}}\right) \Rightarrow \left( \mathrm{{iii}}\right) \Rightarrow \left( \mathrm{{iv}}\right) \Rightarrow \left( \mathrm{i}\right) .\n\]\n\nFor every \( s \in \mathbb{R} \) ,\n\n\[ \{ x \in I \mid f\left( x\right) \leq s\} = I \smallsetm... | Yes |
Theorem 2.1.5. Suppose \( f \) is a Lebesgue measurable function on the interval \( I \) . Let \( c \in \mathbb{R} \) . The following two statements are true:\n\n(i) The function \( f\left( x\right) + c \) is a Lebesgue measurable function on \( I \) .\n\n(ii) The function \( {cf}\left( x\right) \) is a Lebesgue measur... | Proof. Let \( c \in \mathbb{R} \) . Both of these statements are trivial in the case that \( c = 0 \) . Thus, we will assume \( c \neq 0 \) .\n\nTo see that \( f\left( x\right) + c \) is a measurable function, let \( s \in \mathbb{R} \) . Then\n\n\[ \{ x \in I \mid f\left( x\right) + c > s\} = \{ x \in I \mid f\left( x... | Yes |
Theorem 2.1.6. Let \( f \) and \( g \) be Lebesgue measurable functions on I. The following statements hold:\n\n(i) The function \( f\left( x\right) + g\left( x\right) \) is Lebesgue measurable on \( I \) . | Proof. To show (i), let \( s \in \mathbb{R} \) . We will use our earlier observation that\n\n\[ \n\{ x \in I \mid f\left( x\right) + g\left( x\right) > s\} = \{ x \in I \mid f\left( x\right) > s - g\left( x\right) \} .\n\]\n\nTo avoid the difficulty described above, let \( \mathbb{Q} = \left\{ {r}_{k}\right\} \) be a c... | Yes |
Proposition 2.1.9. Suppose \( f \) and \( g \) are two functions defined on the interval \( I \) . If \( f \) is Lebesgue measurable on \( I \) and \( f = g \) a.e. on \( I \) , then \( g \) is Lebesgue measurable on \( I \) . | Proof. Let \( Z = \{ x \in I \mid f\left( x\right) \neq g\left( x\right) \} \) . Then \( Z \) has measure 0 . Moreover, every subset of \( Z \) is a measurable set with measure 0 . Given \( s \in \mathbb{R} \), in order for \( g\left( x\right) > s \), either \( x \notin Z \) (so that \( g\left( x\right) = f\left( x\rig... | Yes |
Theorem 2.1.12. Let \( \left\{ {f}_{n}\right\} \) be a pointwise bounded sequence of Lebesgue measurable functions on an interval \( I \) . Then both \( {f}^{ * } \) and \( {f}_{ * } \) are Lebesgue measurable functions on \( I \) . | Proof. Let\n\n\[ \n{M}_{n}\left( x\right) = \sup \left\{ {{f}_{n}\left( x\right) ,{f}_{n + 1}\left( x\right) ,{f}_{n + 2}\left( x\right) ,\ldots }\right\} , \n\]\n\n\[ \n{m}_{n}\left( x\right) = \inf \left\{ {{f}_{n}\left( x\right) ,{f}_{n + 1}\left( x\right) ,{f}_{n + 2}\left( x\right) ,\ldots }\right\} . \n\]\n\nThe ... | Yes |
Corollary 2.1.13. Let \( \left\{ {f}_{n}\right\} \) be a sequence of Lebesgue measurable functions on \( I \) that converges pointwise to \( f \) . Then the function \( f \) is Lebesgue measurable on \( I \) . | Proof. In this case, \( {f}^{ * } = {f}_{ * } = f \) . Therefore, \( f \) is a measurable function. | No |
Corollary 2.1.14. Let \( \\left\\{ {f}_{n}\\right\\} \) be a sequence of Lebesgue measurable functions on \( I \) . If \( f \) is a function defined on \( I \) with \( \\mathop{\\lim }\\limits_{{n \\rightarrow \\infty }}{f}_{n}\\left( x\\right) = f\\left( x\\right) \) a.e., then \( f \) is Lebesgue measurable on \( I \... | Proof. Let\n\n\[ Z = \\left\\{ {x \\in I \\mid \\mathop{\\lim }\\limits_{{n \\rightarrow \\infty }}{f}_{n}\\left( x\\right) \\neq f\\left( x\\right) }\\right\\} .\n\nFor each \( n \\in \\mathbb{N} \), set\n\n\[ {g}_{n}\\left( x\\right) = \\left\\{ \\begin{array}{ll} {f}_{n}\\left( x\\right) & \\text{ if }x \\notin Z \\... | Yes |
For \( x \in \left\lbrack {0,1}\right\rbrack \), let\n\n\[ \n{\mathcal{X}}_{\mathbb{Q}} = \left\{ \begin{array}{ll} 1 & \text{ if }x \in \mathbb{Q}, \\ 0 & \text{ otherwise } \end{array}\right. \n\]\n\nand \( P \) be the partition \( P = \{ \mathbb{Q} \cap \left\lbrack {0,1}\right\rbrack ,\left\lbrack {0,1}\right\rbrac... | Setting \( {E}_{1} = \mathbb{Q} \cap \left\lbrack {0,1}\right\rbrack \) and \( {E}_{2} = \) \( \left\lbrack {0,1}\right\rbrack \smallsetminus \mathbb{Q} \), we have\n\n\[ \n{M}_{1} = \mathop{\sup }\limits_{{x \in {E}_{1}}}{\mathcal{X}}_{\mathbb{Q}}\left( x\right) = 1 = \mathop{\inf }\limits_{{x \in {E}_{1}}}{\mathcal{X... | Yes |
Lemma 2.2.8. Let \( f \in B\left\lbrack {a, b}\right\rbrack \) .\n\n(i) For any two measurable partitions \( {P}_{1} \) and \( {P}_{2} \) of \( \left\lbrack {a, b}\right\rbrack \) ,\n\n\[ L\left\lbrack {f,{P}_{1}}\right\rbrack \leq U\left\lbrack {f,{P}_{2}}\right\rbrack \]\n\n(ii) Consequently,\n\n\[ {\int }_{a}^{b}f \... | Proof. First we will establish (i). Let \( {P}^{ * } \) be a common refinement of \( {P}_{1} \) and \( {P}_{2} \) . Since \( {P}^{ * } \) is a refinement of both \( {P}_{1} \) and \( {P}_{2} \), by Exercise 12,\n\n\[ L\left\lbrack {f,{P}_{1}}\right\rbrack \leq L\left\lbrack {f,{P}^{ * }}\right\rbrack \;\text{ and }\;U\... | No |
Proposition 2.2.10. Let \( f \in B\left\lbrack {a, b}\right\rbrack \) . If \( f \) is Riemann integrable on \( \left\lbrack {a, b}\right\rbrack \), then \( f \) is Lebesgue integrable on \( \left\lbrack {a, b}\right\rbrack \) . | Proof. For any Riemann partition of \( \left\lbrack {a, b}\right\rbrack \) ,\n\n\[ \n{P}_{R} = \left\{ {a = {x}_{0} < {x}_{1} < {x}_{2} < \ldots < {x}_{n} = b}\right\} , \n\]\n\nwe form a corresponding measurable partition of \( \left\lbrack {a, b}\right\rbrack \) by setting\n\n\[ \n{P}_{L} = \left\{ {\left\lbrack {{x}... | Yes |
Lemma 2.2.11. Let \( f \in B\left\lbrack {a, b}\right\rbrack \) . Then \( f \) is Lebesgue integrable if and only if for every \( \epsilon > 0 \) there is a measurable partition \( P \) such that\n\n\[ U\left\lbrack {f, P}\right\rbrack - L\left\lbrack {f, P}\right\rbrack < \epsilon . \]\n | Proof. Assume first that \( f \) is Lebesgue integrable on \( \left\lbrack {a, b}\right\rbrack \) . Let \( \epsilon > 0 \) be given. By the definition of the lower integral, there is a measurable partition \( {P}_{1} \) of \( \left\lbrack {a, b}\right\rbrack \) such that\n\n\[ {\int }_{a}^{b}f - \frac{\epsilon }{2} < L... | Yes |
Theorem 2.2.12. Let \( f \in B\left\lbrack {a, b}\right\rbrack \) . If \( f \) is measurable on \( \left\lbrack {a, b}\right\rbrack \), then \( f \) is Lebesgue integrable on \( \left\lbrack {a, b}\right\rbrack \) . | Proof. Assume \( f \) is a bounded, measurable function on \( \left\lbrack {a, b}\right\rbrack \) and let \( \epsilon > 0 \) . Because \( f \) is bounded, there is a positive number \( M \) so that \( \left| {f\left( x\right) }\right| < M \) for all \( x \in \left\lbrack {a, b}\right\rbrack \) .\n\nWe will now form a m... | Yes |
Lemma 2.2.14. Let \( f \in B\left\lbrack {a, b}\right\rbrack \) . Suppose \( f \) is measurable with \( f \geq \) 0 a.e. in \( \left\lbrack {a, b}\right\rbrack \) and that \( {\int }_{a}^{b}f = 0 \) . Then \( f = 0 \) a.e. in \( \left\lbrack {a, b}\right\rbrack \) . | Proof. Set\n\n\[ g\left( x\right) = \left\{ \begin{array}{ll} f\left( x\right) & \text{ if }f\left( x\right) \geq 0 \\ 0 & \text{ otherwise } \end{array}\right. \]\n\nso that \( \left( {g - f}\right) = 0 \) a.e. on \( \left\lbrack {a, b}\right\rbrack \) . It is easy to show that \( g \in B\left\lbrack {a, b}\right\rbra... | No |
Example 2.3.2. Let\n\n\\[ \nf\\left( x\\right) = \\left\\{ \\begin{array}{ll} \\frac{1}{x} & \\text{ if }x \\neq 0 \\ 0 & \\text{ if }x = 0 \\end{array}\\right.\n\\]\n\nOur goal is to determine if \\( f \\) is in \\( \\mathcal{L}\\left\\lbrack {0,1}\\right\\rbrack \\) . | For \\( N > 1 \\), on this interval\n\n\\[ \n{}^{N}f\\left( x\\right) = \\left\\{ \\begin{array}{ll} N & \\text{ if }0 < x \\leq \\frac{1}{N}, \\ \\frac{1}{x} & \\text{ if }\\frac{1}{N} < x \\leq 1, \\ 0 & \\text{ if }x = 0, \\end{array}\\right.\n\\]\nso\n\n\\[ \n{\\int }_{0}^{1}{}^{N}f = {\\int }_{0}^{\\frac{1}{N}}N +... | Yes |
Example 2.3.3. Let\n\n\\[ \ng\\left( x\\right) = \\left\\{ \\begin{array}{ll} \\frac{1}{\\sqrt{x}} & \\text{ if }x \\neq 0 \\\\ 0 & \\text{ if }x = 0 \\end{array}\\right.\n\\]\n\nWe will determine if \\( g \\) is in \\( \\mathcal{L}\\left\\lbrack {0,1}\\right\\rbrack \\) . | For \\( N > 1 \\), on this interval\n\n\\[ \n{}^{N}g\\left( x\\right) = \\left\\{ \\begin{array}{ll} N & \\text{ if }0 < x \\leq \\frac{1}{{N}^{2}}, \\\\ \\frac{1}{\\sqrt{x}} & \\text{ if }\\frac{1}{{N}^{2}} < x \\leq 1, \\\\ 0 & \\text{ if }x = 0, \\end{array}\\right.\n\\]\n\nso\n\n\\[ \n{\\int }_{0}^{1}{}^{N}g = {\\i... | Yes |
Theorem 2.3.7. Let \( f \in B\left\lbrack {a, b}\right\rbrack \) . Suppose \( f \in \mathcal{L}\left\lbrack {a, b}\right\rbrack \) and \( f = g \) a.e. on \( \left\lbrack {a, b}\right\rbrack \) . Then \( g \in \mathcal{L}\left\lbrack {a, b}\right\rbrack \) and \[ {\int }_{a}^{b}g = {\int }_{a}^{b}f \] | Proof. We will show that \( g - f \) is in \( \mathcal{L}\left\lbrack {a, b}\right\rbrack \) and \( {\int }_{a}^{b}\left( {g - f}\right) = 0 \) . The result follows by the linearity of the integral. (i) Assume \( \left( {g - f}\right) \in B\left\lbrack {a, b}\right\rbrack \) . Then \( \left( {g - f}\right) = 0 \) a.e. ... | No |
Theorem 2.3.8. Let \( f \in \mathcal{L}\left\lbrack {a, b}\right\rbrack \) . Suppose \( f\left( x\right) \geq 0 \) a.e. on \( \left\lbrack {a, b}\right\rbrack \) . Then \[ {\int }_{a}^{b}f \geq 0 \] | Proof. (i) Assume \( f \in B\left\lbrack {a, b}\right\rbrack \) . Consider \( {f}^{ + } \), the positive part of \( f \) . Then \( f = {f}^{ + } \) a.e. on \( \left\lbrack {a, b}\right\rbrack \) . By Theorem 2.3.7, \( {\int }_{a}^{b}{f}^{ + } = {\int }_{a}^{b}f \), but by Theorem 2.3.6, \[ {\int }_{a}^{b}{f}^{ + } \geq... | Yes |
Theorem 2.3.9. Let \( f \in \mathcal{L}\left\lbrack {a, b}\right\rbrack \) . If \( f\left( x\right) \geq 0 \) a.e. on \( \left\lbrack {a, b}\right\rbrack \) and \( {\int }_{a}^{b}f = 0 \), then \( f = 0 \) a.e. on \( \left\lbrack {a, b}\right\rbrack \) . | Proof. Without loss of generality, we may assume that \( f\left( x\right) \geq 0 \) for all \( x \in \left\lbrack {a, b}\right\rbrack \) .\n\n(i) Assume \( f \) is bounded. This is covered by Lemma 2.2.14,\n\n(ii) Assume \( f \) is unbounded. Then\n\n\[ 0 = {\int }_{a}^{b}f = \mathop{\lim }\limits_{{N \rightarrow \inft... | Yes |
For \( x \in \left\lbrack {0,1}\right\rbrack \) and positive integer \( n \), let \( {f}_{n}\left( x\right) = {x}^{n} \) . Then \( \mathop{\lim }\limits_{{n \rightarrow \infty }}{f}_{n}\left( x\right) = f\left( x\right) \), where | \[ f\left( x\right) = \left\{ \begin{array}{ll} 0 & \text{ if }0 \leq x < 1 \\ 1 & \text{ if }x = 1 \end{array}\right. \] | Yes |
Lemma 2.4.4. Let \( f \in \mathcal{L}\left\lbrack {a, b}\right\rbrack \) . Suppose \( {\left\{ {A}_{k}\right\} }_{k = 1}^{\infty } \) is a countable collection of measurable subsets of \( \left\lbrack {a, b}\right\rbrack \) with\n\n\[ \n{A}_{1} \subseteq {A}_{2} \subseteq {A}_{3} \subseteq \ldots \n\]\n\nand\n\n\[ \n\m... | Proof. Let \( \epsilon > 0 \) be given. Our goal is to find \( N \) so that if \( k > N \) , then\n\n\[ \n\left| {{\int }_{a}^{b}f - {\int }_{{A}_{k}}f}\right| < \epsilon \n\]\n\nLet \( {E}_{k} = \left\lbrack {a, b}\right\rbrack \smallsetminus {A}_{k} \) . Then\n\n\[ \n{\int }_{a}^{b}f - {\int }_{{A}_{k}}f = {\int }_{a... | Yes |
Lemma 2.4.5. Let \( g \in \mathcal{L}\left\lbrack {a, b}\right\rbrack \) . Suppose \( f \) is measurable and \( \left| {f\left( x\right) }\right| \leq \) \( g\left( x\right) \) almost everywhere in \( \left\lbrack {a, b}\right\rbrack \) . Then \( f \in \mathcal{L}\left\lbrack {a, b}\right\rbrack \) . | Proof. Without loss of generality, we may assume that \( \left| {f\left( x\right) }\right| \leq g\left( x\right) \) for all \( x \in \left\lbrack {a, b}\right\rbrack \) . We must show that \( {f}^{ + } \) and \( {f}^{ - } \) are Lebesgue integrable. Since \( \left| {f\left( x\right) }\right| \leq g\left( x\right) \), b... | Yes |
For each positive integer \( n \) and \( x \in \left\lbrack {0,2}\right\rbrack \) define \( {f}_{n}\left( x\right) \) to be\n\n\[ \n{f}_{n}\left( x\right) = \left\{ \begin{array}{ll} 0 & \text{ if }0 \leq x < \frac{1}{n} \\ \sqrt{n} & \text{ if }\frac{1}{n} \leq x \leq \frac{2}{n} \\ 0 & \text{ if }\frac{2}{n} < x \leq... | Let\n\n\[ \ng\left( x\right) = \left\{ \begin{array}{ll} \frac{\sqrt{2}}{\sqrt{x}} & \text{ if }x \neq 0 \\ 0 & \text{ if }x = 0 \end{array}\right. \n\]\n\nThen \( g \in \mathcal{L}\left\lbrack {0,2}\right\rbrack \) and \( \left| {{f}_{n}\left( x\right) }\right| \leq g\left( x\right) \) for all \( x \in \left\lbrack {0... | Yes |
For each positive integer \( n \) and \( x \in \left\lbrack {0,1}\right\rbrack \) define \( {f}_{n}\left( x\right) \) to be\n\n\[ \n{f}_{n}\left( x\right) = \left\{ \begin{array}{ll} 0 & \text{ if }x = 0, \\ \left( {1 - {e}^{-\frac{{x}^{2}}{n}}}\right) \frac{1}{\sqrt{x}} & \text{ if }0 < x \leq 1. \end{array}\right. \n... | If we define \( g \) by\n\n\[ \ng\left( x\right) = \left\{ \begin{array}{ll} 0 & \text{ if }x = 0 \\ \frac{1}{\sqrt{x}} & \text{ if }0 < x \leq 1 \end{array}\right. \n\]\n\nthen \( g \in \mathcal{L}\left\lbrack {0,1}\right\rbrack \) and \( \left| {{f}_{n}\left( x\right) }\right| \leq g\left( x\right) \) for all \( x \i... | Yes |
For each positive integer \( n \) and \( x \in \left\lbrack {0,1}\right\rbrack \) define \( {f}_{n}\left( x\right) \) to be\n\n\[ \n{f}_{n}\left( x\right) = \frac{n\sin x}{1 + {n}^{2}\sqrt{x}} + 2{e}^{x/n}.\n\]\n\nThe pointwise limit is \( \mathop{\lim }\limits_{{n \rightarrow \infty }}{f}_{n}\left( x\right) = 2 \) . | In this case, for \( x \neq 0 \) ,\n\n\[ \n\left| {{f}_{n}\left( x\right) }\right| \leq \frac{n}{1 + {n}^{2}\sqrt{x}} + 2 \leq \frac{1}{n\sqrt{x}} + 2 \leq \frac{1}{\sqrt{x}} + 2.\n\]\n\nTherefore, we may use the dominating function \( g \) where\n\n\[ \ng\left( x\right) = \left\{ \begin{array}{ll} 2 & \text{ if }x = 0... | Yes |
Theorem 2.4.10 (Fatou’s Lemma, preliminary version). Let \( \left\{ {f}_{n}\right\} \) be a sequence of nonnegative functions in \( \mathcal{L}\left\lbrack {a, b}\right\rbrack \) . Suppose \( \mathop{\lim }\limits_{{n \rightarrow \infty }}{f}_{n}\left( x\right) = f\left( x\right) \) a.e. in \( \left\lbrack {a, b}\right... | Proof. Without loss of generality, we may assume \( f\left( x\right) \geq 0 \) for all \( x \in \left\lbrack {a, b}\right\rbrack \) . It follows that for every \( N \) ,\n\n\[ \mathop{\lim }\limits_{{n \rightarrow \infty }}{}^{N}{f}_{n}\left( x\right) = {}^{N}f\left( x\right) \leq N. \]\n\n(Remember, we might be dealin... | No |
Corollary 2.4.12 (Fatou’s Lemma). Let \( \left\{ {f}_{n}\right\} \) be a sequence of nonnegative functions in \( \mathcal{L}\left\lbrack {a, b}\right\rbrack \) . Suppose \( \mathop{\liminf }\limits_{{n \rightarrow \infty }}{f}_{n}\left( x\right) = f\left( x\right) \) a.e. in \( \left\lbrack {a, b}\right\rbrack \) .\n\n... | Proof. For each positive integer \( n \), let \( {g}_{n}\left( x\right) = \mathop{\inf }\limits_{{k \geq n}}{f}_{k}\left( x\right) \) . Then \( {g}_{n} \) is nonnegative and measurable for each \( n \) . Also, by Lemma 2.4.5, for each \( n,{g}_{n} \in \mathcal{L}\left\lbrack {a, b}\right\rbrack \) since \( {g}_{n}\left... | Yes |
Theorem 2.4.13 (Monotone Convergence Theorem). Let \( \\left\\{ {f}_{n}\\right\\} \) be a sequence of nonnegative functions in \( \\mathcal{L}\\left\\lbrack {a, b}\\right\\rbrack \) . Suppose \( \\left\\{ {{f}_{n}\\left( x\\right) }\\right\\} \) is an increasing sequence for almost every \( x \\in \\left\\lbrack {a, b}... | Proof. For each \( n,{f}_{n}\\left( x\\right) \\leq {f}_{n + 1}\\left( x\\right) \) almost everywhere in \( \\left\\lbrack {a, b}\\right\\rbrack \) . Consequently \( {f}_{n} \\leq f \) a.e. in \( \\left\\lbrack {a, b}\\right\\rbrack \) . Therefore, if \( f \\in \\mathcal{L}\\left\\lbrack {a, b}\\right\\rbrack \), we ma... | Yes |
Let \( I = \left\{ {\left( {x, y}\right) \in {\mathbb{R}}^{2} \mid 0 \leq x \leq 1}\right. \) and \( \left. {0 \leq y \leq 1}\right\} \) be the unit square in \( {\mathbb{R}}^{2} \) . Subdivide \( I \) into four congruent squares. Let \( {I}_{1} \) denote the lower left corner square. Take the upper right corner square... | Next, for each \( n \), subdivide \( {I}_{n} \) into four congruent squares. Label these \( {I}_{n}^{1},{I}_{n}^{2},{I}_{n}^{3} \), and \( {I}_{n}^{4} \), starting with the lower left square and moving in a counterclockwise fashion. Next, we will define \( f : {\mathbb{R}}^{2} \rightarrow {\mathbb{R}}^{1} \) by\n\n\[ f... | No |
Example 2.5.3. Define \( f \) on \( \lbrack 0, + \infty ) \) as the piecewise linear function connecting the points \( \left( {0,0}\right) ,\left( {1,1}\right) ,\left( {2,0}\right) ,\left( {3, - \frac{1}{2}}\right) ,\left( {4,0}\right) ,\left( {5,\frac{1}{3}}\right) ,\left( {6,0}\right) \) , \( \left( {7, - \frac{1}{4}... | The improper Riemann integral of \( f \) is\n\n\[ {\int }_{0}^{\infty }f\left( x\right) {dx} = \mathop{\lim }\limits_{{N \rightarrow \infty }}{\int }_{0}^{N}f\left( x\right) {dx} = \mathop{\sum }\limits_{{n = 1}}^{\infty }\frac{{\left( -1\right) }^{n + 1}}{n}. \]\n\nThis is the alternating harmonic series, which conver... | Yes |
Proposition 3.1.2. Define \( \sim \) on \( \mathcal{L}\left\lbrack {a, b}\right\rbrack \) by\n\n\[ f \sim g\;\text{ if and only if }\;f = g\text{ a.e. }\n\]\n\nThen \( \sim \) is an equivalence relation on \( \mathcal{L}\left\lbrack {a, b}\right\rbrack \) . That is, the following three conditions hold:\n\n(i) For all \... | Proof. This is proved in Exercise 3. | No |
Proposition 3.1.5. \( \\parallel \\cdot {\\parallel }_{1} \) is a norm on \( {L}^{1}\\left\\lbrack {a, b}\\right\\rbrack \) . | Proof. The only requirement for a norm that needs to be checked is part (ii) of Definition 3.1.1 If \( \\parallel f{\\parallel }_{1} = 0 \), then\n\n\[ \n{\\int }_{a}^{b}\\left| f\\right| = 0 \n\]\n\nas mentioned earlier. By Theorem 2.3.9, \( f = 0 \) a.e. in \( \\left\\lbrack {a, b}\\right\\rbrack \) . Therefore, in \... | Yes |
Theorem 3.1.8. \( \mathbb{R} \) is a Banach space with respect to the norm \( \left| \cdot \right| \) . | Proof. Let \( \left\{ {a}_{n}\right\} \) be a sequence in \( \mathbb{R} \) that is Cauchy with respect to \( \left| \cdot \right| \) . Thus, for every \( \epsilon > 0 \) there is an \( N \) so that\n\n\[ \left| {{a}_{n} - {a}_{m}}\right| < \epsilon \;\text{ whenever }\;n, m > N. \]\n\nIn order to show that this sequenc... | No |
For \( x \in \left\lbrack {0,1}\right\rbrack \) and positive integer \( n \), let\n\n\[ \n{f}_{n}\left( x\right) = \left\{ \begin{array}{ll} 2{n}^{2}x & \text{ if }0 \leq x \leq \frac{1}{2n} \\ - 2{n}^{2}\left( {x - \frac{1}{n}}\right) & \text{ if }\frac{1}{2n} < x \leq \frac{1}{n} \\ 0 & \text{ otherwise. } \end{array... | \[ \n\mathop{\lim }\limits_{{n \rightarrow \infty }}{\begin{Vmatrix}{f}_{n} - f\end{Vmatrix}}_{1} = \mathop{\lim }\limits_{{n \rightarrow \infty }}{\int }_{0}^{1}\left| {{f}_{n} - f}\right| = \mathop{\lim }\limits_{{n \rightarrow \infty }}{\int }_{0}^{1}{f}^{n} = \mathop{\lim }\limits_{{n \rightarrow \infty }}\frac{1}{... | Yes |
Theorem 3.1.11 (Beppo Levi). Let \( \left\{ {g}_{k}\right\} \) be a sequence of functions in \( {L}^{1}\left\lbrack {a, b}\right\rbrack \) such that \( \mathop{\sum }\limits_{{k = 1}}^{\infty }{\begin{Vmatrix}{g}_{k}\end{Vmatrix}}_{1} \) converges. Then there exists \( g \in {L}^{1}\left\lbrack {a, b}\right\rbrack \) s... | Proof. As with Theorem 3.1.8, the first step is to describe a \ | No |
For \( 0 \leq x < 1 \) , \[ \frac{1}{{\left( 1 - x\right) }^{2}} = \mathop{\sum }\limits_{{k = 1}}^{\infty }k{x}^{k - 1}. \] | Hence, \[ {\left( \frac{\ln x}{1 - x}\right) }^{2} = \frac{1}{{\left( 1 - x\right) }^{2}}{\left( \ln x\right) }^{2} = \mathop{\sum }\limits_{{k = 1}}^{\infty }k{x}^{k - 1}{\left( \ln x\right) }^{2} \] for almost every \( x \in \left\lbrack {0,1}\right\rbrack \) . Setting \( {g}_{k}\left( x\right) = {x}^{k - 1}{\left( \... | Yes |
Theorem 3.1.13. The space \( {L}^{1}\left\lbrack {a, b}\right\rbrack \) is complete with respect to the norm \( \parallel \cdot {\parallel }_{1} \), the \( {L}^{1} \) -norm. | Proof. We must show that if \( \left\{ {f}_{n}\right\} \) is a sequence functions in \( {L}^{1}\left\lbrack {a, b}\right\rbrack \) which is Cauchy with respect to the \( {L}^{1} \) -norm, then there exists a function \( f \in {L}^{1}\left\lbrack {a, b}\right\rbrack \) such that\n\n\[ \mathop{\lim }\limits_{{n \rightarr... | Yes |
Proposition 3.2.3. Let \( f, g \in {L}^{p}\left\lbrack {a, b}\right\rbrack \), and \( c \in \mathbb{R} \) . (i) \( {cf} \in {L}^{p}\left\lbrack {a, b}\right\rbrack \) . (ii) \( f + g \in {L}^{p}\left\lbrack {a, b}\right\rbrack \) . | Proof. (i) If \( f \in {L}^{p}\left\lbrack {a, b}\right\rbrack \), then \( {\left| f\right| }^{p} \) is Lebesgue integrable. As a consequence, \( {\left| c\right| }^{p}{\left| f\right| }^{p} = {\left| cf\right| }^{p} \) is Lebesgue integrable. Hence, \( {cf} \in {L}^{p}\left\lbrack {a, b}\right\rbrack \) . (ii) Since b... | Yes |
Theorem 3.2.7 (Minkowski’s Inequality). Let \( p \geq 1 \) . If \( f, g \in \) \( {L}^{p}\left\lbrack {a, b}\right\rbrack \), then\n\n\[ \parallel f + g{\parallel }_{p} \leq \parallel f{\parallel }_{p} + \parallel g{\parallel }_{p} \] | Proof. We already have this result for the case \( p = 1 \), so assume \( p > 1 \) . This result is trivially true if \( \left| {f + g}\right| = 0 \) a.e. in \( \left\lbrack {a, b}\right\rbrack \) . (Make sure you understand why this is deemed \ | No |
For \( x \in \left\lbrack {0,1}\right\rbrack \) let\n\n\[ f\left( x\right) = \left\{ \begin{array}{ll} {2x} & \text{ if }x \notin \mathbb{Q}, \\ q & \text{ if }x = \frac{p}{q}. \end{array}\right. \]\n\nHere we are assuming \( \frac{p}{q} \) is in the lowest terms. In this case,\n\n\[ \mathop{\sup }\limits_{{x \in \left... | while\n\n\[ \underset{x \in \left\lbrack {0,1}\right\rbrack }{\operatorname{ess}\sup }f = 2 \] | Yes |
Lemma 3.3.1. Let \( p \geq 1 \) and \( f \in {L}^{p}\left\lbrack {a, b}\right\rbrack \) . Given \( \epsilon > 0 \) there is a bounded function \( g \) such that \( \parallel f - g{\parallel }_{p} < \epsilon \) . | Proof. If \( f \) is bounded, we are done, so we will assume \( f \) is unbounded. For \( N > 0 \) set\n\n\[ \n{g}^{N}\left( x\right) = \left\{ \begin{matrix} f\left( x\right) & \text{ if }\left| {f\left( x\right) }\right| \leq N \\ N & \text{ if }f\left( x\right) > N \\ - N & \text{ if }f\left( x\right) < - N \end{mat... | Yes |
Lemma 3.3.2. Let \( p \geq 1 \) and \( f \in {L}^{p}\left\lbrack {a, b}\right\rbrack \) be a bounded function. Then there exists a simple function\n\n\[ \phi = \mathop{\sum }\limits_{{i = 1}}^{n}{a}_{i}{\mathcal{X}}_{{A}_{i}} \]\n\nsuch that \( \parallel f - \phi {\parallel }_{p} < \epsilon \) . | Proof. This proof is reminiscent of Theorem 2.2.12. Since \( f \) is bounded there exists an \( M > 0 \) with \( - M < f\left( x\right) < M \) for all \( x \in \left\lbrack {a, b}\right\rbrack \) . Let\n\n\[ - M = {y}_{0} < {y}_{1} < {y}_{2} < \ldots < {y}_{n} = M \]\n\nwhere \( {y}_{1},{y}_{2},\ldots ,{y}_{n} \) are c... | Yes |
Lemma 3.3.3. Let \( E \subseteq \left\lbrack {a, b}\right\rbrack \) be a measurable set. Given \( \epsilon > 0 \) and \( p \geq 1 \) there is a continous function \( h \) so that \( {\begin{Vmatrix}{\mathcal{X}}_{E} - h\end{Vmatrix}}_{p} < \epsilon \) . | Proof. Let \( \epsilon > 0 \) be given. By the definition of a measurable set, there is an open set \( G \) containing \( E \) such that \( m\left( {G \smallsetminus E}\right) < {\epsilon }^{p}/2 \) . There is also a closed set \( F \) contained in \( E \) such that \( m\left( {E \smallsetminus F}\right) < {\epsilon }^... | Yes |
Corollary 3.3.4. Let \( p \geq 1 \) and \( \phi \) be a simple function\n\n\[ \phi = \mathop{\sum }\limits_{{i = 1}}^{n}{a}_{i}{\mathcal{X}}_{{A}_{i}} \]\n\nwhere \( {A}_{i} \) is a measurable subset of \( \left\lbrack {a, b}\right\rbrack \) for each \( i \) . Then for every \( \epsilon > 0 \) there exists a continuous... | Proof. This is Exercise 23 | No |
Theorem 3.3.5. Let \( p \geq 1 \) and \( f \in {L}^{p}\left\lbrack {a, b}\right\rbrack \) . For every \( \epsilon > 0 \) there is a continuous function \( g \) such that \( \parallel f - g{\parallel }_{p} \leq \epsilon \) . | Proof. Let \( \epsilon > 0 \) be given. By Lemma 3.3.1, there is a bounded function \( {g}^{N} \) with \( {\begin{Vmatrix}f - {g}^{N}\end{Vmatrix}}_{p} < \frac{\epsilon }{2} \) . By Lemma 3.3.2, there is a simple function \( \phi \) such that \( {\begin{Vmatrix}{g}^{N} - \phi \end{Vmatrix}}_{p} < \frac{\epsilon }{4} \)... | Yes |
For \( \overrightarrow{x},\overrightarrow{y} \in {\mathbb{R}}^{3} \) with \( \overrightarrow{x} = \left( {{x}_{1},{x}_{2},{x}_{3}}\right) \) and \( \overrightarrow{y} = \left( {{y}_{1},{y}_{2},{y}_{3}}\right) \) we have the usual dot product, \[ \overrightarrow{x} \cdot \overrightarrow{y} = {x}_{1}{y}_{1} + {x}_{2}{y}_... | As expected, \( {\mathbb{R}}^{3} \) with the dot product is an inner product space. | No |
For \( f, g \in {L}^{2}\left\lbrack {a, b}\right\rbrack \) define \( \langle f, g\rangle \) to be\n\n\[ \langle f, g\rangle = {\int }_{a}^{b}{fg}. \]\n\nOne of the first steps we need to do in order to show that we have created an inner product is to verify that we have defined a function from \( {L}^{2}\left\lbrack {a... | Hölder’s Inequality, Theorem 3.2.5, with \( p = q = 2 \) guarantees that the product \( {fg} \) is in \( {L}^{1}\left\lbrack {a, b}\right\rbrack \), which is precisely what we need. (On the other hand, if \( p \neq 2 \), we cannot use Hölder’s inequality to guarantee that \( {\int }_{a}^{b}{fg} \) always produces a fin... | Yes |
In \( {L}^{2}\left\lbrack {a, b}\right\rbrack \) the induced norm is | \[ \parallel f\parallel = \sqrt{\langle f, f\rangle } = {\left( {\int }_{a}^{b}{f}^{2}\right) }^{\frac{1}{2}} = \parallel f{\parallel }_{2} \] | Yes |
Proposition 3.4.5 (Cauchy-Schwarz Inequality). Let \( V \) be an inner product space with inner product \( \langle \cdot , \cdot \rangle \) . For every \( v, w \in V \) , \[ \left| {\langle v, w\rangle }\right| \leq \parallel v\parallel \parallel w\parallel \] | Proof. The result is easily true if either \( v \) or \( w \) is the zero vector.\n\nHence, we will assume neither \( v \) nor \( w \) is the zero vector.\n\nBy property (iii) of a norm, \[ \langle {tv} - w,{tv} - w\rangle \geq 0 \] for every real number \( t \) . Therefore, \[ {t}^{2}\langle v, v\rangle - {2t}\langle ... | Yes |
Proposition 3.4.6. Let \( V \) be an inner product space with inner product \( \langle \cdot , \cdot \rangle \) and induced norm \( \parallel \cdot \parallel \) . Then for all \( v, w \in V \) , \n\n\[ \n\parallel v + w\parallel \leq \parallel v\parallel + \parallel w\parallel \n\] | Proof. For every \( v, w \in V \), by the definition of the induced norm, properties of the inner product, and the Cauchy-Schwarz Inequality (Proposition 3.4.5), \n\n\[ \n\parallel v + w{\parallel }^{2} = \langle v + w, v + w\rangle \n\] \n\n\[ \n= \langle v, v\rangle + 2\langle v, w\rangle + \langle w, w\rangle \n\] \... | Yes |
Let \( C\left\lbrack {a, b}\right\rbrack \) denote the space of functions that are continuous on the interval \( \left\lbrack {a, b}\right\rbrack \) . We can define an inner product on this space in the same fashion as on \( {L}^{2}\left\lbrack {a, b}\right\rbrack \), that is,\n\n\[ \langle f, g\rangle = {\int }_{a}^{b... | By Theorem 3.3.5 we can approximate \( f \) by continuous functions. That is, we can find a sequence of continuous functions \( \left\{ {g}_{n}\right\} \) that converge to \( f \) in the \( {L}^{2} \) -norm. But this means that \( \left\{ {g}_{n}\right\} \) must be a Cauchy sequence that does not converge to a continuo... | Yes |
Proposition 3.4.10. Let \( V \) be an inner product space with induced norm \( \parallel v\parallel = \sqrt{\langle v, v\rangle } \) . Then for every \( v, w \in V \) , \n\n\[ \n\parallel v + w{\parallel }^{2} + \parallel v - w{\parallel }^{2} = 2\parallel v{\parallel }^{2} + 2\parallel w{\parallel }^{2}.\n\] | Proof. This is Exercise 20, | No |
Consider \( {L}^{1}\left\lbrack {0,1}\right\rbrack \) with the \( {L}^{1} \) -norm. Both \( f\left( x\right) = 1 - x \) and \( g\left( x\right) = x \) are in \( {L}^{1}\left\lbrack {0,1}\right\rbrack \) . | Thus\n\n\[ \parallel f + g{\parallel }_{1} = {\int }_{0}^{1}1 = 1 \]\n\n\[ \parallel f - g{\parallel }_{1} = {\int }_{0}^{1}\left| {1 - {2x}}\right| = \frac{1}{2}, \]\n\n\[ \parallel f{\parallel }_{1} = {\int }_{0}^{1}\left| {1 - x}\right| = \frac{1}{2},\;\text{ and } \]\n\n\[ \parallel g{\parallel }_{1} = {\int }_{0}^... | Yes |
Proposition 3.5.1. Let \( f \in {L}^{2}\left\lbrack {-\pi ,\pi }\right\rbrack \) . For each positive integer \( n \) ,\n\n\[
{\begin{Vmatrix}f - {s}_{n}\end{Vmatrix}}_{2}^{2} = \parallel f{\parallel }_{2}^{2} - \left( {\frac{\pi {a}_{0}^{2}}{2} + \pi \mathop{\sum }\limits_{{k = 1}}^{n}\left( {{a}_{k}^{2} + {b}_{k}^{2}}... | Proof. This is Exercise 26 | No |
Corollary 3.5.2. Let \( f \in {L}^{2}\left\lbrack {-\pi ,\pi }\right\rbrack \) . Let \( {a}_{0},{a}_{1},{b}_{1},{a}_{2},{b}_{2},\ldots \) be the Fourier coefficients for \( f \) . Then \( \mathop{\sum }\limits_{{k = 1}}^{\infty }\left( {{a}_{k}^{2} + {b}_{k}^{2}}\right) \) converges and \[ \frac{\pi {a}_{0}^{2}}{2} + \... | Proof. For each \( n \), by Proposition 3.5.1 \[ \frac{\pi {a}_{0}^{2}}{2} + \pi \mathop{\sum }\limits_{{k = 1}}^{n}\left( {{a}_{k}^{2} + {b}_{k}^{2}}\right) \leq \parallel f{\parallel }_{2}^{2}. \] The result follows by taking \( n \) to infinity. | Yes |
Theorem 3.5.3. Let \( f \in {L}^{2}\left\lbrack {-\pi ,\pi }\right\rbrack \) and let \( {T}_{n} \) be a trigonometric polynomial of degree \( n \) . Then\n\n\[{\begin{Vmatrix}f - {T}_{n}\end{Vmatrix}}_{2} \geq {\begin{Vmatrix}f - {s}_{n}\end{Vmatrix}}_{2}\]\n\nwhere \( {a}_{0},{a}_{1},{b}_{1},\ldots ,{b}_{n} \) are the... | Proof. Let \( {T}_{n} \) be a trigonometric polynomial of degree \( n \), say\n\n\[{T}_{n}\left( x\right) = {A}_{0} + \mathop{\sum }\limits_{{k = 1}}^{n}\left( {{A}_{k}\cos \left( {kx}\right) + {B}_{k}\sin \left( {kx}\right) }\right) .\n\nWe will show that \( {\begin{Vmatrix}f - {T}_{n}\end{Vmatrix}}_{2}^{2} - {\begin{... | Yes |
Corollary 3.5.7. Let \( g \) be continuous on \( \left\lbrack {-\pi ,\pi }\right\rbrack \) with \( g\left( {-\pi }\right) = g\left( \pi \right) \) . For each positive integer \( n \), define \( {\sigma }_{n}\left( x\right) \) to be\n\n\[ \n{\sigma }_{n}\left( x\right) = \frac{1}{n}\mathop{\sum }\limits_{{k = 0}}^{{n - ... | Proof. Uniform convergence allows us to conclude that\n\n\[ \n\mathop{\lim }\limits_{{n \rightarrow \infty }}{\begin{Vmatrix}{\sigma }_{n} - g\end{Vmatrix}}_{2}^{2} = {\int }_{-\pi }^{\pi }\mathop{\lim }\limits_{{n \rightarrow \infty }}\left( {\left| {\sigma }_{n} - g\right| }^{2}\right) = 0.\n\] | Yes |
Theorem 3.5.8. Let \( f \in {L}^{2}\left\lbrack {-\pi ,\pi }\right\rbrack \) . Let \( {s}_{n}\left( x\right) \) equal the \( n \) th partial sum of the Fourier series for \( f \) . Then the sequence \( {s}_{n} \) converges to \( f \) with respect to the \( {L}^{2} \) -norm. That is,\n\n\[ \mathop{\lim }\limits_{{n \rig... | Proof. Given \( \epsilon > 0 \), by Exercise 24 there is a continuous function \( g \) defined on \( \left\lbrack {-\pi ,\pi }\right\rbrack \) with \( g\left( {-\pi }\right) = g\left( \pi \right) \) and\n\n\[ \parallel f - g{\parallel }_{2} < \frac{\epsilon }{2}. \]\n\nBy Corollary 3.5.7, there is a positive integer \(... | No |
Corollary 3.5.9. Let \( f \in {L}^{2}\left\lbrack {-\pi ,\pi }\right\rbrack \) . Let \( {a}_{0},{a}_{1},{b}_{1},{a}_{2},{b}_{2},\ldots \) be the Fourier coefficients for \( f \) . Then\n\n\[ \frac{\pi {a}_{0}^{2}}{2} + \pi \mathop{\sum }\limits_{{k = 1}}^{\infty }\left( {{a}_{k}^{2} + {b}_{k}^{2}}\right) = \parallel f{... | Proof. For each \( n \), by Proposition 3.5.1,\n\n\[ {\begin{Vmatrix}f - {s}_{n}\end{Vmatrix}}_{2}^{2} = \parallel f{\parallel }_{2}^{2} - \left( {\frac{\pi {a}_{0}^{2}}{2} + \pi \mathop{\sum }\limits_{{k = 1}}^{n}\left( {{a}_{k}^{2} + {b}_{k}^{2}}\right) }\right) .\n\nThe result follows by taking \( n \) to infinity a... | Yes |
Proposition 4.1.3. Let \( \mathcal{A} \) be an algebra of sets on \( X \) . If \( A, B \in \mathcal{A} \) , then \( A \cap B \in \mathcal{A} \) . | Proof. By de Morgan's laws,\n\n\[ A \cap B = {\left( {A}^{c} \cup {B}^{c}\right) }^{c}.\n\]\n\nTherefore, this proposition follows from properties (ii) and (iii) of an algebra of sets.\n\nThus, an algebra of sets is also closed under intersection. | Yes |
Proposition 4.1.8. Let \( \mathcal{B} \) be a \( \sigma \) -algebra of sets. If \( \left\{ {A}_{n}\right\} \) is a countable collection of sets in \( \mathcal{B} \), then \( \bigcap {A}_{n} \in \mathcal{B} \) . | Proof. This is Exercise 2. | No |
Let \( \mathcal{B} = \{ C \subseteq \mathbb{R} \mid C \) is finite \( \} \) . Then \( \mathcal{A} \), the \( \sigma \) - algebra generated by \( \mathcal{B} \), must contain all finite sets and their complements. Additionally, \( \mathcal{A} \) must contain countable unions of finite sets. In other words, \( \mathcal{A... | This is left as an exercise (see Exercise 3). | No |
Define \( \mu \) on \( \mathcal{P}\left( \mathbb{R}\right) \), the set of subsets of \( \mathbb{R} \), by\n\n\[ \mu \left( A\right) = \left\{ \begin{array}{ll} 1 & \text{ if }\pi \in A \\ 0 & \text{ otherwise. } \end{array}\right. \] \n\nWe will show that \( \left( {\mathbb{R},\mathcal{P}\left( \mathbb{R}\right) ,\mu }... | First of all, \( \mathcal{P}\left( \mathbb{R}\right) \) is a \( \sigma \) -algebra on \( \mathbb{R} \) and hence \( \left( {\mathbb{R},\mathcal{P}\left( \mathbb{R}\right) }\right) \) is a measurable space. Furthermore, by definition \( \mu \left( \varnothing \right) = 0 \) . Finally, if \( \left\{ {E}_{j}\right\} \) is... | Yes |
Proposition 4.1.17. Let \( \\left( {X,\\mathcal{B},\\mu }\\right) \) be a measure space. If \( A, B \\in \\mathcal{B} \) and \( A \\subseteq B \), then\n\n\[ \n\\mu \\left( A\\right) \\leq \\mu \\left( B\\right) .\n\]\n\nIn addition, if \( \\mu \\left( A\\right) \) is finite, then\n\n\[ \n\\mu \\left( {B \\smallsetminu... | Proof. Since \( \\mathcal{B} \) is a \( \\sigma \) -algebra, \( B \\smallsetminus A = B \\cap {A}^{c} \\in \\mathcal{B} \) . By definition of a measure, \( \\mu \\left( {B \\smallsetminus A}\\right) \\geq 0 \) . Also, \( A \) and \( B \\smallsetminus A \) are disjoint. Therefore, by property (ii) of a measure\n\n\[ \n\... | Yes |
Theorem 4.1.18. Let \( \left( {X,\mathcal{B},\mu }\right) \) be a measure space. If \( \left\{ {E}_{j}\right\} \) is a countable collection of sets in \( \mathcal{B} \), then\n\n\[ \mu \left( {\mathop{\bigcup }\limits_{j}{E}_{j}}\right) \leq \mathop{\sum }\limits_{j}\mu \left( {E}_{j}\right) \] | Proof. Set \( {G}_{1} = {E}_{1} \) . For each \( j > 1 \) let\n\n\[ {G}_{j} = {E}_{j} \smallsetminus \mathop{\bigcup }\limits_{{i = 1}}^{{j - 1}}{E}_{i} \]\n\nThen for each \( j,{G}_{j} \in \mathcal{B} \) and \( {G}_{j} \subseteq {E}_{j} \), and hence \( \mu \left( {G}_{j}\right) \leq \mu \left( {E}_{j}\right) \) by Pr... | Yes |
Corollary 4.1.19. Let \( \left( {X,\mathcal{B},\mu }\right) \) be a measure space. If \( B, C \in \mathcal{B} \) and \( \mu \left( C\right) = 0 \), then \( \mu \left( {B \cup C}\right) = \mu \left( B\right) \) . | Proof. The result follows from the inequalities\n\n\[ \mu \left( B\right) \leq \mu \left( {B \cup C}\right) \leq \mu \left( B\right) + \mu \left( C\right) = \mu \left( B\right) . \] | Yes |
Lemma 4.1.20. Let \( \left( {X,\mathcal{B},\mu }\right) \) be a measure space. If \( \left\{ {A}_{j}\right\} \) is a countably infinite collection of sets in \( \mathcal{B} \) with\n\n\[ \n{A}_{1} \subseteq {A}_{2} \subseteq {A}_{3} \subseteq \ldots \subseteq {A}_{j} \subseteq \ldots ,\n\]\n\nthen\n\n\[ \n\mu \left( {\... | Proof. If \( \mu \left( {A}_{j}\right) = \infty \) for some \( j \), then it must be the case that\n\n\[ \n\mu \left( {\mathop{\bigcup }\limits_{{j = 1}}^{\infty }{A}_{j}}\right) = \infty\n\]\n\nOn the other hand, \( {A}_{j} \subseteq {A}_{n} \) for all \( n \geq j \), and hence by Proposition 4.1.17 \( \mu \left( {A}_... | Yes |
To show that \( f \) is measurable is very similar to Example 2.1.2 We will look at the following cases: | (i) If \( s \geq \pi \), then \( \{ x \in I \mid f\left( x\right) > s\} = \varnothing \), which is a set in \( \mathcal{B} \).\n\n(ii) If \( s < \sqrt{2} \), then \( \{ x \in I \mid f\left( x\right) > s\} = X \), which is a set in \( \mathcal{B} \).\n\n(iii) If \( \sqrt{2} \leq s < 3 \), then \( \{ x \in I \mid f\left(... | Yes |
Theorem 4.2.3. Let \( f : X \rightarrow \overline{\mathbb{R}} \) . The following statements are equivalent:\n\n(i) \( f \) is measurable.\n\n(ii) For every \( s \in \mathbb{R} \), the set \( \{ x \in X \mid f\left( x\right) \leq s\} \) is a measurable set.\n\n(iii) For every \( s \in \mathbb{R} \), the set \( \{ x \in ... | Proof. The proof of this theorem is similar to the proof of Theorem 2.1.4. For example, to show that (ii) implies (iii), note that\n\n\[ \{ x \in X \mid f\left( x\right) < s\} = \mathop{\bigcup }\limits_{{k = 1}}^{\infty }\left\{ {x \in X \mid f\left( x\right) \leq s - \frac{1}{k}}\right\} .\n\]\n\nSince \( \left\{ {x ... | No |
Theorem 4.2.4. Let \( f : X \rightarrow \overline{\mathbb{R}} \) be a measurable function and let \( c \in \mathbb{R} \) . Then the following two statements are true:\n\n(i) The function \( f\left( x\right) + c \) is measurable.\n\n(ii) The function \( {cf}\left( x\right) \) is measurable. | Proof. The proof is similar to the proof of Theorem 2.1.5. | No |
Theorem 4.2.5. Let \( f, g : X \rightarrow \overline{\mathbb{R}} \) be measurable functions. Then\n\n(i) the function \( f\left( x\right) + g\left( x\right) \) is measurable,\n\n(ii) the function \( f\left( x\right) g\left( x\right) \) is measurable, and\n\n(iii) the function \( \frac{f\left( x\right) }{g\left( x\right... | Proof. The proof is similar to the proof of Theorem 2.1.6. | No |
Theorem 4.2.7. Let \( \\left\\{ {f}_{n}\\right\\} \) be a pointwise bounded sequence of measurable functions. Then both \( {f}^{ * } \) and \( {f}_{ * } \) are measurable functions on I. | Proof. As one might expect, the proof is very similar to that for Lebesgue measurable functions. Let\n\n\[ \n{M}_{n}\\left( x\\right) = \\sup \\left\\{ {{f}_{n}\\left( x\\right) ,{f}_{n + 1}\\left( x\\right) ,{f}_{n + 2}\\left( x\\right) ,\\ldots }\\right\\} \\text{and} \n\]\n\n\[ \n{m}_{n}\\left( x\\right) = \\inf \\l... | Yes |
Example 4.2.9. Let \( X = \{ a, b, c, d, e\} \) and\n\n\[ \mathcal{B} = \{ \varnothing, X,\{ a, c\} ,\{ b, d\} ,\{ a, b, c, d\} ,\{ b, d, e\} ,\{ a, c, e\} ,\{ e\} \} .\n\]\n\nDefine \( \mu : \mathcal{B} \rightarrow \overline{\mathbb{R}} \) by\n\n\[ \mu \left( \varnothing \right) = 0,\;\mu \left( {\{ a, c\} }\right) = ... | In this case\n\n\[ \mu \left( {\{ x \in X \mid f\left( x\right) \neq g\left( x\right) \} }\right) = \mu \left( {\{ a, c\} }\right) = 0. \]\n\nTherefore, \( f = g \) a.e. \( \left( \mu \right) \) . | Yes |
Proposition 4.2.10. Let \( \left( {X,\mathcal{B},\mu }\right) \) be a complete measure space. Suppose \( f \) and \( g \) are two functions defined on \( X \) . If \( f \) is measurable and \( f = g \) a.e. \( \left( \mu \right) \), then \( g \) is measurable. | Proof. The proof of this proposition is exactly the same as the proof of Proposition 2.1.9. Let \( Z = \{ x \in X \mid f\left( x\right) \neq g\left( x\right) \} \) . Then \( \mu \left( Z\right) = 0 \) . Since \( \left( {X,\mathcal{B},\mu }\right) \) is a complete measure space, every subset of \( Z \) is measurable (an... | Yes |
Theorem 4.2.12. Let \( \left( {X,\mathcal{B},\mu }\right) \) be a measure space. Given a nonnegative measurable function \( f : X \rightarrow \mathbb{R} \cup \{ + \infty \} \) there exists a sequence of nonnegative simple functions \( \left\{ {\phi }_{n}\right\} \) such that\n\n\[ \mathop{\lim }\limits_{{n \rightarrow ... | Proof. For each \( n \) and \( 1 \leq k \leq {2}^{2n} - 1 \) set\n\n\[ {E}_{k}^{n} = \left\{ {x \in X\left| {\;\frac{k}{{2}^{n}} \leq f\left( x\right) < \frac{k + 1}{{2}^{n}}}\right. }\right\} \]\n\nand set\n\n\[ {E}_{{2}^{2n}}^{n} = \left\{ {x \in X \mid f\left( x\right) \geq {2}^{n}}\right\} . \]\n\nSince \( f \) is ... | Yes |
Corollary 4.2.13. Let \( \\left( {X,\\mathcal{B},\\mu }\\right) \) be a measure space. For any measurable function \( f \), there is a sequence of simple functions \( \\left\{ {\\phi }_{n}\\right\} \) such that\n\n\[ \n\\mathop{\\lim }\\limits_{{n \\rightarrow \\infty }}{\\phi }_{n}\\left( x\\right) = f\\left( x\\right... | Proof. Apply the previous theorem to the positive and negative parts of \( f \) . | No |
We will determine when \( f \) is integrable and, if so, \( {\int }_{\mathbb{R}}{fd\mu } \) . | Suppose \( \phi \left( x\right) \) is a simple function with \( 0 \leq \phi \left( x\right) \leq f\left( x\right) \) for all \( x \in \mathbb{R} \), say,\n\n\[ \phi \left( x\right) = \mathop{\sum }\limits_{{i = 1}}^{n}{a}_{i}{\mathcal{X}}_{{E}_{i}}\left( x\right) \]\n\nwhere \( {a}_{i} \geq 0 \) . Then\n\n\[ {\int }_{\... | No |
Proposition 4.3.5. Let \( f \) and \( g \) be two functions that are integrable with respect to \( \mu \) . If \( f\left( x\right) = g\left( x\right) \) a.e. \( \left( \mu \right) \), then\n\n\[ \int {fd\mu } = \int {gd\mu } \] | Proof. We will first prove this in the case where both \( f \) and \( g \) are nonnegative. Let \( Z = \{ x \in X \mid f\left( x\right) \neq g\left( x\right) \} \) . Then \( \mu \left( Z\right) = 0 \) . Let \( \phi \) be a simple function with \( 0 \leq \phi \leq f \), say\n\n\[ \phi \left( x\right) = \mathop{\sum }\li... | Yes |
Proposition 4.3.6. Let \( f \) and \( g \) be two functions that are integrable with respect to \( \mu \) . If \( 0 \leq f\left( x\right) \leq g\left( x\right) \) a.e. \( \left( \mu \right) \), then\n\n\[ \int {fd\mu } \leq \int {gd\mu } \] | ## Proof. This is Exercise 15, | No |
Corollary 4.3.8 (Fatou’s Lemma). Let \( \left\{ {f}_{n}\right\} \) be sequence of measurable nonnegative functions on the complete measure space \( \left( {X,\mathcal{B},\mu }\right) \) and \( f \) a nonnegative function with \( \mathop{\lim }\limits_{{n \rightarrow \infty }}{f}_{n}\left( x\right) = f\left( x\right) \)... | \[ \int {fd\mu } \leq \mathop{\liminf }\limits_{{n \rightarrow \infty }}\int {f}_{n}{d\mu } \] | Yes |
Theorem 4.3.9 (Monotone Convergence Theorem). Let \( \\left\\{ {f}_{n}\\right\\} \) be a sequence of nonnegative measurable functions with \( {f}_{n}\\left( x\\right) \\leq \) \( {f}_{n + 1}\\left( x\\right) \) a.e. \( \\left( \\mu \\right) \) for every \( n \) . Suppose \( \\mathop{\\lim }\\limits_{{n \\rightarrow \\i... | Proof. By Fatou's Lemma, Corollary 4.3.8,\n\n\[ \n\\int {fd\\mu } \\leq \\mathop{\\liminf }\\limits_{{n \\rightarrow \\infty }}\\int {f}_{n}{d\\mu } \n\]\n\nBy the definitions of lim inf and lim sup,\n\n\[ \n\\mathop{\\liminf }\\limits_{{n \\rightarrow \\infty }}\\int {f}_{n}{d\\mu } \\leq \\mathop{\\limsup }\\limits_{... | Yes |
Proposition 4.3.10. Let \( f \) and \( g \) be nonnegative measurable functions. For any nonnegative numbers \( a \) and \( b \) , \[ \int \left( {{af} + {bg}}\right) {d\mu } = a\int {fd\mu } + b\int {gd\mu }. \] | Proof. Although we will not explicitly show it here, the result is true if both \( f \) and \( g \) are simple functions. This is because the linear combination of simple functions is again a simple function.\n\nAssuming this, we will prove this in the case of more general nonnegative measurable functions.\n\nBy Theore... | No |
Corollary 4.3.11. Let \( \left\{ {f}_{n}\right\} \) be a sequence of nonnegative measurable functions. Then\n\n\[ \int \left( {\mathop{\sum }\limits_{{n = 1}}^{\infty }{f}_{n}}\right) {d\mu } = \mathop{\sum }\limits_{{n = 1}}^{\infty }\left( {\int {f}_{n}{d\mu }}\right) \] | Proof. By Theorem 4.2.7 \( \mathop{\sum }\limits_{{n = 1}}^{\infty }{f}_{n} = \mathop{\lim }\limits_{{N \rightarrow \infty }}\mathop{\sum }\limits_{{n = 1}}^{N}{f}_{n} \) is a measurable function. By Proposition 4.3.10 and the Monotone Convergence Theorem (Theorem 4.3.9),\n\n\[ \mathop{\sum }\limits_{{n = 1}}^{\infty }... | Yes |
Theorem 4.3.12 (Lebesgue Dominated Convergence Theorem). Let \( \\left\\{ {f}_{n}\\right\\} \) be a sequence of measurable functions such that \( \\mathop{\\lim }\\limits_{{n \\rightarrow \\infty }}{f}_{n}\\left( x\\right) = f\\left( x\\right) \) a.e. \( \\left( \\mu \\right) \) . Suppose there exists a \( \\mu \) -int... | Proof. Since \( \\left| {{f}_{n}\\left( x\\right) }\\right| \\leq g\\left( x\\right) \) a.e. \( \\left( \\mu \\right) \) for every \( n,\\left| {f\\left( x\\right) }\\right| \\leq g\\left( x\\right) \) a.e. \( \\left( \\mu \\right) \) . Hence, by Exercise 17 \( {f}_{n} \) is \( \\mu \) -integrable for every \( n \) and... | No |
Theorem 4.4.5. Let \( {\mu }^{ * } \) be an outer measure on \( X \) and \( \mathcal{B} \) be the collection of all \( {\mu }^{ * } \) -measurable sets. Define \( \mu : \mathcal{B} \rightarrow \left\lbrack {0,\infty }\right\rbrack \) by \( \mu \left( E\right) = \) \( {\mu }^{ * }\left( E\right) \) . Then \( \left( {X,\... | Proof. We need to show that \( \mu \) is a measure. Since \( \mathcal{B} \) is a \( \sigma \) -algebra and \( {\mu }^{ * } \) is an outer measure, \( \varnothing \in \mathcal{B} \) and\n\n\[ \mu \left( \varnothing \right) = {\mu }^{ * }\left( \varnothing \right) = 0. \]\n\nNext, suppose \( \left\{ {E}_{j}\right\} \) is... | Yes |
Theorem 4.4.6. Let \( E \subseteq {\mathbb{R}}^{n} \) . \( E \) is Lebesgue measurable if and only if for any \( A \subseteq {\mathbb{R}}^{n} \) , \[ {m}^{ * }\left( A\right) = {m}^{ * }\left( {A \cap E}\right) + {m}^{ * }\left( {A \smallsetminus E}\right) , \] where \( {m}^{ * } \) is Lebesgue outer measure. | Proof. Assume first that \( E \) is Lebesgue measurable. Let \( A \) be a subset of \( {\mathbb{R}}^{n} \) . It will always be the case that \[ {m}^{ * }\left( A\right) \leq {m}^{ * }\left( {A \cap E}\right) + {m}^{ * }\left( {A \smallsetminus E}\right) . \] Our task is to establish equality, or at least, establish the... | No |
Proposition 4.4.8. If \( 0 \leq \alpha < \beta \) and \( {H}_{\alpha }\left( E\right) < + \infty \) for some \( E \subseteq {\mathbb{R}}^{n} \), then \( {H}_{\beta }\left( E\right) = 0 \) . | Proof. Let \( \epsilon > 0 \) be given and suppose \( \left\{ {A}_{k}\right\} \) is a covering of \( E \) by sets with diameter less than \( \epsilon \) . Then\n\n\[ \mathop{\sum }\limits_{k}{\left( \delta \left( {A}_{k}\right) \right) }^{\beta } = \mathop{\sum }\limits_{k}{\left( \delta \left( {A}_{k}\right) \right) }... | Yes |
In \( {\mathbb{R}}^{2} \), let \( A \) be a line segment of length \( l \). \( A \) is a Borel set, and so is Hausdorff measurable. Given \( \epsilon > 0 \), we need roughly \( \frac{l}{\epsilon } \) balls of radius \( \epsilon \) to cover \( A \). For each \( \alpha \), a candidate for \( {H}_{\alpha }^{\epsilon }\lef... | Consequently, \( {H}_{\alpha }^{\epsilon }\left( A\right) = + \infty \) if \( \alpha < 1,{H}_{\alpha }^{\epsilon }\left( A\right) = 0 \) if \( \alpha > 1 \), and \( {H}_{1}^{\epsilon }\left( A\right) = l \). | Yes |
Example 4.5.2. Let \( f \in \mathcal{L}\left\lbrack {a, b}\right\rbrack \) and \( \mathcal{B} \) be the set of all Lebesgue measurable subsets of \( \left\lbrack {a, b}\right\rbrack \) . For any set \( E \in \mathcal{B} \) define \( \nu \left( E\right) \) by\n\n\[ \nu \left( E\right) = {\int }_{E}f \]\n\n\( \nu \) is a... | Verification that \( \nu \) is a signed measure is left as an exercise (see Exercise 26). | No |
Lemma 4.5.5. Let \( \nu \) be a signed measure on \( \left( {X,\mathcal{B}}\right) \). Then:\n\ni) Each measurable subset of a positive set is itself positive.\n\nii) The countable union of positive sets is a positive set. | Proof. Part i) follows directly from the definition of a positive set.\n\nTo show ii), assume \( \left\{ {A}_{j}\right\} \) is a countable collection of positive sets and suppose \( E \) is a measurable set with\n\n\[ E \subseteq \mathop{\bigcup }\limits_{j}{A}_{j} \]\n\nSet \( {E}_{1} = E \cap {A}_{1} \). For each \( ... | Yes |
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