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'''9.9.''' Let <math>f \in L^{1}([0,1])</math> and let <math> F(x)=\int_{0}^{x}f(t)dt </math>. If <math> ~E </math> is a measurable subset of <math>~[0,1] </math>, show that | '''9.9.''' Let <math>f \in L^{1}([0,1])</math> and let <math> F(x)=\int_{0}^{x}f(t)dt </math>. If <math> ~E </math> is a measurable subset of <math>~[0,1] </math>, show that | ||
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'''(a)''' <math> F(E)=\{y: \exist ~x \in E </math> with <math> ~y=F(x)\} </math> is measurable. | '''(a)''' <math> F(E)=\{y: \exist ~x \in E </math> with <math> ~y=F(x)\} </math> is measurable. | ||
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'''Proof.''' | '''Proof.''' | ||
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'''(b)''' <math> m(F(E)) \leq \int_{E}|f(t)| dt </math>. | '''(b)''' <math> m(F(E)) \leq \int_{E}|f(t)| dt </math>. | ||
| − | + | ---- | |
'''Proof.''' | '''Proof.''' | ||
'''(Step 1)''' <math> ~E=(a,b) </math> | '''(Step 1)''' <math> ~E=(a,b) </math> | ||
| − | + | Let <math>~ f=f^{+}-f^{-} </math> and let <math> F^{+}(x)=\int_{0}^{x}f^{+}(t)dt, ~F^{-}(x)=\int_{0}^{x}f^{-}(t)dt </math>, so that <math>~F^{+}</math> and <math>~F^{-}</math> are increasing. | |
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| + | Then, <math>~F(x)=\int_{0}^{x}f(t)dt=F^{+}(x)-F^{-}(x)</math>, and <math>\int_{0}^{x}|f(t)|dt=F^{+}(x)+F^{-}(x) </math>. | ||
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| + | <math> m(F(E)) = \max_{x \in E}F(x)- \min_{x \in E}F(x) \leq \left( \max_{x \in E}F^{+}(x) - \min_{x \in E}F^{-}(t) \right)-\left( \min_{x \in E} F^{+}(x)-\max_{x \in E}F^{-}(x) \right) </math> | ||
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| + | <math>=(F^{+}(b)-F^{-}(a))-(F^{+}(a)-F^{-}(b))=(F^{+}(b)+F^{-}(b))-(F^{+}(a)+F^{-}(a))=\int_{a}^{b}|f(t)|dt =\int_{E}|f(t)|dt </math> | ||
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| + | '''(Step 2)''' <math>~E</math> : open set | ||
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| + | Let <math>E=\bigcup_{n=1}^{\infty}I_{n} </math> when <math> ~I_{n} </math>'s are disjoint open intervals, so that | ||
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| + | <math> m(F(E))=m(F(\bigcup_{n=1}^{\infty}I_{n}))=m(\bigcup_{n=1}^{\infty}F(I_{n})) \leq \sum_{n=1}^{\infty} m(F(I_{n})) \stackrel{(Step 1)}{\leq} \sum_{n=1}^{\infty} \int_{I_{n}}|f(t)| dt \stackrel{\rm LDCT}{=} \int_{E}|f(t)|dt </math> | ||
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| + | '''(Step 3)''' <math>~E</math> is a <math>~G_{\delta}</math>-set | ||
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| + | We can write <math> E= \bigcap_{n=1}^{\infty} G_{i}</math> when <math>~G_{n}</math>'s are nested open sets(<math>~G_{1} \supseteq G_{2} \supseteq ... </math>). Then, | ||
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| + | <math> m(F(E))=m(F(\bigcap_{n=1}^{\infty} G_{i})) \leq m(\bigcap_{n=1}^{\infty}F(G_{n}))=\lim_{n \to \infty}m(\bigcap_{i=1}^{n}F(G_{i}))=\lim_{n \to \infty}m(F(G_{n})) </math> | ||
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| + | <math>\stackrel{(Step 2)}{\leq} \lim_{n \to \infty} \int_{G_{n}}|f(t)|dt=\lim_{n \to \infty} \int_{0}^{1} |f(t)|\chi_{G_{n}}(t) dt \stackrel{\rm MCT} {=} \int_{0}^{1}|f(t)|\chi_{\bigcap_{n=1}^{\infty}G_{n}}(t)dt = \int_{0}^{1}|f(t)| \chi_{E}(t) dt = \int_{E}|f(t)| dt </math> | ||
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| − | + | '''(Step 4)''' <math>~E</math> is a measurable set. | |
| − | <math> | + | From a characterization of measurability, <math>~E=H-Z</math>, where <math>~H</math> is a <math>~G_{\delta}</math> set and <math>~m(Z)=0 </math>. Then, |
| + | <math>m(F(E))=m(F(H-Z)) \leq m(F(H)) \stackrel{(Step 3)}{\leq} \int_{H}|f(t)|dt = \int_{E}|f(t)|dt </math> | ||
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| − | + | - JIN - | |
Latest revision as of 22:10, 21 July 2008
9.9. Let $ f \in L^{1}([0,1]) $ and let $ F(x)=\int_{0}^{x}f(t)dt $. If $ ~E $ is a measurable subset of $ ~[0,1] $, show that
(a) $ F(E)=\{y: \exist ~x \in E $ with $ ~y=F(x)\} $ is measurable.
Proof.
Let $ \int_{0}^{1}|f(t)|dt=M<\infty $.
$ \forall ~ x,y \in [0,1] (x \leq y) $,
$ |F(y)-F(x)|=\int_{x}^{y}f(t)dt=\int_{0}^{1}f(t) \chi_{[x,y]}(t) dt~ \stackrel{\rm Holder} {\leq} ~\left(\int_{0}^{1}|f(t)|dt\right)~||\chi_{[x,y]}||_{\infty} = M|x-y| $.
Hence $ ~F $ is a Lipschitz map, which preserves measurability. This proves (a).
(b) $ m(F(E)) \leq \int_{E}|f(t)| dt $.
Proof.
(Step 1) $ ~E=(a,b) $
Let $ ~ f=f^{+}-f^{-} $ and let $ F^{+}(x)=\int_{0}^{x}f^{+}(t)dt, ~F^{-}(x)=\int_{0}^{x}f^{-}(t)dt $, so that $ ~F^{+} $ and $ ~F^{-} $ are increasing.
Then, $ ~F(x)=\int_{0}^{x}f(t)dt=F^{+}(x)-F^{-}(x) $, and $ \int_{0}^{x}|f(t)|dt=F^{+}(x)+F^{-}(x) $.
$ m(F(E)) = \max_{x \in E}F(x)- \min_{x \in E}F(x) \leq \left( \max_{x \in E}F^{+}(x) - \min_{x \in E}F^{-}(t) \right)-\left( \min_{x \in E} F^{+}(x)-\max_{x \in E}F^{-}(x) \right) $
$ =(F^{+}(b)-F^{-}(a))-(F^{+}(a)-F^{-}(b))=(F^{+}(b)+F^{-}(b))-(F^{+}(a)+F^{-}(a))=\int_{a}^{b}|f(t)|dt =\int_{E}|f(t)|dt $
(Step 2) $ ~E $ : open set
Let $ E=\bigcup_{n=1}^{\infty}I_{n} $ when $ ~I_{n} $'s are disjoint open intervals, so that
$ m(F(E))=m(F(\bigcup_{n=1}^{\infty}I_{n}))=m(\bigcup_{n=1}^{\infty}F(I_{n})) \leq \sum_{n=1}^{\infty} m(F(I_{n})) \stackrel{(Step 1)}{\leq} \sum_{n=1}^{\infty} \int_{I_{n}}|f(t)| dt \stackrel{\rm LDCT}{=} \int_{E}|f(t)|dt $
(Step 3) $ ~E $ is a $ ~G_{\delta} $-set
We can write $ E= \bigcap_{n=1}^{\infty} G_{i} $ when $ ~G_{n} $'s are nested open sets($ ~G_{1} \supseteq G_{2} \supseteq ... $). Then,
$ m(F(E))=m(F(\bigcap_{n=1}^{\infty} G_{i})) \leq m(\bigcap_{n=1}^{\infty}F(G_{n}))=\lim_{n \to \infty}m(\bigcap_{i=1}^{n}F(G_{i}))=\lim_{n \to \infty}m(F(G_{n})) $
$ \stackrel{(Step 2)}{\leq} \lim_{n \to \infty} \int_{G_{n}}|f(t)|dt=\lim_{n \to \infty} \int_{0}^{1} |f(t)|\chi_{G_{n}}(t) dt \stackrel{\rm MCT} {=} \int_{0}^{1}|f(t)|\chi_{\bigcap_{n=1}^{\infty}G_{n}}(t)dt = \int_{0}^{1}|f(t)| \chi_{E}(t) dt = \int_{E}|f(t)| dt $
(Step 4) $ ~E $ is a measurable set.
From a characterization of measurability, $ ~E=H-Z $, where $ ~H $ is a $ ~G_{\delta} $ set and $ ~m(Z)=0 $. Then,
$ m(F(E))=m(F(H-Z)) \leq m(F(H)) \stackrel{(Step 3)}{\leq} \int_{H}|f(t)|dt = \int_{E}|f(t)|dt $
- JIN -
