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| − | '''(Step 3)''' <math>~E is a G_{\delta}-set</math> | + | '''(Step 3)''' <math>~E</math> is a <math>~G_{\delta}-set</math> |
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} \subseteq G_{2} \subseteq ... </math>. Then, | 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} \subseteq G_{2} \subseteq ... </math>. Then, | ||
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| − | '''(Step 4)''' <math>~E is a measurable set </math> | + | '''(Step 4)''' <math>~E</math> is a measurable set </math> |
From a characterization of measurability, <math>E=H-Z</math>, where H is a <math>G_{\delta}</math> set and </math>m(Z)=0 </math>. Then, | From a characterization of measurability, <math>E=H-Z</math>, where H 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{\rm (Step 3)}{\leq} \int_{H}|f(t)|dt \leq \int_{E}|f(t)|dt </math> | <math>m(F(E))=m(F(H-Z)) \leq m(F(H)) \stackrel{\rm (Step 3)}{\leq} \int_{H}|f(t)|dt \leq \int_{E}|f(t)|dt </math> | ||
Revision as of 20:05, 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) $
First assume that $ f \geq 0 $. Since $ ~F $ is monotone and continuous, $ m(F(E))=F(b)-F(a)=\int_{a}^{b}f(t) dt = \int_{E}|f(t)| dt $.
In general, $ m(F(E)) \leq \int_{a}^{b}f^{+}(t) dt + \int_{a}^{b} f^{-}(t)dt =\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{\rm (Step1)}{\leq} \sum_{n=1}^{\infty} \int_{I_{n}}|f(t)| dt= \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} \subseteq G_{2} \subseteq ... $. 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})) $
$ \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 </math>
From a characterization of measurability, $ E=H-Z $, where H is a $ G_{\delta} $ set and </math>m(Z)=0 </math>. Then,
$ m(F(E))=m(F(H-Z)) \leq m(F(H)) \stackrel{\rm (Step 3)}{\leq} \int_{H}|f(t)|dt \leq \int_{E}|f(t)|dt $
