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{(Step 1)}{\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} \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 -
