Difference between revisions of "Jets9.9 OldKiwi" - Rhea

Revision as of 20:44, 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})) \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->\infty}m(\bigcap_{i=1|^{n}F(G_{i})=\lim_{n->\infty}m(F(G_{n})) $