Revision as of 19:30, 21 July 2008 by Kim598 (Talk)

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} (\int_{0}^{1}|f(t)|dt)~||\chi_{[x,y]}||_{\infty} = M|x-y| $

(b) $ m(F(E)) \leq \int_{E}|f(t)| dt $.

Proof.

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