| Line 9: | Line 9: | ||
<math>\forall ~ x,y \in [0,1] (x \leq y)</math>, | <math>\forall ~ x,y \in [0,1] (x \leq y)</math>, | ||
| − | <math> |F(y)-F(x)|=\int_{x}^{y}f(t)dt=\int_{0}^{1}f(t) \chi_{[x,y]}(t) dt \stackrel{ | + | <math> |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| </math> |
'''(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.''' | ||
Revision as of 18:32, 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} (\int_{0}^{1}|f(t)|dt)~||\chi_{[x,y]}||_{\infty} = M|x-y| $
(b) $ m(F(E)) \leq \int_{E}|f(t)| dt $.
Proof.
