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'''(Step 2)''' <math>~E</math> : open set | '''(Step 2)''' <math>~E</math> : open set | ||
| − | We can write <math>E=\bigcup_{n=1}^{\infty}I_{n}</math> when <math> I_{1} \ | + | We can write <math>E=\bigcup_{n=1}^{\infty}I_{n} </math> when <math> I_{n}'s </math> are open intervals, so that |
| + | |||
| + | <math> 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})) </math> \leq \sum_{n=1}^{\infty} \int_{I_{n}}|f(t)| dt= \int_{E}|f(t)|dt </math> | ||
Revision as of 19:33, 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)|=\left|\int_{a}^{b}f(t) dt \right| \leq \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
We can write $ E=\bigcup_{n=1}^{\infty}I_{n} $ when $ I_{n}'s $ are 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 </math>
