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Since <math> ~F</math> is monotone and continuous, <math> m(F(E))=|F(b)-F(a)|=\left|\int_{a}^{b}f(t) dt \right| \leq \int_{E}|f(t)| dt </math>. | Since <math> ~F</math> is monotone and continuous, <math> m(F(E))=|F(b)-F(a)|=\left|\int_{a}^{b}f(t) dt \right| \leq \int_{E}|f(t)| dt </math>. | ||
| − | In general, <math> m(F(E)) \leq \int_{a}^{b}f^{+}(t) dt + \int_{a}^{b} f^{-}(t)dt =\int_{a}^{b}|f(t)|dt </math>. | + | In general, <math> m(F(E)) \leq \int_{a}^{b}f^{+}(t) dt + \int_{a}^{b} f^{-}(t)dt =\int_{a}^{b}|f(t)|dt </math>=\int_{E}|f(t)|dt </math>. |
'''(Step 2)''' <math>~E</math> : open set | '''(Step 2)''' <math>~E</math> : open set | ||
Revision as of 19:11, 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 </math>.
(Step 2) $ ~E $ : open set
