(New page: 9.9. Let <math>f \in L^{1}([0,1])</math> and let <math> F(x)=\int_{0}^{x}f(t)dt </math>. If <math>E </math> is a measurable subset of <math>[0,1] </math>, show that (a) <math> F(E)=\{y...) |
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| − | 9.9. Let <math>f \in L^{1}([0,1])</math> and let <math> F(x)=\int_{0}^{x}f(t)dt </math>. If <math>E </math> is a measurable subset of <math>[0,1] </math>, show that | + | '''9.9.''' Let <math>f \in L^{1}([0,1])</math> and let <math> F(x)=\int_{0}^{x}f(t)dt </math>. If <math> ~E </math> is a measurable subset of <math>~[0,1] </math>, show that |
| − | (a) <math> F(E)=\{y: \exist x \in E | + | '''(a)''' <math>~~ F(E)=\{y: \exist ~x \in E </math> with <math> ~y=F(x)\} </math> is measurable. |
| − | (b) <math> m(F(E)) \leq \int_{E}|f(t)| dt </math>. | + | '''Proof.''' |
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| + | '''(b)''' <math>~~ m(F(E)) \leq \int_{E}|f(t)| dt </math>. | ||
| + | |||
| + | '''Proof.''' | ||
Revision as of 18:24, 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.
(b) $ ~~ m(F(E)) \leq \int_{E}|f(t)| dt $.
Proof.
