| Line 1: | Line 1: | ||
'''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)''' | + | '''(a)''' <math> F(E)=\{y: \exist ~x \in E </math> with <math> ~y=F(x)\} </math> is measurable. |
'''Proof.''' | '''Proof.''' | ||
| − | '''(b)''' | + | Let <math>\int_{0}^{1}|f(t)|dt=M<\infty </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 \{leq} | ||
| + | |||
| + | '''(b)''' <math> m(F(E)) \leq \int_{E}|f(t)| dt </math>. | ||
'''Proof.''' | '''Proof.''' | ||
Revision as of 19:28, 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 \{leq} '''(b)''' <math> m(F(E)) \leq \int_{E}|f(t)| dt $.
Proof.
