Measure Theory Problem

**everk** · October 5th 2010, 07:43 PM

Hi everyone
I cannot find the solution to this problem,
Let A a Lebesgue Measurable Set; Prove that given $0 \leq b \leq m(A)$ then exists $B \subseteq A$ Lebesgue Measurable such that m(B) = b

I appreciate your help
Thanks
everk.

**Opalg** · October 6th 2010, 10:13 AM

Originally Posted by everk

Let A a Lebesgue Measurable Set; Prove that given $0 \leq b \leq m(A)$ then exists $B \subseteq A$ Lebesgue Measurable such that m(B) = b

Let ${\mathstrut\chi}_A$ denote the characteristic function of $A$ (so that ${\chi\mathstrut}_A(t) =1$ if $t\in A$ , and ${\mathstrut\chi}_A(t) = 0$ if $t\notin A$ ). For each real number $x$ , let $\displaystyle f(x) = \int_{-\infty}^x\!\!{\mathstrut\chi}_A(t)\,dt = m(A\cap(-\infty,x]).$ Show that $f$ is continuous and use the intermediate value theorem to deduce that there exists $c$ such that $f(c)=b.$

**Bruno J.** · October 6th 2010, 11:36 AM

Does that work if $A=(-\infty, 0)$ ?

**Opalg** · October 6th 2010, 12:10 PM

Originally Posted by Bruno J.

Does that work if $A=(-\infty, 0)$ ?

It will certainly need a bit of adjustment if $m(A) = \infty$ . I think that you would then have to start by looking at $\displaystyle\int_x^X\!\!{\mathstrut\chi}_A(t)\,dt$ , which can be made arbitrarily large if $x$ is small enough and $X$ is large enough. Then reduce $X$ until the integral reaches the value $b$ .

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