# Measure Theory Problem

• Oct 5th 2010, 08:43 PM
everk
Measure Theory Problem
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

Thanks
everk.
• Oct 6th 2010, 11:13 AM
Opalg
Quote:

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 $f(c)=b.$
• Oct 6th 2010, 12:36 PM
Bruno J.
Does that work if $A=(-\infty, 0)$?
• Oct 6th 2010, 01:10 PM
Opalg
Quote:

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 $b$.