1. ## 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.

2. 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 $f(c)=b.$

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

4. 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 $b$.