Show that there is a measurable subset A such that $\mu(A) = \frac{1}{2}\mu(E)$

Problem Statement: If E is a Lebesgue measurable subset of $[0,1]$ and $\mu$ Lebesgue measure. Then there is a measurable subset $A\subset E$ such that $\mu(A) = \frac{1}{2} \mu(E)$.

I have a proof that there is a measurable subset of A of $\mathbb{R}$ such that for each open interval $(a,b)$ we have that $\mu(A \cap (a,b)) = \frac{1}{2}a-b$. I'm not sure how to use this to help me though. Any tip on how to begin would be appreciated.

1 Answer

Hint: define $f(x)=\mu(E\cap[0,x])$. This function is continuous (why?) and has $f(0)=0$ and $f(1)=\mu(E)$. Then apply the intermediate value theorem.

• I'm sure my reasoning for why the function is continuous is wrong, but I said its because the Lebesgue measure is defined as the inf of the sum of the length of the intervals that cover E. The length function is continuous, and the sum of continuous functions is continuous, therefore f(x) is continuous. Is this the right direction (and just needs to be formalized) or am I off the mark? Sep 26, 2016 at 2:27
• @BenRay I would argue like this. Let $\varepsilon>0$ and consider $x_0\in [0,1]$. Let $\delta=\varepsilon$. Then if $|x-x_0|<\delta$ (WLOG let $x<x_0$) then $|f(x)-f(x_0)|=|\mu(E\cap[0,x])-\mu(E\cap[0,x_0])|=\mu(E\cap[x,x_0])\leq \mu([x,x_0])<\delta=\varepsilon$.
– user223391
Sep 26, 2016 at 2:31
• Ah, of course. Thank you for your help!. Sep 26, 2016 at 2:33