$|T\cap E| = |T|$ for closed set $T$ and measurable set $E$

Question

Let $E \subset \mathbb{R}$ be a measurable set with $0 < |E| < \infty$. Does there exist a closed interval $T$ with positive measure such that $|T\cap E| = |T|$. ($|E|$ denotes the measure of the set $E$.)

Answer 1

In the following, I presume 'measure' refers to the Lebesgue measure. The questions asked in the title and body are different, and have correspondingly different answers. It is quite easy if you just want to find a closed set with this property: by inner regularity of the Lebesgue measure, $E$ has a compact subset $K$ with positive measure, therefore $0<|K|=|K\cap E|$ and you are done. If however you want to find a closed interval, then in general it cannot be done. The following is one such counterexample:

Let $K = \cap_n K_n$ be a fat Cantor set, where $K_0 = [0,1]$, $K_1 = [0,\frac{3}{8}] \cup [\frac{5}{8},1]$, and in general $K_{n+1}$ is formed by removing the middle $\frac{1}{4^{n+1}}$ interval of each of the $2^{n}$ subintervals of $K_n$. $K$ is compact hence measurable, and has measure $1-\sum_{n\geq 0} \frac{2^n}{4^{n+1}} = \frac{1}{2}$. If $A=[a,b]$ is a closed interval, then $A\cap K \subset A\cap K_n$ for each $n$. Convince yourself that for $n$ sufficiently large, there is some sub-interval of $K_n$ whose endpoints lie in $A$, in which case $K_{n+1}$ removes an interval in $A$, so $|A\cap K| \leq |A\cap K_{n+1}|<|A|$, hence $|A|\neq|A\cap K|$.