Lebesgue measure: $\lim_{t\to 0} m((B+t)\cap B)=m(B) $?

Question

Here is a question. If $B$ is a set in $\mathbb{R}$ of finite measure, do we have that:

$$\lim_{t\to 0} m(B_t∩B)=m(B),$$

here $B_t$ denotes the t-translated set: $B_t= B+t$ and $t$ is a real number.

I tried to use the dominated convergence theorem on $X_{B_t\cap B}<X_B$, but then we don't have the point-wise convergence. So I guess we must go back to the basic theory of measurable sets to solve it ?

Answer 1

I had the right idea to use the Lebesgue convergence theorem. I was not able to show the point-wise converge because of the sets were originally arbitrary. However, by assuming first that $B$ is open, we can show that indeed we have the point-wise convergence, hence the result is true for open intervals. Once we have that, we use the original definition of Lebesgue measure to find open intervals that approximate our general measurable set $B$ and then prove the result in general.

Answer 2

Hint: for each $\varepsilon > 0$ there is a finite union of open intervals $I = I_1 \cup I_2 \cup \ldots \cup I_n$ such that $\mu( B \mathbin{\Delta} I ) \leqslant \varepsilon$. So first show that the claim holds for intervals, then for finite unions of intervals, then use the above approximation.