Steinhaus-like problem

Question

I know there are similar problems on here, but I believe this is not a duplicate.

Let $E \subset \mathbb{R}$ be a measurable set of positive finite measure. Define $f:[0,\infty) \rightarrow \mathbb R$ by $$f(t)= m(E \cap E_t),$$ where $E_t=\{t+x:x\in E\}$. Prove that $f$ is continuous on $[0,\infty)$.

I wanted to rewrite $f$ as a convolution of two sufficiently nice functions (in this case $L^1$ and $L^\infty$) which we know to be continuous:

$$f(t)=\int_{E\cap E_t}1 dx= \int_E 1_{E_t} dx= \int_E 1_{E}(x-t) dx=\int_{\mathbb R} 1_{-E}(t-x) 1_E dx= 1_{-E}*1_E(t), $$

and $1_{-E}$ is $L^1$ and $1_E$ is $L^\infty$.

Alternative solutions (assuming this actually is one) are welcome, too.

Also, a good reference for convolution results like the one used here would be much appreciated.

Answer 1

\begin{align} |f(t+\Delta t)-f(t)|&=\left|\int_{E\cap E_{t+\Delta t}}1 dx-\int_{E\cap E_{t}}1dx\right| \\ &= \left|\int_E 1_{E_{t+\Delta t}}dx-\int_E 1_{E_{t}}dx\right| \\ &= \left|\int_E (1_{E}(x-t-\Delta t)-1_{E}(x-t))dx\right| \\ &=\left|\int_E 1_{E_{\Delta t}}dx\right| \\ &\leqslant m(E)|\Delta t| \\ &\to 0 \hspace{5mm} as \hspace{5mm}t\to0 \end{align} So $f$ is continuous.

Answer 2

For any $s\in\mathbb R$, $E\cap E_s\subset E$, so $\chi_{E\cap E_s}\leqslant \chi_E$, and $$\int \chi_E\mathsf dm = m(E)<\infty. $$ If $t_n\to t$, then $\chi_{E\cap E_{t_n}}\to\chi_{E\cap E_t}$, so by dominated convergence, \begin{align} \lim_{n\to\infty}f(t_n) =& \lim_{n\to\infty} \int \chi_{E\cap E_{t_n}} \mathsf dm\\ &= \int \lim_{n\to\infty}\chi_{E\cap E_{t_n}}\mathsf dm\\ &= \int \chi_{E\cap E_t}\mathsf dm\\ &= m(E\cap E_t)\\ &= f(t), \end{align} which shows that $f$ is continuous.