proving that the function $f(x) = \lambda(S ∩ (S + x))$ is a continuous and $\lim_{x\to\infty} f(x) = 0$ Let $S \subset \mathbb{R}$ be a Borel measurable subset with finite Lebesgue
measure. Show that the function $f(x) = \lambda(S ∩ (S + x))$ is a continuous
function of $x$ and that $\lim_{x\to\infty} f(x) = 0$.
$\textbf{My attempt:}$ I tried this but not sure and stuck (though I think in general I am in the right direction).
I have to show that $\forall \epsilon>0, \forall x\in S$, $\exists \delta>0$ s.t. if $|x-x_n|<\delta$ , then $|f(x)-f(x_n)|<\epsilon$. 
So; assume that $|x-x_n|<\delta$, fix $x\in S$. I can write
\begin{align}
f(x) 
& = \lambda(S ∩ (S + x)) =\int_\mathbb{R} \chi_{S ∩ (S + x)}(t)d\lambda(t)=\int_\mathbb{R} \chi_{S}(t) \chi_{(S + x)}(t)d\lambda(t)\\
& = \int_\mathbb{R} \chi_{S}(t) \chi_{S}(t-x)d\lambda(t)
\end{align}
so 
\begin{align}
|f(x)-f(x_n)|
& =|\int_\mathbb{R} \chi_{S}(t) \chi_{S}(t-x)d\lambda(t)-\int_\mathbb{R} \chi_{S}(y) \chi_{S}(y-x_n)d\lambda(y)|\\
& = ...
\end{align}
for the second part;
\begin{align}
\lim_{x\to\infty}f(x)
& = \lim_{x\to\infty}\int_\mathbb{R} \chi_{S}(t) \chi_{S}(t-x)d\lambda(t)\\
& =\int_\mathbb{R} \chi_{S}(t) \lim_{x\to\infty}\chi_{S}(t-x)d\lambda(t) ~~~~ {DCT}\\
& = 0 ~~~~~~ \text{since} ~~~~ \chi_{S}(t-x)\to 0 ~~~~ \text{as} ~~~~x\to\infty
\end{align}
 A: There is a useful fact: For $\varphi\in L^{1}$, then $\displaystyle\int|\varphi(x+t)-\varphi(t)|dt\rightarrow 0$ as $x\rightarrow 0$.
First consider a compactly supported continuous function, apply Lebesgue Dominated Convergence Theorem to this continuous function, and then the general case follows by the density of all such functions in $L^{1}$.
The second part is not correct, for $\chi_{S}(t-x)$ need no converge to zero as $x\rightarrow\infty$. But this will do for bounded set $S$. To this end, choose a compact set $K$ such that $|S-K|$ is small, then
\begin{align*}
\int\chi_{S}(t)\chi_{S}(t-x)=\int\chi_{S}(t)\chi_{S}(t-x)-\chi_{K}(t)\chi_{K}(t-x)+\int\chi_{K}(t)\chi_{K}(t-x),
\end{align*}
the term $\displaystyle\int\chi_{K}(t)\chi_{K}(t-x)$ goes to zero by Lebesgue Dominated Convergence Theorem, and 
\begin{align*}
&\left|\int\chi_{S}(t)\chi_{S}(t-x)-\chi_{K}(t)\chi_{K}(t-x)\right|\\
&=\left|\int(\chi_{S}(t)-\chi_{K}(t))\chi_{S}(t-x)+\int\chi_{K}(t)(\chi_{S}(t-x)-\chi_{K}(t-x))\right|\\
&\leq|S-K|+\int\left|\chi_{S}(t-x)-\chi_{K}(t-x)\right|\\
&=2|S-K|,
\end{align*}
which is also small.
A: Instead of using the $\epsilon$-$\delta$ definition of continuity, show that if $x_n\rightarrow x$ then $f(x_n)\rightarrow f(x)$. Such a proof should be similar to the proof you gave for the second part.
A: Alternatively you can see $ f(t) = \lambda(S ∩ (S + x)) = 1_{S} * 1_{-S}(x) $ and use that the colvolution of 2 functions in $ L^2(\mathbb{R}) $ gives a $C_0(\mathbb{R}) $ function. 
