A continuous function defined by Lebesgue measure Let $A,B\in\mathcal A_{\Bbb R}^*$ given with $\overline{\lambda}(A)<\infty$ and $\overline{\lambda}(B)<\infty$. Lets define $\; \overline{\lambda}_{A,B}:\Bbb R\to\Bbb R$ as follows:
$$\overline{\lambda}_{A,B}(x)=\overline{\lambda}(A\cap(B+x))$$
where $B+x=\{b+x:b\in B\}$. 
So what I want to prove is that $\overline{\lambda}_{A,B}$ is continuous.
Proof:
Let $c\in\Bbb R$ fixed and let $x_n=c-\frac{1}{n}\;\forall n\in\Bbb N$. Clearly $x_n\in\Bbb R\;\forall n\in\Bbb N$.
Then, let $B_n=B+x_n\;\forall n\in\Bbb N\Rightarrow\ B_n=\{b+c-\frac{1}{n}:b\in B\}\;\forall n\in\Bbb N$

Lemma 1: $$B_n\subseteq B_{n+1}\;\forall n\in\Bbb N$$

Let $y\in B_n\Rightarrow\ y=b+c-\frac{1}{n}=b+c-\big(\frac{1}{n(n+1)}+\frac{1}{n+1}\big)=b+c-\frac{1}{n(n+1)}-\frac{1}{n+1}\;\forall n\in\Bbb N$. Thus $ y=b+c'-\frac{1}{n+1}\;\forall n\in\Bbb N$ with $c'=c-\frac{1}{n(n+1)}\Rightarrow\ y\in B_{n+1}$.
Lemma 2: $$\lim_{n\to\infty}(A\cap B_n)=A\cap (B+c)$$

Since $B_n\subseteq B_{n+1}\;\forall n\in\Bbb N$ by Lemma 1, the limit of $(B_n)_{n\in\Bbb N}$ exists and $\lim_{n\to\infty}(B_n)=\bigcup_{n=1}^\infty B_n,\ $but $\bigcup_{n=1}^\infty B_n=\bigcup_{n=1}^\infty \{b+c-\frac{1}{n}:b\in B\}=\{b+c:b\in B\}=B+c$. Now, clearly $A\cap B_n\subseteq A\cap B_{n+1}\;\forall n\in\Bbb N\Rightarrow\ \lim_{n\to\infty}(A\cap B_n)=\bigcup_{n=1}^\infty (A\cap B_n)=A\cap\bigcup_{n=1}^\infty B_n=A\cap (B+c)$.

So by Lemma 2 we get that:
$$\lim_{n\to\infty}\overline{\lambda}_{A,B}(x_n)=\lim_{n\to\infty}\overline{\lambda}(A\cap(B+x_n))=\lim_{n\to\infty}\overline{\lambda}(A\cap B_n)=\overline{\lambda}\big(\lim_{n\to\infty}(A\cap B_n)\big)=\overline{\lambda}(A\cap (B+c))=\overline{\lambda}_{A,B}(c)$$
And clearly $\lim_{n\to\infty}x_n=c$, thus $\overline{\lambda}_{A,B}$ is continuous.
Did I miss something?
 A: So continuing with Nate River's hint, let $(x_n)_{n\in\Bbb N}$ be any sequence in $\Bbb R$ s.t. $\lim_{n\to\infty} x_n=c\;\ \forall c\in\Bbb R$ fixed.
Then, let $f_n=\chi_{A\cap (B+x_n)}\;\ \forall n\in\Bbb N\Rightarrow\ (f_n)_{n\in\Bbb N}\subset M(\Bbb R,\mathcal A_{\Bbb R}^*)\;\;\forall n\in\Bbb N$. And let $f=\chi_{A\cap (B+c)}\in M(\Bbb R,\mathcal A_{\Bbb R}^*)$

Lemma 1: $f_n\to f$ as $n\to\infty$
Let $E_n=A\cap (B+x_n)\;\;\forall n\in\Bbb N$. So, clearly $\; \underline{\lim}_{n\to\infty}(E_n)\subseteq \overline{\lim}_{n\to\infty}(E_n)$. Then, if $x\in\overline{\lim}_{n\to\infty}(E_n)$ we know that:
$$\Rightarrow\ x\in E_n\;\;\text{for infite values of $n$}\\
\Rightarrow\ x\in A\cap (B+x_n)\;\;\text{for infite values of $n$}\\
\Rightarrow\ x\in B+x_n\;\;\text{for infite values of $n$} \text{ and } x\in A\\
\Rightarrow\ \exists\ b\in B,\;\ x=b+x_n\;\;\text{for infite values of $n$} \text{ and } x\in A\\
\Rightarrow\ \exists\ b\in B,\;\ x=\lim_{n\to\infty}b+x_n=b+c \text{ and } x\in A\\
\Rightarrow\ x\in B+c \text{ and } x\in A$$
thus $\overline{\lim}_{n\to\infty}(E_n)\subseteq A\cap (B+c)$.
Now, if $x\in A\cap (B+c)$ we know that:
$$\Rightarrow\ \exists\ b\in B,\;\ x=b+c \text{ and } x\in A$$
and, since $\lim_{n\to\infty} x_n=c,$ given $\epsilon>0\ \exists\ n_0\in\Bbb N$ s.t. $\forall n\ge n_0\;\; x=b+x_n$
$$\Rightarrow\ \exists\ b\in B,\;\ x=b+x_n\;\forall n\ge n_0\;\;\text{and}\; x\in A\\
\Rightarrow\ x\in B+x_n\;\;\forall n\ge n_0\;\;\text{and}\; x\in A\\
\Rightarrow\ x\in A\cap (B+x_n)\;\;\forall n\ge n_0\\
\Rightarrow\ x\in E_n\;\;\forall n\ge n_0\Rightarrow\ x\in\underline{\lim}_{n\to\infty}(E_n)$$
thus $A\cap (B+c)\subseteq \underline{\lim}_{n\to\infty}(E_n)$.
Thereby $\underline{\lim}_{n\to\infty}(E_n)=\lim_{n\to\infty}(E_n)=\overline{\lim}_{n\to\infty}(E_n)\Rightarrow\ (\chi_{E_n})_{n\in\Bbb N}$ converges, where:
$$A\cap (B+c)\subseteq \underline{\lim}_{n\to\infty}(E_n)=\lim_{n\to\infty}(E_n)=\overline{\lim}_{n\to\infty}(E_n)\subseteq A\cap (B+c)\\
\text{so}\;\; \lim_{n\to\infty}(E_n)=A\cap (B+c)$$
and thus $(\chi_{E_n})_{n\in\Bbb N}$ converges to $\chi_{\lim_{n\to\infty}(E_n)}=\chi_{A\cap (B+c)}$. 
Thus $f_n=\chi_{A\cap(B+x_n)}\to\chi_{A\cap (B+c)}=f$ as $n\to\infty$

Now let $g=\chi_A$ so, since $A\cap (B+x_n)\subseteq A\;\;\forall n\in\Bbb N$, we get that $|f_n|\le g\;\;\forall n\in\Bbb N$ where: $\int gd\overline{\lambda}=\overline{\lambda}(A)<\infty\Rightarrow\ g\in\mathcal L_1(\overline{\lambda})$. So applying LDCT (Lebesgue Dominated Convergence Theorem) to $(f_n)_{n\in\Bbb N}$ we get that:
$$\lim_{n\to\infty}\int f_n d\overline{\lambda}=\int fd\overline{\lambda}\\
\Leftrightarrow\ \lim_{n\to\infty}\int \chi_{A\cap (B+x_n)} d\overline{\lambda}=\int \chi_{A\cap (B+c)}d\overline{\lambda}\\
\Leftrightarrow\ \lim_{n\to\infty} \overline{\lambda}(A\cap (B+x_n))=\overline{\lambda}(A\cap (B+c))\\
\Leftrightarrow\ \lim_{n\to\infty} \overline{\lambda}_{A,B}(x_n)=\overline{\lambda}_{A,B}(c)$$
thereby $\overline{\lambda}_{A,B}$ is continuous.
