Showing that the function is measurable. I have that $\mu$ is a $\sigma$-finite Borel measure on $\mathbb{R}$, and $E \in \mathcal{B}$. I need to show that the function $\mu(E-x)$ is Borel-measurable. That is, the inverse image of any open set of real numbers of this function is in $\mathcal{B}$.
If I have an open set O, and look at $\mu^{-1}(O)$, I get a collection: $L=\{K \in \mathcal{B}| \mu(K) \in O\}$. I also have that for any Borel-measurable set, we also have that we get a borel measurable set by translation. If I could show that $\cup_{K \in L}K=K'$ is a borel measurable set I would get a little further, but the problem is that the collection might be bigger than countable, and a sigma-algebra is not closed under arbitrary unions.
Do you guys have any hints?
 A: HINT: Let's do a simple case: $\mu( A) = \int_{A} f(t) dt$ for some bounded Borel function $f$ and $E=[0,1]$. We have 
$$x\mapsto \phi(x)\colon = \mu(E+x) = \int_x^{x+1} f(t) dt$$
If $f$ is continoous it is easy to show that $\phi(x)$ is continous ( in fact $C^1$). Now for every bounded Borel function $f$ there exists a sequence of continous functions $f_n{\longrightarrow } f$ ae and moreover $|f_n| \le M $ ( same constant bounding $f$). By dominated convergence theorem
$$\int_x^{x+1} f_n(t) dt \to \int_x^{x+1} f(t) dt$$
for every $x$. We may cover in this way the case of absolutely continous measures.
We'll now give a proof that works in general. Let $\mu$ a sigma-finite Borel measure on $\mathbb{R}$. Let $K$ a compact. Let's show that the function
$x\mapsto \mu(K+x)$ is upper semicontinous. Indeed, let $m > \mu(K)$. There exists an open subset containing $K$ such that $mu(U) < m$. Now, for $|x|$ small we'll have $K+x\subset U$. It follows that $\mu(K+x) < m$. It follows that the function is also Borel measurable. 
Let now $U$ open. There exists a sequence of compact subsets $K_n$ so that $K_n \subset U$ and $K_n$ exhaust $U$. It follows that the functions $x \mapsto \mu(K_n + x)$ converge pointwise to $x \mapsto \mu (U+x)$. Hence this function is also Borel measurable. 
Let's consider now the set of all Borels $E$ so that the corresponding function is Borel. This forms a Dynkin system (not hard to show). It follows that this set contains the sigma algebra generated by the open sets, thus all the Borels. 
