Is $f(x)=\mu(V+x)$ continuous? 
Suppose $V$ is open in $R^k$ and $\mu$ is a finite postive Borel measure on $R^k$. Is the function $f(x)=\mu(V+x)$ continuous? lower semicontinuous? upper semicontinuous?

Clearly $f$ cannot be guaranteed to be continuous or even upper semicontinuous. If we let $$\mu_{x_0}(E)=\begin{cases} 1 & x_0\in E \\ 0 & x_0\not\in E \end{cases}$$ and $V= B_r(x_0)$, we then find that it is strictly lower semicontinuous.  
But I am finding it hard to:


*

*Find an example where $f$ isn't lower semicontinuous.

*Find a proof that $f$ must be lower semicontinuous.


Any Ideas as to how .
 A: Lower-semicontinuity of $f$ can be thought as a consequence of the portmanteau theorem. Here is a self-contained proof:
Since $V$ is open, there exists a sequence $(\varphi_n)$ of compactly supported continuous functions on $\Bbb{R}^k$ such that $0 \leq \varphi_1 \leq \varphi_2 \leq \varphi_3 \leq \cdots$ and $\varphi_n \uparrow \mathbf{1}_V$ pointwise. From this, for any $x \in \Bbb{R}^k$ and for any $n = 1, 2, \cdots$ we have
$$ f(x)
= \mu(x + V)
= \int_{\Bbb{R}^k} \mathbf{1}_V (t-x) \, \mu(dt)
\geq \int_{\Bbb{R}^k} \varphi_n(t-x) \, \mu(dt). $$
Now for each fixed $n$, we find that $\varphi_n(\cdot - x)$ converges pointwise to $\varphi_n(\cdot - x_0)$ as $x \to x_0$. Thus by the bounded convergence theorem,
$$ \liminf_{x\to x_0} f(x)
\geq \liminf_{x\to x_0} \int_{\Bbb{R}^k} \varphi_n(t-x) \, \mu(dt)
= \int_{\Bbb{R}^k} \varphi_n(t-x_0) \, \mu(dt). $$
Taking $n \to \infty$ together with the monotone convergence theorem shows that
$$ \liminf_{x\to x_0} f(x) \geq f(x_0), $$
which implies lower-semicontinuity of $f$.
A: It can be shown that $x\mapsto\mu(x+V)$ is continuous for all open $V$ if and only if $\mu$ is absolutely continuous with respect to Lebesgue measure. See  A note on translation continuity of probability measures by S. Zabell.
