Measures of translates of an open set Suppose $\mu$ is a positive Borel measure on $\mathbb{R}^n$ and $V$ is an open set. Define the map $\varphi\colon \mathbb{R}^n\to [0,\infty]$ by $\varphi(x)=\mu(x+V)$, where $x+V$ is the translation of $V$ by $x$.
I am told that $\varphi$ is lower semi-continuous, provided that $\mu$ is finite. I have produced a proof of this, but my proof doesn't seem to use the finiteness assumption. This goes roughly as follows: suppose $x_0\in\varphi^{-1}((\alpha,\infty))$ isn't an interior point. This gives a sequence $y_n$, converging to $x_0$, with $\varphi(y_n)\leq\alpha$. Since $V$ is open, each $x\in x_0+V$ lies in the sets $y_n+V$ from some point onward. Write $x_0+V$ as an increasing union, based on the index at which points land in the sequence $y_n+V$. But then $\varphi(x_0)$ is the limit of a sequence, bounded above by $\alpha$. Contradiction.
I have spent some time looking for an oversight on my part in the proof, but I can't find one. Could someone please confirm that the finiteness assumption is unnecessary, or alternatively, point to a flaw in my proof or give a counterexample?

For the sake of having an answer, I'll note that Mike's comment about Fatou's lemma works.
Take a sequence $x_n \to x$.  Note that $\liminf 1_{V + x_n}(y) \ge 1_{V + x}(y)$ for every $y$.  In other words, if $y \in V + x$, then $y \in V + x_n$ for all but finitely many $n$.  The converse need not hold, so the inequality can be strict for some $y$, but that is okay.
Now Fatou's lemma says
$$\liminf \int 1_{V + x_n} \ge \int \liminf\, 1_{V + x_n} \ge \int 1_{V + x}$$
which is to say $\liminf \varphi(x_n) \ge \varphi(x)$.
This does not require any assumptions on the measure, except that it be positive.
