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?