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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?

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By the way, what do you mean by an infinite measure? Is it one that assigns infinite measure to some sets? For example, like Lebesgue measure on the line. – Patrick Mar 7 '12 at 22:45
@Patrick Yes, by an infinite measure I mean a measure that assigns the whole of $\mathbb{R}^n$ infinite measure. Try saying that three times fast. – Miha Habič Mar 7 '12 at 22:50
Your argument should begin with $x_0\in\varphi^{-1}((\alpha,\infty])$, but I think it is OK for $\mathbb R^n$. Maybe the assertion you saw was for some more general setting? – GEdgar May 7 '12 at 21:48
@GEdgar You're right, $x_0+V$ could have infinite measure. I don't think it matters in any case, since the boundary points of the upper level sets have low $\varphi$-value. – Miha Habič May 8 '12 at 6:14
The poser shows that the indicator functions converge ptwise, and then he could quote Fatou, which does not require finiteness. – mike May 10 '12 at 16:57
up vote 5 down vote accepted

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.

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Thanks to Mike and Nate for this solution. It's very elegant; I admit Fatou's lemma hadn't crossed my mind, but then again I don't recall ever using it for anything other than proving dominated convergence. – Miha Habič May 14 '12 at 15:48

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