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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 .

2 Answers 2

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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 $f$.

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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.