# Prove that $x\mapsto \mu(V+x)$ is lower semicontinuous (Big Rudin, Problem $2.23$)

Problem $$2.23$$: Suppose $$V$$ is open in $$\mathbb R^k$$ and $$\mu$$ is a finite positive Borel measure on $$\mathbb R^k$$. Is the function $$f(x)=\mu(V+x)$$ continuous? lower semicontinuous? upper semicontinuous?

The problem above already has an answer here, which I do not understand quite well. Moreover, I'm primarily looking for help to complete my work on the above problem, or get solutions along the lines of what I have thought.

My work: I have already guessed that $$f$$ is lower semicontinuous. It is easy to see that $$f$$ need not be continuous or upper semicontinuous, by considering Dirac measures and appropriate open sets. Consider some $$V$$ and $$\mu$$ which satisfy the hypothesis. We want to show that $$A = \{x: f(x) > \alpha\}$$ is open for every real $$\alpha$$. Consider $$x\in A$$. It suffices to find $$r_x > 0$$ such that $$B(x,r_x) \subset A$$, where $$B(x,r_x)$$ denotes the open ball of radius $$r_x$$ centered at $$x$$. Since $$x\in A$$, $$\mu(V+x) > \alpha$$. Also, $$V+x$$ is open due to the openness of $$V$$. We can write $$V$$ as a disjoint, at most countable union, of open cubes $$\{Q_j\}$$ in $$\mathbb R^k$$, i.e. $$V + a = \bigcup_{j=1}^\infty Q_j$$ How do I find a required $$r_x$$?

Thank you!

Update: Thanks to the current answer, I have at least one solution that works - but I still need help in completing my attempt.

• I just read your attempt more carefully. If you want to proceed this way, I think you should use the outer regularity of Borel measures in $\mathbb{R}^d$. so that you can take points in an open set that intersects $V+x$; also you may need to consider cases where $\alpha\leq0$ and $\alpha>0$. In my opinion, in this problem it is easier to used results of integration such as dominated convergence of Fatou's Lemma. Commented Jun 17, 2021 at 1:55

Let $$g(t):=\mathbb{1}_V(t)$$. If $$V$$ is open, then $$\phi$$ is lower semicontinuous and so, for any $$x_n\rightarrow x$$ $$g(t-x)=\mathbb{1}_{V+x}(t)\leq\liminf_n\mathbb{1}_{V+x_n}(t)=\liminf_ng(t-x_n)$$

An application of Fatou's lemma gives $$\phi(x):=\int\mathbb{1}_{V+x}\,d\mu\leq \int\liminf_n\mathbb{1}_{V+x_n}\leq\liminf_n\int\mathbb{1}_{V+x_n}\,d\mu=\liminf_n\phi(x_n)$$ This means that $$\phi$$ is lower semicontinuous.

Here we used the fact that a function $$f:X\subset \mathbb{R}^d\rightarrow\overline{\mathbb{R}}$$ is lower semicontinuous (l.s.c.) iff $$f(x)\leq \liminf_n f(x_n)$$ for all $$x\in X$$ and $$(x_n:n\in\mathbb{N})\subset X$$ with $$x_n\xrightarrow{n\rightarrow\infty}x$$.

Sketch of a Proof:

Suppose $$f$$ is l.s.c. on $$X$$. Fix $$x\in X$$, and let $$x_n$$ be a sequence in $$X$$ that converges to $$x$$. For any $$a, $$V_a:=\{f>a\}$$ is an open set (in $$X$$) and so, there is $$N\in\mathbb{N}$$ such that $$f(x_n)\in V_a$$ for all $$n\geq N$$. This means that $$a for all $$n\geq N$$. Consequently, $$a\leq\liminf_nf(x_n)$$. Since this holds for any $$a, we conclude that $$f(x)\leq\liminf_nf(x_n)$$.

Conversely, suppose $$f$$ satisfies the property $$f(x)\leq\limsup_nf(x_n)$$ for any $$x\in X$$ and sequence $$x_n$$ in $$X$$ such that $$x_n\xrightarrow{n\rightarrow\infty}x$$. Let $$a\in\mathbb{R}$$ and define $$F_a:=\{f\leq a\}$$. If we show that $$F_a$$ is closed (in $$X$$) then the desired conclusion will follow. Suppose $$x_n$$ is a sequence in $$F_a$$ that converges to some $$x\in X$$. Then $$f(x_n)\leq a$$ for all $$n$$ and so, $$f(x)\leq \liminf_nf(x_n)\leq a$$ Hence $$x\in F_a$$.

• A similar characterization exists for upper semicontinuous functions on $$X\subset\mathbb{R}^d$$.
• If one considers nets in place of sequences, the corresponding characterizations carry over to general $$T_1$$-separable topological spaces.
• I think you should clarify what you mean by $\phi$. Besides, I have suggested some edits - please take a look. Follow-up question: Is there a similar result for upper semicontinuity and $\limsup$ as well? Commented Jun 17, 2021 at 7:24
• @epsilon-emperor: $\phi$ in the context of my solution is defined as the function $x\mapsto \int \mathbb{1}_{V+x}\,d\mu=\mu(V+x)$. The note at the end of my answer is a common characterization of lower semicontinuity. The strategy in my answer works well because of Fatou's lemma. As for upper semicontinuity, under some integrability or dominated assumptions, you can have a reversed Fatou's lemma where $\limsup$ is in place of $\liminf$ and the direction of the inequality is flipped. Commented Jun 17, 2021 at 13:24
If $$(x_{n})_{n \in \mathbb{N}}$$ satisfies $$x_{n} \to x$$, then $$\begin{equation*} V + x \subseteq \bigcup_{n = 1}^{\infty} \bigcap_{m = n}^{\infty} V + x_{m}. \end{equation*}$$ This follows from the fact that $$V$$ is open. Thus, by continuity of measure, $$\begin{equation*} \mu(V+ x) \leq \mu \left(\bigcup_{n = 1}^{\infty} \bigcap_{m = n}^{\infty} V + x_{m} \right) = \lim_{n \to \infty} \mu \left( \bigcap_{m = n}^{\infty} V + x_{m}\right). \end{equation*}$$ For each $$n \in \mathbb{N}$$, we can write $$\begin{equation*} \mu\left( \bigcap_{m = n}^{\infty} V + x_{m}\right) \leq \mu(V + x_{n}). \end{equation*}$$ That leads us to the conclusion that $$\begin{equation*} \mu(V + x) \leq \lim_{n \to \infty} \mu \left( \bigcap_{m = n}^{\infty} V + x_{m}\right) \leq \liminf_{n \to \infty} \mu(V + x_{n}). \end{equation*}$$
This proves $$x \mapsto \mu(V + x)$$ is a lower semi-continuous function in $$\mathbb{R}^{k}$$. (Note that we do not need $$\mu(\mathbb{R}^{k}) < \infty$$ here.)