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.

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    $\begingroup$ 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. $\endgroup$
    – Mittens
    Commented Jun 17, 2021 at 1:55

2 Answers 2


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 $a<f(x)$, $V_a:=\{f>a\}$ 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 $a<f(x_n)$ for all $n\geq N$. Consequently, $a\leq\liminf_nf(x_n)$. Since this holds for any $a<f(x)$, 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 $x\in F_a$.


  • The characterization of l.s.c. functions describe above is very useful in applications. It is similar in nature to that of continuity in terms of sequences.
  • A similar characterization exists for upper semicontinuous functions on $X\subset\mathbb{R}^d$.
  • These characterizations carry over to functions defined on metric spaces (or first countable topological spaces).
  • If one considers nets in place of sequences, the corresponding characterizations carry over to general $T_1$-separable topological spaces.
  • $\begingroup$ 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? $\endgroup$ Commented Jun 17, 2021 at 7:24
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    $\begingroup$ @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. $\endgroup$
    – Mittens
    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.)


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