# Semicontinuity

We will recall the definition of semicontinuity of a function and some of related properties. (pdf)

Throughout this note, we will assume that ${\mathbb X}$ is a metric space and ${f: \mathbb X \mapsto \mathbb R}$ is a Borel measurable function. We say that

1. ${f}$ is upper semicontinuous (USC) at ${x_{0} \in \mathbb X}$, if

$\displaystyle \forall \epsilon >0, \ \exists \delta >0, \ s.t. \ \sup\{f(x): x\in B_{\delta}(x_{0}) \}\le f(x_{0}) + \epsilon.$

2. ${f}$ is lower semicontinuous (LSC) at ${x_{0} \in \mathbb X}$, if ${-f}$ is USC at ${x_{0}}$.
3. ${f}$ is USC if ${f}$ is USC everywhere, and ${f}$ is LSC if it is LSC everywhere.

It is equivalent to

Proposition 1

${f}$ is USC at ${x_{0}}$ if and only if ${\lim\sup_{x\rightarrow x_{0}} f(x) \le f(x_{0}).}$

–QED–

Example 1

${f: \mathbb R \rightarrow \mathbb R}$ is USC and increasing if and only if it is RCLL.

— QED —

It’s well known that composition of two continuous functions is continuous. However, it’s not the case for semicontinuity as shown in the next example.

Example 2

It is false to say “A composition of two USC functions is USC again.” as shown below.
Let ${A}$ be a closed set in ${\mathbb X}$.

1. The indicator function ${f_{1} = I_{A}}$ is USC.
2. Let ${f_{2}(x) = -x: \mathbb R \mapsto \mathbb R}$. Then ${f_{2} \circ f_{1}}$ is LSC.

Now, we can generalize the above example.

Example 3

If

1. ${f: \mathbb X \mapsto \mathbb R}$ is USC at ${x_{0}}$;
2. ${g: \mathbb R \mapsto \mathbb R}$ is an RCLL increasing function,

then ${g\circ f}$ is USC at ${x_{0}}$.

Proof: Let’s denote ${h = g \circ f}$. Fix arbitrary ${\epsilon >0}$, and we look for ${\delta>0}$, such that

$\displaystyle h (x_{0}) \ge h(x) - \epsilon, \ \forall x\in B_{\delta}(x_{0}).$

Since ${g}$ is RCLL and increasing, there exists ${\epsilon_{1}>0}$ such that ${g(f(x_{0}) + \epsilon_{1}) \le g(f(x_{0})) + \epsilon}$, or equivalently

if ${f(x) \le f(x_{0}) + \epsilon_{1}}$, then ${h(x) \le h(x_{0}) + \epsilon}$.

On the other hand, since ${f}$ is USC at ${x_{0}}$, there exists ${\delta>0}$ such that ${f(x) \le f(x_{0}) + \epsilon_{1}}$ for all ${x\in B_{\delta}(x_{0})}$. Thus, the above ${\delta}$ fulfills the requirement. $\Box$