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

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