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Upper semicontinuous functions are Borel and their circle averages are defined

Statement

Let $Ω \subseteq C$ be open and let $u : Ω \to [- \infty, \infty)$ be upper semicontinuous. Then:

$u$ is Borel measurable;
for every circle $C = {a + r e^{i t} : 0 \leq t \leq 2 π} \subseteq Ω$ , the boundary function $t \mapsto u (a + r e^{i t})$ is Borel measurable and bounded above, so its average $\frac{1}{2 π} \int_{0}^{2 π} u (a + r e^{i t}) d t$ is a well-defined element of $[- \infty, \infty)$ .

Facts & Assumptions

Given: An upper semicontinuous function $u : Ω \to [- \infty, \infty)$ and a circle $C = {a + r e^{i t} : 0 \leq t \leq 2 π} \subseteq Ω$ .

[A1]

The function $u$ is upper semicontinuous on $Ω$ , and the circle $C$ lies in $Ω$ .

Proof

technique · direct

1.1

For every real $α$ , the set ${z \in Ω : u (z) < α}$ is open because $u$ is upper semicontinuous. Hence the sets ${u \geq α}$ are closed, and therefore $u$ is Borel measurable.

1.2

The circle $C$ is compact. If $u$ is not identically $- \infty$ on $C$ , upper semicontinuity gives a point of maximum and therefore a finite upper bound $M$ on $C$ ; if $u \equiv - \infty$ on $C$ , then $- \infty$ is already an upper bound. So the boundary function on $[0, 2 π]$ is Borel measurable and bounded above.

2.1

The parametrization $t \mapsto a + r e^{i t}$ is continuous, so composing it with the Borel function from step 1.1 makes $t \mapsto u (a + r e^{i t})$ Borel measurable on $[0, 2 π]$ .

step 1.1

3.1

A Borel measurable function bounded above on a finite interval has an extended-real integral in $[- \infty, \infty)$ , so the displayed circle average is well defined.

step 2.1step 1.2∎

Used by

Dependency tree · 0 levels

Nothing. This result depends on no other item in the library.

Sources

Harold P. Boas, Class Notes Math 618: Complex Variables II, Spring 2016 (standard reference, not scraped)