Integrability in Definition of Subharmonic functions

Question

This might be a couple of very elementary measure theory questions, so please don't assume too much knowledge on my side.

I want to understand a definition of subharmonic functions and for this I read about upper semi-continuity. Here are my questions:

1. For an upper semi-continuous function $u$ the set sets $u^{-1}([-\infty,a))$ are open for all $a \in \mathbb{R}$. This implies that $u$ is Borel-measureable. Why is this, is this just the definitionn of a Borel-measurable function?
2. Does 1. also mean that $u$ is a (Lebesgue-) measurable function?
3. We look at the definition of a subharmonic function:

A map $u : Ω → [−∞, ∞)$ is said to be subharmonic in $Ω$ if

(a) $u$ is upper semi-continuous and not identically −∞.

(b) For each $z ∈ Ω$ there is a ball $B(z, R_z) ⊆ Ω$ such that for all $r \in (0,R_z)$ $$ u(z) \leq \frac1{2\pi}\int_0^{2\pi}u(z + re^{i \theta}) d \theta$$

Why can the integral in (b) be calculated, is this a consequence of $u$ being a Borel-measurable function? Is $u$ is (Lebesgue-) integrable? And why?

Answer 1

Note first that this definition is wrong, unless we're assuming that $\Omega$ is connected! (a) should say that $u$ is not identically $-\infty$ in any nonempty open subset of $\Omega$. (Given (b), this turns out to be equivalent to assuming that none of the integrals in (b) equal $-\infty$.)

Yes, (1) is immediate from the definition.

In general it's trivial that Borel measurable implies Lebesgue measurable. Say $B$ and $L$ are the Borel and Lebesgue $\sigma$-algebras:

If $O$ is open then $O\in L$.

You prove that somehow, depending on how $L$ was defined.

Corollary. $B\subset L$

Proof: By definition $B$ is the intersection of all the $\sigma$-algbras containing every open set; $L$ is one of these $\sigma$-algebras.

Corollary${}^2$. If $u$ is Borel measurable then $u$ is Lebesgue measurable.

Note it's important that $u$ be Borel measurable, not just Lebesgue measurable:

Say $S_{z,r}=\{z+re^{it}\}$ and $u_{z,r}(t)=u(z+re^{it})$. In (b) we need to know that $u_{z,r}$ is Lebesgue measurable. If we assume just that $u$ iis Lebesgue measurable this does not follow: Since $S_{z,r}$ is a null set and Lebesgue measure is complete, $u_{z,r}$ can be any ($2\pi$-periodic) function whatever. But here since $u$ is usc it follows that $u_{z,r}$ is usc, hence Borel measurable, hence Lebesgue measurable.