How can a boundary measure of a function be absolutely continuous? I'm studying firsts tools in several complex variables. In my book I found what follows:

It can be proved that if $\varphi$ is strongly subharmonic and has a finite majorant in the unit ball, then its boundary measure is absolutely continuous.

First of all $\varphi$ is defined on an open $\Omega\subseteq\Bbb C$, and from we deduce $\Omega$ must contain the unit ball.
But my real question is: what is a boundary measure of a function?
I'd say we're dealing with the integral of the function over $\partial B(0,1)$... but this last one is a number, whereas the book speaks of an object which can be absolutely continuous (meaningless for numbers).
Can someone help me? Many thanks.
 A: People have commented on what it means for a measure to be absolutely continuous, but I don't see anything actually addressing your question regarding what the boundary measure of a function is. What it is is a slightly long story:
First, I'd read it as saying the function is defined in the ball, period.
This is a generalization of the F&M Riesz theorem.
Say $B$ is the unit ball (in $\mathbb R^n$ or $\mathbb C^n$, makes no difference since I'm leaving out details.). Say $S$ is the boundary of the unit ball and let $d\sigma$ denote integration on $S$ with respect to surface area. Say $u$ is harmonic in $B$. If $$\sup_{0\le r<1}\int_S|u(r\zeta)|d\sigma(\zeta)<\infty\quad(*)$$then $u$ is the Poisson integral of some Borel measure $\mu$ on $S$, which says that $$u(z)=P[\mu]=\int_SP(z,\zeta)\,d\mu(\zeta).$$Here $P(z,\zeta)$ is the Poisson kernel for the ball, which Google knows all about. This $\mu$ is the "boundary measure" of $u$.
The classical F&M Riesz theorem says that if $B$ is the unit ball in the plane and a holomorphic function $u$ satisfies (*) then $\mu$ is absolutely continuous (that is, $\mu<<\sigma$). People have commented on the meaning of "absolutely continuous"; you should look at a book on real analysis, in the neighborhood of words like "Radon-Nikodym".
That's for harmonic functions. Now if $v$ is subharmonic there may or may not exist a harmonic function $u$ with $u\ge v$. If such a $u$ exists there is a least such $u$, the "least harmonic majorant" of $v$. And if $u$ then satisfies (*), the measure $\mu$ such that $u=P[\mu]$ is the "boundary measure" of $v$.
Which raises the question of what $\mu$ has to do with $v$. For that see the Riesz Decomposition Theorem somewhere...
