So, I'm working on a proof of the lemma: If $u$ is subharmonic, then $\int_0^{2 \pi} u(a+r_1 e^{i \theta}) d \theta \leq \int_0^{2 \pi} u(a+r_2 e^{i \theta}) d \theta$, whenever $0<r_1 < r_2$.
The first statement of this proof is:
Since $u$ is upper semicontinuous, then for all $\epsilon >0$, there exists a function h, continuous on $\partial D (a, r_2)$ such that $u < h < u \ + \ \epsilon$.
I'm a bit confused as to how this would work. Since h is continuous, it would be bounded above and below on $\partial D(a,r_2)$, while $u$ and $u \ + \ \epsilon$ would only be bounded above. So, how could such a continuous function exist? Any help would be greatly appreciated.
(put it as a comment but it got too long) Where is the proof from? Maybe there are some extra assumptions - in general usc functions are inferior limits of a decreasing sequence of continuous functions so one can pick a sequence of continuos functions $f_n \to u$ on $C(a,r_2)$ with decreasing convergence ($u \le f_n, f_{n+1}\le f_n$) for which (by the monotone convergence theorem) the integral satisfies the inequality:
$\int_0^{2 \pi} u(a+r_2 e^{i \theta}) d \theta \le \int_0^{2 \pi} f_n(a+r_2 e^{i \theta}) d \theta \le \int_0^{2 \pi} u(a+r_2 e^{i \theta}) d \theta + \epsilon$, for $n \ge n(\epsilon)$,
but it is not necessary true pointwise; then we take $u_n$ the Poisson transform of $f_n$ and since $u_n$ is harmonic and by defintion of sh function $u(z) \le u_n(z), z \in \bar D(a,r_2)$ so:
$\int_0^{2 \pi} u(a+r_1 e^{i \theta}) d \theta \le \int_0^{2 \pi} u_n(a+r_1 e^{i \theta}) d \theta = \int_0^{2 \pi} u_n(a+r_2 e^{i \theta}) d \theta \le \int_0^{2 \pi} u(a+r_2 e^{i \theta}) d \theta + \epsilon$
for $n \ge n(\epsilon)$ and let $\epsilon \to 0$