Subharmonic functions and monotonically increasing integrals Let $U \subset \mathbb{C}$ be open and $f : U \to \mathbb{R}\cup\{ - \infty\}$ be a subharmonic function. Define $\varphi(r) := \frac{1}{2\pi} \int_{0}^{2\pi} f(a + re^{i\theta}) \ d\theta$. Show that $\varphi$ is monotonically increaing by first showing that it is monotonically increasing if $f$ is a $C^2$ function and then proceed to the general case where we assume no differentiability of $f$. 
Note the definition I'm using for subharmonic is that it is upper-semicontinuous and satisfies the sub-mean-value property. 
It is easy to show in the $C^2$ case - just apply Green's theorem and use the fact that subharmonicity is equivalent to the Laplacian being non-negative when the function is $C^2$. I'm struggling with the general case. I thought to do the following:
We want to show that for $f$ subharmonic and $r_1 < r_2$ then $\varphi(r_1) \leq \varphi(r_2)$. If $f_{\epsilon}$ denotes $f$ mollified with the standard mollifier then for each $\epsilon > 0$ we have that $f$ is smooth and subharmonic so the $C^2$ case applies and we get that $\frac{1}{2\pi} \int_{0}^{2\pi} f_{\epsilon}(a + r_2e^{i\theta}) - f_{\epsilon}(a + r_1 e^{i\theta}) \ d\theta \geq 0$. What I really would like to do is take the limit as $\epsilon \to 0$ of the above and interchange the integral and limit but I can't quite figure out how to justify it (that is, make sure the hypotheses of the dominated convergence theorem hold) while still making sure that the inequality is maintained (i.e., without taking absolute values). 

Since, $f$ is u.s.c it is infimum of a monotone decreasing sequence $\{g_k\}_{k \in \mathbb{N}}$ of continuous functions, that is $g_k \searrow f$ monotonically. Then we kave $f \le g_k$ on $\partial D_{r_2}(a)$ and consider the unique harmonic function $h$ in $D_{r_2}(a)$ s.t., $h\rvert_{\partial D_{r_2}(a)} = g_k$. Then by mean value property we have $\displaystyle h(a) = \frac{1}{2\pi} \int_0^{2\pi} g_k(a + r_2e^{i\theta})\,d\theta$. 
Also by comparison theorem for subharmonic functions we know $f \le h$ in $D_{r_2}(a)$ and in particular for $r_1 < r_2$ we have $$\frac{1}{2\pi} \int_0^{2\pi} f(a + r_1e^{i\theta})\,d\theta \le \frac{1}{2\pi} \int_0^{2\pi} h(a + r_1e^{i\theta})\,d\theta = h(a) = \frac{1}{2\pi} \int_0^{2\pi} g_k(a + r_2e^{i\theta})\,d\theta.$$
Now letting $k \to \infty$ by monotone convergence theorem we have the desired conclusion $$\frac{1}{2\pi} \int_0^{2\pi} f(a + r_1e^{i\theta})\,d\theta \le \frac{1}{2\pi} \int_0^{2\pi} f(a + r_2e^{i\theta})\,d\theta, \forall \, r_1 < r_2.$$
