Sub-mean value property

Remarks

Proof idea: since $u$ is upper semicontinuous, there are continuous functions $\varphi_k$ on $S(a, r)$ that decrease to $u$ there. Let $h_k$ be the harmonic function in $B(a, r)$ with boundary values $\varphi_k$, given by the Poisson integral formula. The function $u - h_k$ again satisfies the local sub-mean inequality, so the maximum principle for subharmonic functions gives $u \le h_k$ in the ball. Hence $u(a) \le h_k(a) = \int \varphi_k \, d\sigma_{a,r}$, and monotone convergence gives part 1. Part 2 follows by integrating part 1 over the radii from $0$ to $r$.

So a locally defined subharmonic function satisfies the inequality on every ball whose closure lies in $\Omega$, not only on small ones. Example: $u(x) = \abs{x}^2$ on $\R^n$ has sphere mean $\abs{a}^2 + r^2$ and ball mean $\abs{a}^2 + \frac{n}{n+2} r^2$, both at least $\abs{a}^2$. For a harmonic function both inequalities become equalities, which is the mean value property. The ball $\overline{B(a, r)}$ must lie in $\Omega$: for $u(x) = -\log\abs{x}$ on $\R^2 \setminus \set{0}$, which is harmonic there, the circle mean about $a = (1, 0)$ with $r = 2$ is $-\log 2 < 0 = u(a)$.

Going deeper

For a student

The global inequality is what makes subharmonic functions useful: it gives the maximum principle, local integrability, and comparison with harmonic functions on balls. The averages over spheres of radius $r$ are increasing in $r$ and tend to $u(a)$ as $r \to 0$. For $C^2$ functions the condition is equivalent to $\Delta u \ge 0$.