I am reading a proof in a paper and it basically says:
$u$ is subharmonic so we can assume for all $x\in \Omega$: $$u(x) = \lim_{r \rightarrow 0}\frac{1}{|B_R(x)|}\int_{B_R(x)}u$$
Shouldn't the $=$ sign be an $\leq$? I thought this was a typo, but I've found it now in two different places. For context, I've included part of the proof below.
Notice that there is a limit in front of the integral! I am happy to add more details if needed.
So $u$ is called subharmonic if $\Delta u \geq 0$. From this follows (or it can be shown) that for any point $a$ of the domain of $u$ and any positive radius $R$ one has $$ u(a) \leq \frac{1}{B_R(a)} \int_{B_R(a)} u \,dx $$ as well that for $R_1 < R_2$ $$ \frac{1}{B_{R_1}(a)} \int_{B_{R_1}(a)} u \,dx \leq \frac{1}{B_{R_2}(a)} \int_{B_{R_2}(a)} u \,dx $$ holds. So one can take the limit to obtain $$ u(a) = \lim_{R\searrow 0}\frac{1}{B_R(a)} \int_{B_R(a)} u \,dx. $$
This is true since we have (for continuous $u$ at $a$)
$$ \left| u(a) - \frac{1}{B_r(a)}\int_{B_r(a)} u dx \right| \leq \frac{1}{B_r(a)}\int_{B_r(a)} |u(a) - u| dx \leq \sup_{x\in B_R(a)} |u(a) - u(x)| $$ and that will go to $0$ as $R$ goes to zero (from above).
Now, if you actually asking about the generalized version of a subharmonic function one can obtain a similar result at least almost everywhere.