I'm reading 'Partial Differential Equations' by Evans, and on page 31 he writes
Assume $u$ is harmonic in $U$. Then $u$ is analytic in $u$.
[...]
$$\frac{1}{\alpha(n)}\|u\|_{L^1(B(x_0,2r))}\leq \infty.$$
He provides no justification for the fact that the given $L^1$ norm of $u$ is finite, so I assume it's supposed to be fairly obvious.
We know that if a function is harmonic over $U$, then $$u(x)=\frac{1}{\alpha(n)r^n}\int_{B(x,r)}u(y)\mathrm{d}y,$$ where $\alpha(n)$ is the volume of the unit $n$-dimensional ball, for every ball $B(x,r)\subset U$. Seeing as $u(x)$ is finite, $\int_{B(x,r)}u\mathrm{d}y$ must also be finite.
However, in order for the $L^1$ norm to be finite, we would need $\int_{B(x,r)}|u(y)|\mathrm{d}y\leq\infty.$ Why should this hold?
At this point in his book, he talks about classical harmonic functions which are required to be $C^2$ and in particular continuous. Hence, $|u|$ is bounded on the ball $B(x,r)$ so the integral is finite.