Referring to this question on harmonic function in which is proved the sequent
Let $u(x)$ a harmonic function in $\mathbb{R}^n$ such as: \begin{equation} \int_{\mathbb{R}^n}|u(x)|dx =K< \infty \end{equation} Show thtat $u(x)=0$, $\forall x \in \mathbb{R}^n$.
Is there a heuristic argument to prove it or an explanation about why harmonic function that gives a finite value when integrated must be 0, without doing the rigorous proof?
Yes there is. Heuristically, the "mean value over $\mathbb{R}^n$" of our function should give the value of our function at any particular point. Since the integral is finite and $\mathbb{R}^n$ is not, the function vanishes necessarily. This can be made rigorous as follows.
Supposing $u(p) = y \neq 0$ for some $p \in \mathbb{R}^n$, then the mean value theorem for harmonic functions and the triangle inequality give $$ |y| \cdot |B(p,R)| = \left|\int_{|x-p|\leq R} u(x) dx \right|< \int_{|x-p|\leq R} |u(x)| dx, $$ where $|B(p,R)|$ denotes the volume of the ball of radius $R$ around $p$. Thus, $$\int_{\mathbb{R^n}} |u(x)| dx = \lim_{R\rightarrow \infty} \int_{|x-p|\leq R} |u(x)| dx \geq \lim_{R\rightarrow \infty} y \cdot |B(p,R)| = \infty,$$ that is a contradiction.