First let’s look at the most general definition.

Note that by the above, the function which is identically $-\infty$ is subharmonic, but some authors exclude this function by definition. We can define superharmonic functions in a similar fashion to get that $\varphi$ is superharmonic if and only if $-\varphi$ is subharmonic.

If we restrict our domain to the complex plane we can get the following definition.

Intuitively what this means is that a subharmonic function is at any point no greater than the average of the values in a circle around that point. This implies that a non-constant subharmonic function does not achieve its maximum in a region $G$ (it would achieve it at the boundary if it is continuous there). Similarly for a superharmonic function, but then a non-constant superharmonic function does not achieve its minumum in $G$. It is also easy to see that $\varphi$ is subharmonic if and only if $-\varphi$ is superharmonic.

Note that when equality always holds in the above equation then $\varphi$ would in fact be a harmonic function. That is, when $\varphi$ is both subharmonic and superharmonic, then $\varphi$ is harmonic.

It is possible to relax the continuity statement above to take $\varphi$ only upper semi-continuous in the subharmonic case and lower semi-continuous in the superharmonic case. The integral will then however need to be the Lebesgue integral rather than the Riemann integral which may not be defined for such a function. Another thing to note here is that we may take ${\mathbb{R}}^{2}$ instead of ${\mathbb{C}}$ since we never did use complex multiplication. In that case however we must rewrite the expression $z+re^{{i\theta}}$ in terms of the real and imaginary parts to get an expression in ${\mathbb{R}}^{2}$.

It is also possible generalize the range of the functions as well. A subharmonic function could have a range of ${\mathbb{R}}\cup\{-\infty\}$ and a superharmonic function could have a range of ${\mathbb{R}}\cup\{\infty\}$. With this generalization, if $f$ is a holomorphic function then $\varphi(z):=\log\lvert f(z)\rvert$ is a subharmonic function if we define the value of $\varphi(z)$ at the zeros of $f$ as $-\infty$. Again it is important to note that with this generalization we again must use the Lebesgue integral.