linear algebra - Monotone Convergence Theorem for nonnegative functions (not quite a decreasing sequence)

This question suggests that the MCT for functions is to be applied, but I can't see how this could be done.

Assume $g_n: X \to \bar{\mathbb{R}}$ is a sequence of nonnegative measurable functions satisyfing $$\int g_nd\mu < \frac{1}{n^2}$$
for each $n\geq1$. Using the Monotone Convergence Theorem for nonnegative functions, or otherwise, prove that $$\sum_{N=1}^{\infty}g_n(x)<+\infty$$ $\mu$-almost everywhere.

It seems to me that $g_n$ is similar to a decreasing sequence, although not necessarily for all $x$; and that as $n$ tends to infinity, the integral must tend to $0$, so somehow the 'limit' of $g_n$ must be equal to $0$ $\mu$-almost everywhere. But since $g_{n+1}$ is not necessarily less than $g$ for all $x$, I can't see that the pointwise limit even exists.

Answer

Let $f_n=\sum_{k=1}^ng_k$. Then the sequence $\{f_n\}$ is increasing because the $g_k$ are non-negative, hence by the monotone convergence theorem
$$0\leq \int \sum_{k=1}^{\infty}g_k\;d\mu=\sum_{k=1}^{\infty}\int g_k\;d\mu\leq \sum_{k=1}^{\infty}\frac{1}{k^2}<\infty$$

Since $\sum_{k=1}^{\infty}g_k$ is a non-negative function with finite integral, it must be finite almost everywhere.