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.
 A: 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.
A: Hint: The sequence of partial sums $\Big(\sum_{n=1}^{N}g_{n}\Big)_{N=1}^{\infty}$ is monotonically increasing because each $g_{n}$ is non-negative.
A: You need to know (or to prove) that $\sum \frac 1 {n^2}$ is finite, with sum $S$ (you don't need to know what the sum is - that is a much harder problem than just knowing the sum is finite). Now if $f_M = \sum_1^M g_n$, the $f_M$ form a monotonically increasing sequence (why?) and the integral of $f_M$ is bounded by S, so the integral of the limit (that is, the integral of the infinite sum of $g_n$) exists and is finite. So the pointwise sum of $g_n(x)$ may only be infinite on a set of measure zero.
A: Hint: 
$$\sum \int g_n = \int \sum g_n$$
Why?
Also, if $\int f < \infty$ where $f \ge 0$, then $f < \infty$ a.e. (this is well-known)
