$g_n$ is a sequence satifying$\int g_n d\mu <\frac{1}{n^2}$ for each $n\geq 1$. Prove that $\Sigma_{n=1}^{\infty}g_n(x) \leq + \infty$. I am expected to use the Monotone convergence theorem for non-negative functions in order to solve this question. I.e. LEt $(X, \Sigma, \mu)$ me a measure space. Suppose $fn:X \rightarrow \bar{\mathbb{R}}$ is a sequence of non-decreasing, non-negative measurable functions. Then $\int \lim_{n\rightarrow \infty}f_nd\mu = \lim_{n\rightarrow \infty} \int f_n d\mu$.
Assume $g_n$ is a sequence of non-negative measurable functions satifying$\int g_n d\mu < \frac{1}{n^2}$ for each $n\geq 1$. Prove that $\Sigma_{n=1}^{\infty}g_n(x) \leq + \infty$. 
My thinking was that $\Sigma_{n=0}^{\infty} \int g_n < \Sigma_{n=0}^{\infty} \frac{1}{n^2}$, but because the integral amounts to summation anway we could swap the sum and the integral around... This seems dodgy and it doesn't use the monotone convergence theorem though! Help!!!
 A: Here, it is not the question of integral amounting to summation. Notice that
\begin{equation}
\sum\limits_{n=1}^{\infty}g_{n}(x)=\lim\limits_{k\to \infty}\sum\limits_{n=1}^{k}g_{n}(x).
\end{equation}
The key idea is that $\sum\limits_{n=1}^{k}g_{n}(x)\uparrow \sum\limits_{n=1}^{\infty}g_{n}(x)$ as $k\to \infty$, which is why the limit on the right hand side of the above equation I have written always exists (could perhaps be $+\infty$) since it is a limit of a sequence of nondecreasing positive numbers. Now, monotone convergence theorem tells us that
\begin{align}
\int \sum\limits_{n=1}^{\infty}g_{n}(x)\,d\mu(x)&=\int \lim\limits_{k\to \infty}\sum\limits_{n=1}^{k}g_{n}(x)\,d\mu(x)\stackrel{MCT}{=}\lim\limits_{k\to \infty}\int\sum\limits_{n=1}^{k}g_{n}(x)\,d\mu(x)\\&=\lim\limits_{k\to \infty}\sum\limits_{n=1}^{k}\int g_{n}(x)\,d\mu(x)\leq \sum\limits_{n=1}^{\infty}\frac{1}{n^{2}}<\infty.
\end{align}
So, what we have shown is that $\int \sum\limits_{n=1}^{\infty}g_{n}(x)\,d\mu(x)<\infty$. Now, using the fact that for a nonnegative random variable $X$, $\int X\,d\mu<\infty$ implies $X<\infty$ a.e., we arrive at the conclusion that $\sum\limits_{n=1}^{\infty}g_{n}<\infty$ a.e.
