Monotone convergence theorem for series (basic proof) My question is how to prove monotone convergence theorem for infinite series without more advanced technique like counting measure. I see this used a lot. But looking through books like Rudin, the theorem for series or elementary proof is not to be found.
The theorem is: 
For a sequence $x_{mn} \geq 0$ if $\lim_{m \to  \infty} x_{mn} = y_n$ (monotonically increasing for $m$) then
$$\lim_{m \to \infty} \sum_{n=1}^\infty x_{mn} = \sum_{n=1}^\infty 
 \lim_{m \to \infty} x _{mn}$$

Here's a constructive $\epsilon$ proof.
Let $\epsilon > 0$ be arbitrary. Consider the the smallest value of $N$ such that the partial sum $$\sum_{n=1}^N 
 \lim_{m \to \infty} x _{mn}$$ is within $\epsilon/2$ of the actual sum.
Denote $\lim_{m \to \infty} x _{mn}$ by $x _{\infty,n}$ from now on.
Now let $M$ be such that whenever $m \geq M$ and $n \leq N$: $x_{\infty,n} - x_{mn} < \frac\epsilon{2N}$.
It follows that we are done, because for any $m\geq M$, $$\sum_{n=1}^\infty x_{\infty,n} \geq \sum_{n=1}^\infty x_{mn} \geq \sum_{n=1}^\infty x_{\infty,n} - \epsilon,$$
which can be proved in the following way:
$$\begin{aligned}
x_{\infty,n} &> x_{mn}\\
\therefore \sum_{n=1}^\infty x_{\infty,n} &\geq \sum_{n=1}^\infty x_{mn}\\
&\geq \sum_{n=1}^N (x_{mn} )\\
&\geq \sum_{n=1}^N (x_{\infty,n} - \frac\epsilon{2N})\\
&\geq \sum_{n=1}^N (x_{\infty,n}) - \frac\epsilon{2}\\
&\geq \sum_{n=1}^\infty (x_{\infty,n}) - \frac\epsilon{2} - \frac\epsilon{2}\\
&= \sum_{n=1}^\infty (x_{\infty,n}) - \epsilon.
\end{aligned}$$
