# 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

$$\lim_{m \to \infty} \sum_{n=1}^\infty x_{mn} = \sum_{n=1}^\infty \lim_{m \to \infty} x _{mn}$$

This is a fairly easy consequence of Fatou’s Lemma for series, stating if $x_{mn}$ is nonnegative, then

$$\sum_{n=1}^\infty \,\liminf_{m \to \infty} \,\,x_{mn} \leqslant \,\liminf_{m \to \infty}\,\sum_{n=1}^\infty \, x_{mn}.$$

Consequently, if $x_{mn} \uparrow y_n$ then $x_{mn} \leqslant y_n$ for all $n$ and

$$\limsup_{m \to \infty} \, \sum_{n=1}^\infty x_{mn} \leqslant \sum_{n=1}^\infty y_n = \sum_{n=1}^\infty \,\liminf_{m \to \infty} \,\,x_{mn} \leqslant \,\liminf_{m \to \infty}\,\sum_{n=1}^\infty \, x_{mn}.$$

Thus, $\lim_{m \to \infty} \sum_{n=1}^\infty x_{mn} = \sum_{n=1}^\infty y_n$.

To prove Fatou's lemma, note that for all $k \geqslant m$ we have $\inf_{j \geqslant m} x_{jn} \leqslant x_{kn}$ and for every positive integer $N$,

$$\sum_{n=1}^N \inf_{j \geqslant m} x_{jn} \leqslant \inf_{k \geqslant m} \sum_{n=1}^N x_{kn} \leqslant \inf_{k \geqslant m} \sum_{n=1}^\infty x_{kn}$$

Taking the limit of boths sides as $m \to \infty$ followed by the limit as $N \to \infty$ gives us the result.

• How do you get first inequality in second and third lines? – SAS Jan 23 '18 at 0:53
• Assume you mean $\limsup_{m \to \infty} \, \sum_{n=1}^\infty x_{mn} \leqslant \sum_{n=1}^\infty y_n$ by second line. Since $0 < x_{mn} \leqslant y_n$ for all n we have $\sum_{n=1}^\infty x_{mn} \leqslant \sum_{n=1}^\infty y_n$ and taking the $\limsup$ as $m \to \infty$, we get the inequality since the RHS does not depend on $m$. – RRL Jan 23 '18 at 1:03
• For $\sum_{n=1}^N \inf_{j \geqslant m} x_{jn} \leqslant \inf_{k \geqslant m} \sum_{n=1}^N x_{kn}$ -- again summing both sides of $\inf_j x_{jn } \leqslant x_{kn}$ we get the inequality without $\inf$ on the RHS. But this is true for all $k \geqslant m$, so taking the $\inf$ on the RHS preserves the inequality. – RRL Jan 23 '18 at 1:07

Cribbing from the way Rudin writes these things, one can give a proof that's the same as what Daniel Schepler said, except there's no need to split it into two cases:

Suppose that $$\alpha<\sum y_n.$$There exists $N<\infty$ so that $$\alpha<\sum_{n=1}^N y_n\le\sum y_n.$$

Now we certainly have $\lim_{m\to\infty}\sum_{n=1}^nx_{mn}=\sum_1^Ny_n$, so there exists $M$ such that $$\alpha<\sum_{n=1}^Nx_{mn}\le\sum_1^Ny_n,\quad(m>M).$$Hence for every $m>M$ we have $$\alpha<\sum_{n=1}^\infty x_{mn}\le\sum y_n.$$

Here's the outline of one simple proof:

First, suppose $\sum_{n=1}^\infty y_n$ converges. Then, for each $\epsilon > 0$, we can choose $N$ such that $\sum_{n=N+1}^\infty y_n < \frac{\epsilon}{2}$. Now, for each $m$, $\sum_{n=N+1}^\infty (y_n - x_{mn}) \le \sum_{n=N+1}^\infty y_n < \frac{\epsilon}{2}$. Whereas for $n=1, \ldots, N$, we can choose $M$ such that $y_n - x_{mn} < \frac{\epsilon}{2 N}$ whenever $m \ge M$. Therefore, whenever $m \ge M$, we have $0 \le \sum_{n=1}^\infty (y_n - x_{mn}) < \epsilon$, which implies the desired result. (So, the idea is: by being uniformly bounded by a convergent series, the tail of $\sum_{n=1}^\infty x_{mn}$ past some uniform $N$ doesn't matter that much, and then the rest of the terms are just a finite sum for which each term approaches $y_n$.)

Similarly, in the case $\sum_{n=1}^\infty y_n = \infty$ diverges, for any $R$ we can choose $N$ such that $\sum_{n=1}^N y_n > R + 1$. Now, we can choose $M$ such that whenever $m \ge M$, for each $n=1, \ldots, N$, we have $y_n - x_{mn} < \frac{1}{N}$. This implies that $\sum_{n=1}^\infty x_{mn} \ge \sum_{n=1}^N y_n - \sum_{n=1}^N (y_n - x_{mn}) > (R+1)-1 = R$. Thus, $\sum_{n=1}^\infty x_{mn} \to \infty$ as $m \to \infty$. (So, the idea in this case is that $\sum_{n=1}^N y_n$ can be made as large as we want, and then $\sum_{n=1}^N x_{mn}$ can be made close to this large value.)

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}