# The Unapologetic Mathematician

## The Monotone Convergence Theorem

We want to prove a strengthening of the dominated convergence theorem. If $\{f_n\}$ is an a.e. increasing sequence of extended real-valued, non-negative, measurable functions, and if $f_n$ converges to $f$ pointwise a.e., then

$\displaystyle\lim\limits_{n\to\infty}\int f_n\,d\mu=\int f\,d\mu$

If $f$ is integrable, then $f$ dominates the sequence $\{f_n\}$, and so the dominated convergence theorem itself gives us the result we assert. What we have to show is that if $\int f\,d\mu=\infty$, then the limit diverges to infinity. Or, contrapositively, if the limit doesn’t diverge then $f$ must be integrable.

But this is the limit of a sequence of real numbers, and so if it converges then it’s Cauchy. That is, we can conclude that

$\displaystyle\lim\limits_{m,n\to\infty}\left\lvert\int f_m\,d\mu-\int f_n\,d\mu\right\rvert=0$

Our assumption that $f_n$ is a.e. increasing tells us that for any fixed $m$ and $n$, the difference $f_m-f_n$ is either a.e. non-negative or a.e. non-positive. That is,

$\displaystyle\left\lvert\int f_m\,d\mu-\int f_n\,d\mu\right\rvert=\int\lvert f_m-f_n\rvert\,d\mu$

And thus the sequence $\{f_n\}$ is mean Cauchy, and thus mean convergent to some integrable function $g$, which must be equal to $f$ almost everywhere.

One nice use of this is when talking about series of functions. If $\{f_n\}$ is a sequence of integrable functions so that

$\displaystyle\sum\limits_{n=1}^\infty\int\lvert f_n\rvert\,d\mu<\infty$

then I say that the series

$\displaystyle\sum\limits_{n=1}^\infty f_n(x)$

converges a.e. to an integrable function $f$, and further that

$\displaystyle\int f\,d\mu=\int\sum\limits_{n=1}^\infty f_n\,d\mu=\sum\limits_{n=1}^\infty\int f_n\,d\mu$

To see this, we let $S_n(x)$ be the partial sum

$\displaystyle S_n(x)=\sum\limits_{i=1}^n\lvert f_i(x)\rvert$

which gives us a pointwise increasing sequence of non-negative measurable functions. The monotone convergence theorem tells us that these partial sums converge pointwise to some $S(x)$ and that

$\displaystyle\int S\,d\mu=\lim\limits_{n\to\infty}\int S_n\,d\mu=\lim\limits_{n\to\infty}\int\sum\limits_{i=1}^n\lvert f_i\rvert\,d\mu=\lim\limits_{n\to\infty}\sum\limits_{i=1}^n\int\lvert f_i\rvert\,d\mu=\sum\limits_{i=1}^\infty\int\lvert f_i\rvert\,d\mu$

But this is exactly the sum we assumed to converge before. Thus the function $S$ is integrable and the series of the $f_n$ is absolutely convergent. That is, since $S$ must be a.e. finite, the series

$\displaystyle\sum\limits_{n=1}^\infty f_n(x)$

is absolutely convergent for almost all $x$, and so it must be convergent pointwise almost everywhere. Since $S$ dominates the partial sums

$\displaystyle S(x)=\sum\limits_{i=1}^\infty\lvert f_i(x)\rvert\geq\sum\limits_{i=1}^n\lvert f_i(x)\rvert\geq\left\lvert\sum\limits_{i=1}^nf_i(x)\right\rvert$

the bounded convergence theorem tells us that limits commute with integrations here, and thus that

$\displaystyle\begin{aligned}\int f\,d\mu&=\int\sum\limits_{i=1}^\infty f_i\,d\mu\\&=\int\lim\limits_{n\to\infty}\sum\limits_{i=1}^nf_i\,d\mu\\&=\lim\limits_{n\to\infty}\int\sum\limits_{i=1}^nf_i\,d\mu\\&=\lim\limits_{n\to\infty}\sum\limits_{i=1}^n\int f_i\,d\mu\\&=\sum\limits_{i=1}^\infty\int f_i\,d\mu\end{aligned}$

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Posted by John Armstrong | Analysis, Measure Theory

## 10 Comments	»

1. [...] means we can bring the monotone convergence theorem to bear. This tells us [...]

Pingback by | June 16, 2010 | Reply

2. [...] functions overall are handled by using their positive and negative parts. Then you can prove the monotone convergence theorem, followed by Fatou’s lemma, and then the Fatou-Lebesgue theorem, which leads to dominated [...]

Pingback by | June 18, 2010 | Reply

3. [...] We define to be the limit of the — is the maximum of all the — and use the monotone convergence theorem to tell us [...]

Pingback by | July 7, 2010 | Reply

4. [...] let be an increasing sequence of non-negative simple functions converging pointwise to . Then monotone convergence tells us [...]

Pingback by | July 12, 2010 | Reply

5. [...] a monotone class. That is a monotone class follows from the dominated convergence theorem and the monotone convergence theorem, and so we have only to show that the assertions hold for measurable rectangles with finite-measure [...]

Pingback by | July 22, 2010 | Reply

6. [...] we have used the monotone convergence theorem to exchange the sum and the [...]

Pingback by | July 23, 2010 | Reply

7. [...] function , we can find an increasing sequence of simple functions converging pointwise to . The monotone convergence theorem tells us [...]

Pingback by | July 28, 2010 | Reply

8. [...] for the case of for a measurable set . Linear combinations will extend it to simple functions, the monotone convergence theorem extends to non-negative measurable functions, and general functions can be decomposed into positive [...]

Pingback by | August 2, 2010 | Reply

9. [...] monotone convergence theorem now tells us that the limiting [...]

Pingback by | August 31, 2010 | Reply

10. [...] the monotone convergence theorem as we find that [...]

Pingback by | September 3, 2010 | Reply

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