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

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

June 15, 2010 - Posted by John Armstrong | Analysis, Measure Theory

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