# The Unapologetic Mathematician

## Equicontinuity, Convergence in Measure, and Convergence in Mean

First off we want to introduce another notion of continuity for set functions. We recall that a set function $\nu$ on a class $\mathcal{E}$ is continuous from above at $\emptyset$ if for every decreasing sequence of sets $E_n\in\mathcal{E}$ with $\lim_nE_n=\emptyset$ we have $\lim_n\nu(E_n)=0$. If $\{\nu_m\}$ is a sequence of set functions, then, we say the sequence is “equicontinuous from above at $\emptyset$” if for every sequence $\{E_n\}\subseteq\mathcal{E}$ decreasing to $\emptyset$ and for every $\epsilon>0$ there is some number $N$ so that if $n\geq N$ we have $\lvert\nu_m(E_n)\rvert<\epsilon$. It seems to me, at least, that this could also be called “uniformly continuous from above at $\emptyset$“, but I suppose equicontinuous is standard.

Anyway, now we can characterize exactly how convergence in mean and measure differ from each other: a sequence $\{f_n\}$ of integrable functions converges in the mean to an integrable function $f$ if and only if $\{f_n\}$ converges in measure to $f$ and the indefinite integrals $\nu_n$ of $\lvert f_n\rvert$ are uniformly absolutely continuous and equicontinuous from above at $\emptyset$.

We’ve already shown that convergence in mean implies convergence in measure, and we’ve shown that convergence in mean implies uniform absolute continuity of the indefinite integrals. All we need to show in the first direction is that if $\{f_n\}$ converges in mean to $f$, then the indefinite integrals $\{\nu_n\}$ are equicontinuous from above at $\emptyset$.

For every $\epsilon>0$ we can find an $N$ so that for $n\geq N$ we have $\lVert f_n-f\rVert_1<\frac{\epsilon}{2}$. The indefinite integral of a nonnegative a.e. function is real-valued, countably additive, and nonnegative, and thus is a measure. Thus, like any measure, it’s continuous from above at $\emptyset$. And so for every sequence $\{E_m\}$ of measurable sets decreasing to $\emptyset$ there is some $M$ so that for $m\geq M$ we find

\displaystyle\begin{aligned}\int\limits_{E_m}\lvert f_n-f\rvert\,d\mu&<\frac{\epsilon}{2}\\\int\limits_{E_m}\lvert f\rvert\,d\mu&<\frac{\epsilon}{2}\end{aligned}

the first for all $n$ from $1$ to $N$. Then if $m\geq M$ we have

$\displaystyle\lvert\nu_n(E_m)\rvert=\int\limits_{E_m}\lvert f_n\rvert\,d\mu\leq\int\limits_{E_m}\lvert f_n-f\rvert\,d\mu+\int\limits_{E_m}\lvert f\rvert\,d\mu<\epsilon$

for every positive $n$. We control the first term in the middle by the mean convergence of $\{f_n\}$ for $n\geq N$ and by the continuity from above of $\int_E\lvert f_n-f\rvert\,d\mu$ for $n\leq N$. And so the $\nu_n$ are equicontinuous from above at $\emptyset$.

Now we turn to the sufficiency of the conditions: assume that $\{f_n\}$ converges in measure to $f$, and that the sequence $\{\nu_n\}$ of indefinite integrals is both uniformly absolutely continuous and equicontinuous from above at $\emptyset$. We will show that $\{f_n\}$ converges in mean to $f$.

We’ve shown that $N(f_n)=\{x\in X\vert f(x)\neq0\}$ is $\sigma$-finite, and so the countable union

$E_0=\bigcup\limits_{n=1}^\infty N(f_n)$

of all the points where any of the $f_n$ are nonzero is again $\sigma$-finite. If $\{E_n\}$ is an increasing sequence of measurable sets with $\lim_nE_n=E_0$, then the differences $F_n=E_0\setminus E_n$ form a decreasing sequence $\{F_n\}$ converging to $\emptyset$. Equicontinuity then implies that for every $\delta>0$ there is some $k$ so that $\nu_n(F_k)<\frac{\delta}{2}$, and thus

$\displaystyle\int\limits_{F_k}\lvert f_m-f_n\rvert\,d\mu\leq\int\limits_{F_k}\lvert f_m\rvert\,d\mu+\int\limits_{F_k}\lvert f_n\rvert\,d\mu=\nu_m(F_k)+\nu_n(F_k)\leq\frac{\delta}{2}+\frac{\delta}{2}=\delta$

For any fixed $\epsilon>0$ we define

$\displaystyle G_{mn}=\left\{x\in X\big\vert\lvert f_m-f_n\rvert\geq\epsilon\right\}$

and it follows that

\displaystyle\begin{aligned}\int\limits_{E_k}\lvert f_m-f_n\rvert\,d\mu&\leq\int\limits_{E_k\cap G_{mn}}\lvert f_m-f_n\rvert\,d\mu+\int\limits_{E_k\setminus G_{mn}}\lvert f_m-f_n\rvert\,d\mu\\&\leq\int\limits_{E_k\cap G_{mn}}\lvert f_m-f_n\rvert\,d\mu+\epsilon\mu(E_k)\end{aligned}

By convergence in measure and uniform absolute continuity we can make the integral over $E_k\cap G_{mn}$ arbitrarily small by choosing $m$ and $n$ sufficiently large. We deduce that

$\displaystyle\limsup\limits_{m,n\to\infty}\int\limits_{E_k}\lvert f_m-f_n\rvert\,d\mu\leq\epsilon\mu(E_k)$

and, since $\epsilon>0$ was arbitrary, we conclude

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

Now we can see that

$\displaystyle\int\lvert f_m-f_n\rvert\,d\mu=\int\limits_{E_0}\lvert f_m-f_n\rvert=\int\limits_{E_k}\lvert f_m-f_n\rvert+\int\limits_{F_k}\lvert f_m-f_n\rvert\,d\mu$

and thus

$\displaystyle\limsup\limits_{m,n\to\infty}\int\lvert f_m-f_n\rvert\,d\mu<\delta$

and, since $\delta>0$ is arbitrary,

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

That is, the sequence $\{f_n\}$ is Cauchy in the mean. But we know that the $L^1$ norm is complete, and so $\{f_n\}$ converges in the mean to some function $g$. But this convergence in mean implied convergence in measure, and so $\{f_n\}$ converges in measure to $g$, and thus $f=g$ almost everywhere.

June 9, 2010 Posted by | Analysis, Measure Theory | 5 Comments