The Unapologetic Mathematician

Mathematics for the interested outsider

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


  1. […] functions converge (in the mean) to integrable limits. Yes, we have the definition and the characterization in terms of convergence in measure, but this theorem is often easier to […]

    Pingback by Lebesgue’s Dominated Convergence Theorem « The Unapologetic Mathematician | June 10, 2010 | Reply

  2. […] sequence of characteristic functions must converge in mean to some function . But mean convergence implies convergence in measure, which is equivalent to a.e. convergence on sets of finite measure, which is […]

    Pingback by Completeness of the Metric Space of a Measure Space « The Unapologetic Mathematician | August 9, 2010 | Reply

  3. someone ask me to define equicontinuous sequence of function

    Comment by clemmy | June 20, 2012 | Reply

  4. About combining equicontinuity and measures, let me put a question.

    Let $f: X\to X$ be a measurable map of a measurable space $X$.
    Define a measure $\mu$ to be {\bf absolutely equicontinuous w.r.t. $f$}
    if for every $\epsilon>0$ there is $\delta>0$ such that
    $\mu(A)<\delta$ implies $\mu(f^{-m}(A))<\epsilon$ for all $m\in \mathbb{N}$.

    For example, every invariant measure is equicontinuous w.r.t. $f$.

    My question is:
    Are there expansive homeomorphisms $f$ on decent compact metric spaces
    (e.g. ones without isolated points) exhibiting NON-INVARIANT Borel probability measures
    which are absolutely equicontinuous w.r.t. $f$?

    Thanks for watching!

    Comment by Chivo | October 24, 2014 | Reply

  5. Honestly, I don’t know offhand. I never was primarily an analyst, and I haven’t been around a professional math department in years.

    Comment by John Armstrong | October 24, 2014 | Reply

Leave a Reply

Fill in your details below or click an icon to log in: Logo

You are commenting using your account. Log Out /  Change )

Twitter picture

You are commenting using your Twitter account. Log Out /  Change )

Facebook photo

You are commenting using your Facebook account. Log Out /  Change )

Connecting to %s

%d bloggers like this: