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 on a class
is continuous from above at
if for every decreasing sequence of sets
with
we have
. If
is a sequence of set functions, then, we say the sequence is “equicontinuous from above at
” if for every sequence
decreasing to
and for every
there is some number
so that if
we have
. It seems to me, at least, that this could also be called “uniformly continuous from above at
“, but I suppose equicontinuous is standard.
Anyway, now we can characterize exactly how convergence in mean and measure differ from each other: a sequence of integrable functions converges in the mean to an integrable function
if and only if
converges in measure to
and the indefinite integrals
of
are uniformly absolutely continuous and equicontinuous from above at
.
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 converges in mean to
, then the indefinite integrals
are equicontinuous from above at
.
For every we can find an
so that for
we have
. 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
. And so for every sequence
of measurable sets decreasing to
there is some
so that for
we find
the first for all from
to
. Then if
we have
for every positive . We control the first term in the middle by the mean convergence of
for
and by the continuity from above of
for
. And so the
are equicontinuous from above at
.
Now we turn to the sufficiency of the conditions: assume that converges in measure to
, and that the sequence
of indefinite integrals is both uniformly absolutely continuous and equicontinuous from above at
. We will show that
converges in mean to
.
We’ve shown that is
-finite, and so the countable union
of all the points where any of the are nonzero is again
-finite. If
is an increasing sequence of measurable sets with
, then the differences
form a decreasing sequence
converging to
. Equicontinuity then implies that for every
there is some
so that
, and thus
For any fixed we define
and it follows that
By convergence in measure and uniform absolute continuity we can make the integral over arbitrarily small by choosing
and
sufficiently large. We deduce that
and, since was arbitrary, we conclude
Now we can see that
and thus
and, since is arbitrary,
That is, the sequence is Cauchy in the mean. But we know that the
norm is complete, and so
converges in the mean to some function
. But this convergence in mean implied convergence in measure, and so
converges in measure to
, and thus
almost everywhere.
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someone ask me to define equicontinuous sequence of function
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!
Honestly, I don’t know offhand. I never was primarily an analyst, and I haven’t been around a professional math department in years.