## 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.

[...] 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 [...]

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[...] 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 [...]

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someone ask me to define equicontinuous sequence of function

Comment by clemmy | June 20, 2012 |