## Convergence in Measure II

Sorry, I forgot to post this before I left this morning.

The proposition we started with yesterday shows us that on a set of finite measure, a.e. convergence is equivalent to convergence in measure, and a sequence is Cauchy a.e. if and only if it’s Cauchy in measure. We can strengthen it slightly by removing the finiteness assumption, but changing from a.e. convergence to almost uniform convergence: almost uniform convergence implies convergence in measure. Indeed, if converges to almost uniformly then for any two positive real numbers and there is a measurable set with and for all and sufficiently large . Thus we can make the set where and are separated as small as we like, as required by convergence in measure.

We also can show some common-sense facts about sequences converging and Cauchy in measure. First, if converges in measure to , then is Cauchy in measure. We find that

because if both and are within of the same number , then they’re surely within of each other. And so if we have control on the measures of the sets on the right, we have control of the measure of the set on the left.

Secondly, if also converges in measure to , then it only makes sense that and should be “the same”. It wouldn’t do for a convergence method to have many limits for a convergent sequence. Of course, this being measure theory, “the same” means a.e. we have . But this uses almost the same relation:

Since we can make each of the sets on the right arbitrarily small by choosing a large enough , we must have for every ; this implies that almost everywhere.

Slightly deeper, if is a sequence of measurable functions that is Cauchy in measure, then there is some subsequence which is almost uniformly Cauchy. For every positive integer we find some integer so that if

We define , and to be the larger of or , to make sure that is a strictly increasing sequence of natural numbers. We also define

If then for every x not in we have

That is, the subsequence is uniformly Cauchy on the set . But we also know that

and so is almost uniformly Cauchy, as asserted.

Finally, we can take this subsequence that is almost uniformly Cauchy, and see that it must be a.e. Cauchy. We write at all where this sequence converges. And then for every ,

The measure of the first set on the right is small for sufficiently large and by the assumption that is Cauchy in measure. The measure of the second approaches zero because almost uniform convergence implies convergence in measure.

And thus we conclude that if is Cauchy in measure, then there is some measurable function to which converges in measure. The topology of convergence in measure may not come from a norm, but it is still complete.

[…] sequences of integrable simple functions, then they’re both also Cauchy in measure, which implies that they each converge in measure to some function. If they converge to the same function (a.e.) , […]

Pingback by Indefinite Integrals and Convergence II « The Unapologetic Mathematician | June 1, 2010 |

a counter example of a sequence of functions on the real interval E=[0,1] would be the sequence of

indicator functions on the intervals [0,1/2] [1/2,1] [0,1/3] [1/3,2/3] [2/3,1] [0,1/4]…

it is convergent in measure to the null function but not pointwise convergent almost everywhere.

or am i wrong?

Comment by tom | September 13, 2011 |

I’m sorry, there’s a problem somewhere in here, but this is an ancient post, I don’t even remember the reference I was using, and I have a real job I have to attend to; I can’t fix this at the moment.

Comment by John Armstrong | September 13, 2011 |

Just for those who are (still) interested: Convergence in measure and a.e. convergence are not equivalent (even on sets of finite measure), as tom’s example shows. But the following holds:

If E is of finite measure and f_n converges a.e. to f, then f_n converges to f in measure.

Conversely, if f_n converges to f in measure, there exists a subsequence of f_n which converges to f a.e.

The existence of a subsequence in the latter implication is the critical point here.

Comment by Daniel | February 16, 2014 |