Unlike our other methods of convergence, it’s not necessarily apparent that convergence in measure plays nicely with algebraic operations on the algebra of measurable functions. All our other forms are basically derived from pointwise convergence, and so the limit laws clearly hold; but it takes some work to see that the same is true for convergence in measure. So, for the rest of this post assume that and are sequences of finite-valued measurable functions converging in measure to and , respectively.
First up: if and are real constants, then converges in measure to . Indeed, we find that
Thus if and , then . That is,
Since and converge in measure to and , we can control the size of each of these sets by choosing a sufficiently large , and thus converges in measure to .
Next, if a.e., the sequence converges in measure to . Indeed, the set differs negligibly from the set . This, in turn, is exactly the same as the set , which differs negligibly from the set . Thus control on the measure of one of these sets is control on all of them.
Now we’ll add the assumption that the whole space is measurable, and that is finite (that is, the measure space is “totally finite”). This will let us conclude that the sequence converges in measure to . As the constant increases, the measurable set gets larger and larger, while its complement gets smaller and smaller; this complement is measurable because is measurable.
In fact, the measure of the complement must decrease to zero, or else we’d have some set of positive measure on which is larger than any finite , and thus on a set of positive measure. But then couldn’t converge to in measure. Since is totally finite, the measure must start at some finite value and decrease to zero; if were infinite, these measures might all be infinite. And so for every there is some so that .
In particular, we can pick a so that . On , then, we have . Convergence in measure tells us that we can pick a large enough so that
has measure less than as well. The set must be contained between these two sets, and thus will have measure less than for sufficiently large .
Now we can show that converges in measure to for any , not just ones that are a.e. zero. We can expand , and thus rewrite . Our first result shows that converges to , and our second result then shows that also converges to . Our third result shows that converges to . We use our first result to put everything together again and conclude that converges to as we asserted.
And finally we can show that converges in measure to . We can use the same polarization trick as we’ve used before. Write ; we’ve just verified that the squares converge to squares, and we know that linear combinations also converge to linear combinations, and so converges in measure to .
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.