Convergence in Measure and Algebra
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 .