## Associated Metric Spaces and Absolutely Continuous Measures II

Yesterday, we saw that an absolutely continuous finite signed measure on a measure space defines a continuous function on the associated metric space , and that a sequence of such finite signed measures that converges pointwise is actually uniformly absolutely continuous with respect to .

We’re going to need to assume that is nonnegative. We’d usually do this by breaking into its positive and negative parts, but it’s not so easy to get ahold of the positive and negative parts of in this case. However, we *can* break each into and . Then we can take the limits and , which will still satisfy . The only difference between this decomposition and the positive and negative parts is that this pair of set functions might have some redundancy that gets cancelled off in this subtraction. And so, without loss of generality, we will assume that all the are nonnegative, and that their limit is as well.

Now, given such a sequence, define the limit function . I say that is itself a finite signed measure, and that . Indeed, is finite by assumption, and additivity is easy to check. As for absolute continuity, if , then each since , and so as the limit of the constant zero sequence.

What we need to check is continuity. We know that it suffices to show that is continuous from above at . So, let be a decreasing sequence of measurable sets whose limit is . We must show that the limit of is zero. But we know that the limit of is zero, and thus for a large enough we can make for any given . And since we know that for any there is some so that if then . Thus we can always find a large enough to guarantee that , and so the limit is zero, as asserted.

Finally, what happens if we remove the absolute continuity requirement from the ? That is: what can we say if is a sequence of finite signed measures on so that exists and is finite for each . I say that is a signed measure. What we need is to find some measure so that all the , and then we can use the above result.

Since is a finite signed measure, we can pick some upper bound . Then we define

If any , then , and so . And thus for all , as desired.

No comments yet.

## Leave a Reply