## Measurable Covers

Again, let be a ring, let be the smallest -ring containing , and let be the smallest hereditary -ring containing . Given a measure on , it induces an outer measure on , which restricts to a measure on .

The sets in are easily described — they’re everything that can be countably covered by sets in — but we don’t have a measure on this collection. Instead, we need good approximations of these sets by sets in , where we *do* have a measure. And so we say that a set is a “measurable cover” of if , and if for every with we have . That is, may be larger than , but only negligibly.

If has -finite outer measure, then it has a measurable cover with . Indeed, our hypothesis is that can be written as the countable disjoint union of sets of finite outer measure, so if we can show this for sets of finite measure the -finite case will follow.

By the comparisons we showed last time, for every there is some set so that and . Taking to be the intersection of this sequence, we must have

Since is arbitrary, we must have . If fits into , then . And so:

Since we can subtract it off and find .

In fact, any measurable cover must have . Further, any two of them have a negligible difference.

Indeed, if we have two measurable covers, and , then , which tells us that . But since is a measurable cover, we conclude that , and similarly that . But is the union of these two differences, and so .

If , then any measurable cover must have as well. On the other hand, if is finite, then there exists at least one measurable cover with . Then any other measurable cover differs negligibly, and so has the same measure.

And so if the measure on is -finite, then so are the measures on and on . We’ve already seen that must be -finite in this situation, and so any -measurable set can be covered by a countable sequence of -finite sets. Applying the above results to each of the sets in this sequence gives our result.