Measurable Subspaces III
To recap: we’ve got a measure space and we’re talking about what structure we get on a subset . If is measurable — if — then we can set =. Since each of these subsets is itself measurable as a subset of , we can just define . On the other hand, if is nonmeasurable but thick we can use the same definition for . This time, though, the subsets may not themselves be in , and so we can’t do the same thing. We saw, though, that we can define .
So what if is neither measurable nor thick? It turns out that if we want to use this latter method of defining , must be thick! In particular, in order to prove that is well-defined we had to show that if and are two measurable subsets of with , then . I say that if implies for any two measurable sets and , then must be thick.
To see this, first take any measurable set and pick another one . Then we see that , and so our hypothesis tells us that . Since , the subtractivity of tells us that , and we conclude that . That is, every measurable set that fits into must have measure zero, and thus — as the supremum of the measures of all these sets — must be zero as well.
And so we see that in order for this to be unambiguously defined, we must require that be thick.
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