The Unapologetic Mathematician

Mathematics for the interested outsider

Measurable Subspaces III

To recap: we’ve got a measure space (X,\mathcal{S},\mu) and we’re talking about what structure we get on a subset X_0. If X_0 is measurable — if X_0\in\mathcal{S} — then we can set \mathcal{S}_0=\mathcal{S}\cap X_0=\{M\cap X_0\vert M\in\mathcal{S}\}. Since each of these subsets M\cap X_0 is itself measurable as a subset of X, we can just define \mu_0(M\cap X_0)=\mu(M\cap X_0). On the other hand, if X_0 is nonmeasurable but thick we can use the same definition for \mathcal{S}_0. This time, though, the subsets M\cap X_0 may not themselves be in \mathcal{S}, and so we can’t do the same thing. We saw, though, that we can define \mu_0(M\cap X_0)=\mu(M).

So what if X_0 is neither measurable nor thick? It turns out that if we want to use this latter method of defining \mu_0, X_0 must be thick! In particular, in order to prove that \mu_0 is well-defined we had to show that if M_1 and M_2 are two measurable subsets of X with M_1\cap X_0=M_2\cap X_0, then \mu(M_1)=\mu(M_2). I say that if M_1\cap X_0=M_2\cap X_0 implies \mu(M_1)=\mu(M_2) for any two measurable sets M_1 and M_2, then X_0 must be thick.

To see this, first take any measurable set E and pick another one F\subseteq E\setminus X_0. Then we see that (E\setminus F)\cap X_0=E\cap X_0, and so our hypothesis tells us that \mu(E)=\mu(E\setminus F). Since F\subseteq E\setminus X_0\subseteq E, the subtractivity of \mu tells us that \mu(E)=\mu(E\setminus F)=\mu(E)-\mu(F), and we conclude that \mu(F)=0. That is, every measurable set F that fits into E\setminus X_0 must have measure zero, and thus \mu_*(E\setminus X_0) — as the supremum of the measures of all these sets — must be zero as well.

And so we see that in order for this \mu_0 to be unambiguously defined, we must require that X_0 be thick.

April 29, 2010 Posted by | Analysis, Measure Theory | Leave a comment