# The Unapologetic Mathematician

## 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