The Unapologetic Mathematician

Measurable Subspaces II

Last time we discussed how to define a measurable subspace $(X_0,\mathcal{S}_0)$ of a measurable space $(X,\mathcal{S})$ in the easy case when $X_0$ is itself a measurable subset of $X$: $X_0\in\mathcal{S}$.

But what if $X_0$ isn’t measurable as a subset of $X$? To get at this question, we introduce the notion of a “thick” subset. We say that a subset $X_0$ of a measure space $(X,\mathcal{S},\mu)$ is thick if $\mu_*(E\setminus X_0)=0$ for all measurable $E\in\mathcal{S}$. If $X$ is itself measurable (as it often is), this condition reduces to asking that $\mu_*(X\setminus X_0)=0$. If, further, $\mu(X)<\infty$, then we ask that $\mu^*(X_0)=\mu(X)$. As an example, the maximally nonmeasurable set we constructed is thick.

Now I say that if $X_0$ is a thick subset of a measure space $(X,\mathcal{S},\mu)$, if $\mathcal{S}_0=\{M\cap X_0\vert M\in\mathcal{S}\}$ consists of all intersections of $X_0$ with measurable subsets of $X$, and if $\mu_0$ is defined by $\mu_0(M\cap X_0)=\mu(M)$, then $(X_0,\mathcal{S}_0,\mu_0)$ is a measure space. This definition of $\mu_0$ is unambiguous, since if $M_1$ and $M_2$ are two measurable subsets of $X$ with $M_1\cap X_0=M_2\cap X_0$, then $(M_1\Delta M_2)\cap X_0=\emptyset$. The thickness of $X_0$ implies that $\mu_*((M_1\Delta M_2)\cap X_0^c)=0$, and we know that

$\displaystyle\mu_*((M_1\Delta M_2)\cap X_0^c)+\mu^*((M_1\Delta M_2)\cap X_0)=\mu(M_1\Delta M_2)$

Since $(M_1\Delta M_2)\cap X_0=\emptyset$, the second term must be zero, and so $\mu(M_1\Delta M_2)=0$. Therefore, $\mu(M_1)=\mu(M_2)$, and $\mu_0$ is indeed unambiguously defined.

Now given a pairwise disjoint sequence $\{F_n\}$ of sets in $\mathcal{S}_0$, define $\{E_n\}\subseteq\mathcal{S}$ to be measurable sets so that $F_n=E_n\cap X_0$. If we define

$\displaystyle\tilde{E}_n=E_n\setminus\bigcup\limits_{1\leq i

then we find

\displaystyle\begin{aligned}(\tilde{E}_n\Delta E_n)\cap X_0&=\left(\left(E_n\setminus\bigcup\limits_{1\leq i

and so $\mu(\tilde{E}_n\Delta E_n)=0$. Therefore

\displaystyle\begin{aligned}\sum\limits_{n=1}^\infty\mu_0(F_n)&=\sum\limits_{n=1}^\infty\mu(E_n)\\&=\sum\limits_{i=1}^\infty\mu(\tilde{E}_n)\\&=\mu\left(\bigcup\limits_{i=1}^\infty\tilde{E}_n\right)\\&=\mu\left(\bigcup\limits_{i=1}^\infty E_n\right)\\&=\mu_0\left(\bigcup\limits_{i=1}^\infty F_n\right)\end{aligned}

which shows that $\mu_0$ is indeed a measure.

April 28, 2010 - Posted by | Analysis, Measure Theory

1. You mean “If $X_0$ is itself measurable (as it often is)”
2. No, I mean $X$. Go back to where I defined a measurable space and notice that I specifically did not require that $X$ is itself measurable.