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

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

then we find

$\displaystyle\begin{aligned}(\tilde{E}_n\Delta E_n)\cap X_0&=\left(\left(E_n\setminus\bigcup\limits_{1\leq i<n}E_i\right)\cap X_0\right)\Delta\left(E_n\cap X_0\right)\\&=\left(\left(E_n\cap X_0\right)\setminus\bigcup\limits_{1\leq i<n}\left(E_i\cap X_0\right)\right)\Delta F_n\\&=\left(F_n\setminus\bigcup\limits_{1\leq i<n}F_n\right)\Delta F_n\\&=F_n\Delta F_n=0\end{aligned}$

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.

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

3 Comments »

You mean “If $X_0$ is itself measurable (as it often is)”

Comment by Tom Ellis | April 28, 2010 | Reply
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.

Comment by John Armstrong | April 28, 2010 | Reply
[...] 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 [...]

Pingback by Measurable Subspaces III « The Unapologetic Mathematician | April 29, 2010 | Reply

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