The Unapologetic Mathematician

Mathematics for the interested outsider

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.

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


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

    Comment by Tom Ellis | April 28, 2010 | Reply

  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.

    Comment by John Armstrong | April 28, 2010 | Reply

  3. […] 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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