## Measurable Subspaces II

Last time we discussed how to define a measurable subspace of a measurable space in the easy case when is itself a measurable subset of : .

But what if isn’t measurable as a subset of ? To get at this question, we introduce the notion of a “thick” subset. We say that a subset of a measure space is thick if for all measurable . If is itself measurable (as it often is), this condition reduces to asking that . If, further, , then we ask that . As an example, the maximally nonmeasurable set we constructed is thick.

Now I say that if is a thick subset of a measure space , if consists of all intersections of with measurable subsets of , and if is defined by , then is a measure space. This definition of is unambiguous, since if and are two measurable subsets of with , then . The thickness of implies that , and we know that

Since , the second term must be zero, and so . Therefore, , and is indeed unambiguously defined.

Now given a pairwise disjoint sequence of sets in , define to be measurable sets so that . If we define

then we find

and so . Therefore

which shows that is indeed a measure.

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

Comment by Tom Ellis | April 28, 2010 |

No, I mean . Go back to where I defined a measurable space and notice that I specifically did not require that is itself measurable.

Comment by John Armstrong | April 28, 2010 |

[…] 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 […]

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