To recap: we’ve got a measure space 
and we’re talking about what structure we get on a subset 
. If is measurable — if 
— then we can set 
=
. Since each 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 so we can’t do the same thing. We saw, though, that we can define 
.
So what if is neither measurable nor thick? It turns out that if we want to use this latter method of defining 
, must be thick! In particular, in order to prove that is well-defined we had to show that if 
and 
are two measurable subsets of with 
, then 
. I say that if implies for any two measurable sets and , then must be thick.
To see this, first take any measurable set 
and pick another one 
. Then we see that 
, and so our hypothesis tells us that 
. Since 
, the subtractivity of 
tells us that 
, and we conclude that 
. That is, every measurable set 
that fits into 
must have measure zero, and thus 
— as the supremum of the measures of all these sets — must be zero as well.
And so we see that in order for this to be unambiguously defined, we must require that be thick.
April 29, 2010
Posted by John Armstrong |
Analysis, Measure Theory |
Leave a Comment
