The attentive reader will note that in our study of Lebesgue measure we’ve defined it on some complete -algebra . In general, there’s no reason to believe that such a -algebra would be all of , and so there’s some interesting structure worth investigating. Our intuition is amazingly bad at conceiving of just how bizarre and perverse a “general” subset of the real line can be, and so the facts we learn about Lebesgue measurable sets seem sort of opaque when we first see them, but they will turn out to be useful.
First, let’s recall where they come from. There’s an outer measure associated with Lebesgue measure , and then there is the collection of sets measurable by . These are the sets so that for any subset we have
If is a Lebesgue measurable set of positive, finite measure, and if is a real number, then there exists some open interval so that
That is, Lebesgue measurable sets do a pretty good job of covering sections of the real line. If we think of as a measure of efficiency, we can always find some interval so that covers at least that portion of it.
Let is the class of open sets. We already know that
So we can find some open containing and satisfying . This will be the disjoint union of a sequence of open intervals , and so it follows that
We must have for at least one of these summands, and this gives us the we were looking for.
If is again a Lebesgue measurable set with positive measure, then there exists an open interval containing and entirely contained in the set of “differences” in . That is, the set defined by
If contains an open interval itself, then this is trivially true. We can just take the desired interval to be , which clearly falls into the difference set . But general Lebesgue measurable sets don’t contain open intervals. And yet this is not just a collection of scattered points, because then it would have measure zero. We have points clustered closely enough to fill up area, but not enough to actually fill out any open intervals. This is, in fact, supremely weird if you really think about it.
But even if doesn’t fill up an interval, it comes close! The above result tells us that we can find a so that
Now pick some . Take the set , slide it over by adding to every point in the set, and call the result . The set is now contained in an open interval — — whose length is less than . If and were disjoint, then they’d have to have the same measure, and their union would have measure greater than and wouldn’t fit into this interval. So and must overlap a bit.
But this tells us that there’s a point that’s equal to the point . Thus is a point in . And so contains the whole interval , as we wanted.