## Real Analysis Primer, Part 5: Measurable Sets are Pretty Close To Stuff We Like.

### June 27, 2011

I was going to write about measurable functions, but after a little discussion about measurable sets with a friend, I decided to put that off for a bit.  The discussion had to do with the nature of measurable sets and how weird they could look.  He notes (paraphrased!):

"I mean, open sets are measurable.  Those are nice.  And closed sets are measurable.  Those are nice."

"Uh-huh."

"But there’s all sorts of sets which aren’t open or closed which are measurable.  I mean, there’s sorts of sets which don’t even look like finite unions of open and closed sets."

"Uh-huh."

"So to say that measurable sets are ‘nice’ in some way is really not an accurate statement.  We can make a measurable set as not-nice as we want it!"

And the reader may also feel this way: yes, measurable sets may look crazy, but up to an arbitrary degree, we can make them look like open or closed sets, or things which we can describe with unions and intersections of open or closed sets.  Saying that some measurable sets are ‘not nice’ to some degree is even implicitly saying that measurable sets must be ‘nice’ to some other degree!  The degree to which they are nice can be made precise, in fact, and this is what we’ll do in this short post.

## A Set Is Measurable If…

The following theorem is one which comes up in nearly every book with measure theory in it.  It states a bunch of equivalent definitions for a set $A$ to be (Lebesgue) measurable.  I’m going to break it up a bit, because all of the equivalent definitions can look intimidating to someone who is reading it for the first time!

Theorem.  The following are equivalent.

1. $A$ is Lebesgue measurable with measure $\mu(A) < \infty$.
2. There exists an open set $U$ such that $A\subseteq U$ such that $\mu^{\ast}(U) = \mu^{\ast}(A) + \epsilon$.
3. Call a set $V$ an $G_{\delta}$ set if it is a countable intersection of open sets (note that this is not necessarily open!).  There exists an $G_{\delta}$ set $V$ such that $\mu^{\ast}(V - A) = 0$

A few things to note.  First, $G_{\delta}$ sets appear elsewhere in mahematics (and the "G" stands for the German word "Gebiet", meaning neighborhood, and the $\delta$ stands for the German word "Durchschnitt" which means intersection).  The idea here is that we may arbitrarily closely approximate $A$ with some open set, but in order to get a really, really close approximation (one where our set and the approximation differ by a set of measure zero) we need to allow countable intersections instead of just finite ones.

Proof.  The first part, that (1) implies (2) is relatively easy.  In fact, this is how we defined the measure essentially.  If $A$ is measurable, then it is the infimum of some elementary sets.  Let each element of this elementary set be an open interval (since, at most, this will differ by a set of measure zero by some other elementary set with non-open intervals) and then we have that for any $\epsilon > 0$ there will be some elementary set $U$ which is open, contains $A$, and such that $\mu^{\ast}(U) = \mu^{\ast}(A) + \epsilon$

The second part, that (2) implies (3), is a little trickier.  But it’s not that bad once you know the trick.  Set $\epsilon = \frac{1}{n}$.  Now find $U_{n}$ such that $\mu^{\ast}(U_{n}) = \mu^{\ast}(A) + \frac{1}{n}$.  Taking $\bigcap_{i=1}^{\infty}U_{n} = V$, we note $V$ is a $G_{\delta}$ set, and, moreover (this takes a bit of work, but not much.  In particular, it can be done by approximating the intersection up to $N$ and noting that we have a nested sequence) that the intersection of these will leave us with $\mu^{\ast}(V - A) = 0$, mainly because $\frac{1}{n}\rightarrow 0$

The last part, that (3) implies (1) is a bit trickier.  We note that sets of measure zero are measurable, so $V - A$ is measurable.  We also note the difference of measurable sets (which we may not have shown yet, but it is true) is also measurable, and you should also be able to reason that ever $G_{\delta}$ set is measurable.  Hence, $V$ is measurable, and so $V - (V - A) = A$ is measurable.  $\Box$

There are, as you’d expect, corresponding closed set stuff that you can work out yourself (it’s really just taking compliments), and there’s a neat one about symmetric difference if the measure is finite — it’s in Royden, Kolmogorov, and many more, so just look in those texts for it.  Note that above we never really assumed that the measure was finite, so this works for sets regardless of measure.  Neat.

The point, then, is that measurable sets are nice.  They’re not all that crazy!  Next time, we’ll delve down deep into measurable functions.  For anyone who is interested in topology, they will seem like familiar beasts with a slight twist!  Exciting.