## Mild Introduction.

Last time, we described some facts about Lebesgue measure and showed that despite being a nice measure of sets on the real line there are some sets which cannot be measured by the Lebesgue measure.  Intuitively, this gives us a big contrast with “real world” objects — there are very few things (except for the very, very small and the very, very large, perhaps) which we cannot hold a ruler next to and measure or, at least, approximate.  The previous set we constructed could not even be approximated — or, at least, it is not obvious how to approximate such a beast!

On the other hand, we will show that there are sets which, informally speaking, the Lebesgue measure “doesn’t care” about.  These sets are called sets of measure zero, and they’re quite easy to describe.

Definition.  We say that $A$ is a set of measure zero if $\mu(A) = 0$.

These sets will act like a “zero” element with measures.  If we have some set $B$, then it will turn out that if $A$ has measure zero, we will show that $\mu(B) = \mu(B\cup A)$.  This is more or less what I mean when I say that the Lebesgue measure does not care about these sets.

[Note: from now on, I will use the term “measurable” to mean “Lebesgue measurable” and I will qualify when I don’t mean this.]

## Last time, and some Equivalent Statements.

Last time we talked about the Lebesgue measure in terms of outer and inner measure.  Recall that outer measure of a set is this "almost-measure" that we made up by taking the infimum of sets that looked like disjoint intervals which covered the set we’re measuring.  Recall that we worked in the unit interval $[0,1]$ and recall that $E\subseteq [0,1]$ is an elementary set if it is a countable union of intervals (half-open, open, or closed).  Then we can better write outer measure as

$\displaystyle \mu^{\ast}(A) = \inf_{A\subseteq E}\{\mu^{\ast}(E)\}$

and we also had a corresponding "inner measure" which was another almost-measure, defined in the following way:

$\mu_{\ast}(A) = 1 - \mu^{\ast}([0,1]-A)$

and we pieced these together to make an actual measure, called the Lebesgue measure,  which is defined on every set where the inner and outer measures match up.  In other words, if $\mu^{\ast}(A) = \mu_{\ast}(A)$, then $\mu(A)$ exists and $\mu(A) = \mu^{\ast}(A)$

## Review and Motivation!

Last time we constructed the notion of a measure in a pretty general setting.  What was it?  It was a function from a nice set of subsets (a sigma-algebra) to the non-negative reals union infinity that satisfied a relatively reasonable condition: if we were to union up a bunch of disjoint measurable sets, the measure of the union should be the sum of the measure of the sets.  That’s a pretty general kind of function.  But even the most unobservant of readers wouldn’t be able to help but notice my lack of creativity when talking about measures — they all related to the real line!  The reader should note at this point that there are other kinds of measures out there; in fact, there are measures which are wild and crazy and have nothing to do with the intuitive notion of "length"!  We may be touching on some of these measures in later posts.

The real line is comfortable.  We know a lot about it.  We can picture it.  We can even apply it to things in the real world.  Thus, if we start talking about measures, a good place to start building them would be on the real line.  And if measures are supposed to, somehow, be this generalization of "length" or "size", then what better a way to make a measure on the subsets of the real line than to have our measure give us the total length of the subset!

And this is the general idea behind Lebesgue measure.  So what’s the big deal?  Well, remember when we defined measures before, we needed to define them on a sigma-algebra.  It turns out that this kind of measure doesn’t work so well on the powerset of the reals: indeed, there are subsets of the real line which are not measurable (!) in the way that we just vaguely defined it.  The first time I learned about this, it blew my mind.  And it still does.  How could something so simple go wrong?

Maybe we are expecting too much.  Let’s start modestly.  At the very least, we can talk about intervals — we know what length we’d like to assign to them.

[In this post, I have made liberal use of additional notes, which I will put in-between braces and in italics, just like the sentence you are reading right now!  These additional notes can be skipped with no major loss of understanding the subject, and only provide additional rigor or verify statements which are necessary but break the flow of the post.  As I’ve mentioned before in this blog, my intention is NOT to write the Great American Mathematics text book, but simply give motivation and a general argument as to how we do what we do.  Last, this post, for the curious reader, follows Kolmogorov’s Introduction to Real Analysis text, section 25. My exposition is slightly different, but I don’t think there should be any problems with it.]

## Introduction.

Every so often we do things not because we want to, but because we must.  It is not a secret that I do not like Analysis.  It may be because, as some of my peers suggest, I don’t truly understand it.  I don’t deny this claim.  It may be that I don’t see "the point" to much of it.  I’m not sure.  Analysis, to me, is like a rigid building where everything must be just so — I contrast this with topology, where things seem somehow more fluid and less strict…

But enough whining.  These posts will go through the various topics in analysis, perhaps stopping along the way to offer solutions to various related problems.  To be perfectly honest with the reader, the reason I am doing this is because I need to take a Qualifying exam in Analysis in the Fall and I need to review a lot of this material.  Therefore, if you find an error — and there may be some! — do not hesitate to tell me.

## Measures: The Size of Things.

In Topology, there is not usually a notion of size, in that even very "large" looking things can be homeomorphic to very "small" looking things — the real line and the unit open interval, for example, are homeomorphic but the real line is "much bigger looking" than the unit interval.  If topology ruled the world then things might get messy: if some fabric costs $5 for one yard, then via some homeomorphism you could potentially get an infinite number of yards of fabric for$5.  This is not a great way to run a business.  But notice that even in this example there is some notion of measurement: a yard is an interval of a specific size which (relativity and such aside) which does not change from place to place; indeed, a yard of fabric is the same "length" as a yard of fish.

## Cute Proofs: Closure of a Connected Subspace is Connected.

### May 16, 2011

[Big Edit: Thank you to the commenters below (and especially the latest one) who pointed out some of my previous statements only held true in a specific case.  I’ve streamlined the proof, and I think everything is kosher now.  I’ve also added some counterexamples to the end.]

I was reading through some of the earlier chapters of Bredon and I got to one of those questions that should be easy but I couldn’t recall the argument off of the top of my head.  The actual question is slightly more general, but the question we’ll answer first is:

Problem: If $A$ is connected, show $\bar{A}$ is connected.

There are several nice definitions of connected to choose from.  In particular, we have:

A subset $A\subseteq X$ is connected if one (and hence all) of the following hold:

• There DO NOT exist disjoint non-empty closed (with respect to $A$) sets $C,D$ such that $C\cap A\neq\emptyset$ and $D\cap A\neq \emptyset$ but $C\cup D = A$.
• The only subsets of $A$ which are both open and closed in the relative topology are $\emptyset$ and $A$.

[Note:  In a former draft of this post, I had (incorrectly!) stated that a subset $A$ is connected if we did not have two disjoint open sets which do not intersect $A$ trivially, but which union to $A$.  This can be shown to be false (which was done by a clever commenter below), but it will work if $X$ (and therefore $A$ in the subspace topology) is a metric space.]

To solve this, we will use the first statement above but, for fun, try to prove it yourself using the second statement.  We’ll be using the contrapositive, meaning we will prove if $\bar{A}$ is not connected, then $A$ is not connected.  This is equivalent to proving the theorem above (why?).

Solution.  Let it be the case that $\bar{A}$ is not connected, so that there exists $C,D$ disjoint closed sets neither of which intersect $\bar{A}$ trivially.   Suppose that $A$ is connected; then for any two disjoint closed sets $C,D$ with $A\subseteq C\cup D$ we have that one of these closed sets intersects $A$ trivially (or else it would not be connected).  In other words, wlog we have $A\subseteq C$.  Taking the closure of both sides, we have $\bar{A}\subseteq \bar{C} = C$, but since $C$ and $D$ are disjoint this implies that $\bar{A}\cap D = \emptyset$, contradicting the fact that neither $C$ nor $D$ may intersect $\bar{A}$ nontrivially.  $\Box$

In fact, nearly the same kind of argument shows that if there is some $B$ such that $A\subseteq B\subseteq \bar{A}$ with $A$ connected, then $B$ is also connected.  This is pretty nice; it tells us we can add limit points to our connected space (even if we only add some limit points and not others!) and it remains connected.

### Some counter examples…

One might ask, "Hey, what if $\bar{A}$ is connected?  Then is $A$ connected?"  And a minute of thinking should provide a counter-example.  If not, then just think about $A = {\mathbb Q}$ and its closure.

Additionally, one might think that if a set $A$ is connected, then its interior (if it is nonempty)  would be connected.  This is true in the case of the closed disk $D\subset {\mathbb R}^{2}$, but if we were to put two disjoint disks in the plane and connect them together with a line (this will look like a barbell, or a pair of eyeglasses) the interiors will be two disjoint open disks, which are clearly not connected.

Maybe, though, the boundary of a connected set $A$ is always connected (if it is nonempty).  To do away with this idea, just think about the interval in ${\mathbb R}$.

A question to think about (assume all of these sets are non-empty):

• If the boundary of some set $A$ is connected, is $A$ connected?  (For this one, think about a figure-8 in the plane.)
• If the interior of some set $A$ is connected, is $A$ connected?  (For this one, think about what the interior of a point is, and then let $A$ be some nice set with non-empty interior union some point.)
• If the interior and boundary of some set $A$ are connected, is $A$?  (For this one, you might have to think a bit.  I’ll try to hide an answer a little bit so you can work on it.  Consider the set of positive rational numbers, ${\mathbb Q}_{+}$ union the interval $[0,1]$.  The interior will be $(0,1)$ which is connected, and the boundary will be $[0,\infty)$ which is connected,  but the set itself is not connected (for the same reason the rationals are not connected).