Real Analysis Primer, part 1: Measures and Sigma-Algebras.

May 26, 2011


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.

To begin talking about analysis, we must talk about measures. 

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. 

On the real line it is not complicated to make a measure that corresponds with what our expectation of a measure should be.  What is the measure of the closed interval between 0 and 1?  Well, that is 1 unit long.  What about between 0 and 2?  Well, that’s 2 units long.  And so forth.  But perhaps something not so obvious: what is the measure of (0,1)?  What’s the measure of [0,1)?  Can these measures be the same, even if there is just a "little bit more" on the second one?  We may be tempted to say that it’s such a small amount that it really doesn’t matter, and just say that every point has length 0.  But then isn’t [0,1] just an uncountable number of points?  So isn’t this also of length 0?  What’s the deal?  We need to be more careful.

To begin constructing a measure, we need to think about what it should be.  We would really like the measure of some set to be a real number — in fact, for now, we’d really like it to be a non-negative real number (though we may extend this later).  So our measure \mu should take a set A and spit out \mu(A) which is the "measure of A".  Note that we will also allow A to have infinite measure and we will write \mu(A) = \infty in this case.  But what is the domain of this function \mu?  To be absolutely general here we must mention \sigma-algebras.



If we had a measure on some set, what would we want?  If we could measure some set A and some other set B, then we’d want to be able to measure both of them together (in other words, A\cup B) and if we knew the measure of the entire set X that A was sitting inside, we’d like to be able to measure X without A; in other words, we’d like to be able to measure the compliment of A.  We need to be careful, though, since taking uncountable unions (say, of points on the real line) will give us some not-so-nice properties in our measure, so we’ll stop at countable unions.  Using these rules, let’s take some subsets of X that would be satisfy these things.  This kind of structure actually has a name, too: it’s called a \sigma-algebra.

Definition.  A subset \Sigma  of the powerset of the set X is called a \sigma-algebra if it satisfies the following properties:

  1. It is nonempty (this is just so that we don’t wind up with anything trivial).
  2. \Sigma is closed under taking compliments.  If A\in \Sigma then so is A^{C} = X - A
  3. \Sigma is closed under countable unions.  If A_{1}, A_{2}, \dots are in \Sigma, then so is their union. 


Notice that we’ve said nothing about intersections.  In fact, countable intersections ARE in the \sigma-algebra, but it is redundant to require them to be:  if A_{i}\in \Sigma for i\in {\mathbb N} then so are A_{i}^{C} their compliments, and hence \bigcup_{i}A_{i}^{C}\in \Sigma but by de Morgan’s laws this is the same as \left(\bigcap_{i} A_{i}\right)^{C} the compliment of the intersection, and the compliment of this is just \bigcap_{i}A_{i} which gives us countable intersections.

Notice also that to make a sigma algebra, the "best" we can do is to take the powerset of the set.  This will give us the "biggest" \sigma-algebra over our set.

The \sigma-algebra we will most often take will be the following: let X be a topological space (think about the reals if you’d like) and take the set of open sets in X.  There is a smallest such \sigma-algebra that contains these open sets: this is the one we will often refer to.


Back to Measures.

Since we’d like the measure to be defined on a whole bunch of subsets, it makes sense that we’d take our set X and then take an associated \sigma-algebra over X to be the domain of the measure.  For example, in the reals we could take the smallest \sigma-algebra generated by the open sets in {\mathbb R} with the standard topology (and we will use this example for most of the rest of Lebesgue theory —).

Thus, so far we have for X is a set and \Sigma is a \sigma-algebra over X, a measure must be a function \mu:\Sigma\rightarrow {\mathbb R}_{+,0}\cup\{\infty\} where the latter space is the non-negative reals along with \infty.  But we’d like our measure to have a bit more structure: as is it, we could make it whatever we wanted so long as it was a function.  A measure ought to have certain properties that are nice and help us differentiate the sizes of sets.

The first property is non-negativity.  We’ve already talked about this.  Also, we need to say that the empty-set has measure 0 (since what other measure would we want to give it?).  Last, if we have a countable number of disjoint sets A_{i} for i\in {\mathbb N}, then we’d want their measures to sum up nicely, right?  In other words, the measure of the union should be the sum of their measures.  (If you have a 3 yard piece of string and a 5 yard piece of string, together you’d want to have 8 yards.)

At this point, we can finally define a measure in some legit way.


Definition. Let X be a set and \Sigma is a \sigma-algebra over X. A measure is a function \mu:\Sigma\rightarrow {\mathbb R}_{+,0}\cup\{\infty\} such that the following  properties hold:

  1. \mu(A) \geq 0 for every A\in \Sigma and \mu(\emptyset) = 0.
  2. If \{A_{i}\}_{i\in {\mathbb N}} are a countable collection of (pairwise) disjoint sets in \Sigma then \displaystyle \mu\left(\bigcup_{i}A_{i}\right) = \sum_{i}\mu(A_{i}).


Note that we do not say that if the measure of some set is equal to 0 then that set must be the empty set.  The converse of the first property isn’t true, in other words.

Also, one last definition so I don’t have to keep writing \Sigma.  These sets in our \sigma-algebra are exactly the sets we can define a measure on, and so we call them measurable sets.


Definition.  If X is a set and \Sigma is a \sigma-algebra over X, then we call A\in \Sigma a measurable set.



Before we finish up this post, there are a few properties we can derive just from what we have here.  Let’s do these.


Proposition.  \mu is monotonic; that is, if A\subseteq B are measurable, then \mu(A)\leq \mu(B)


Proof.  Note that B = A\cup (A^{C}\cap B).  These sets are disjoint, so we have that

\displaystyle \mu(B) = \mu(A \cup (A^{C}\cap B)) = \mu(A) + \mu(A^{C} \cap B).

We don’t even need to figure out what the value of the last term here is, since we know that \mu(A^{C} \cap B) \geq 0.  Then we have \mu(A) plus a non-negative term is equal to \mu(B); this implies directly that \mu(A) \leq \mu(B)


Proposition.  \mu is a countably subadditive function.  That is, if \{A_{i}\}_{i} is a countable not-necessarily-disjoint collection of measurable sets, then

\displaystyle\mu\left(\bigcup_{i=1}^{\infty}A_{i}\right) \leq \sum_{i=1}^{\infty}\mu(A_{i}).

Proof.  The idea of the proof is really just to make these sets disjoint sets.  Do this in the following way (and remember the process, since it comes up a bunch):

Let B_{1} = A_{1}.

Let \displaystyle B_{n} = A_{n} - \bigcup_{i=1}^{n-1}A_{i}.

Then if r > s without loss of generality,

\displaystyle B_{r} \cap B_{s} = \left(A_{r} - \bigcup_{i=1}^{r-1}A_{i}\right) \cap \left(A_{s} - \bigcup_{i=1}^{s-1}A_{i}\right) = \emptyset

so we have that the B_{i} are pairwise disjoint and moreover,

\displaystyle\bigcup_{i=1}^{\infty}B_{i} = \bigcup_{i=1}^{\infty} \left(A_{i} - \bigcup_{j=1}^{i-1}A_{j}\right) = \bigcup_{i=1}^{\infty}A_{i}.

So the point of that construction was to make a pairwise disjoint collection of measurable sets that union to the same set as the not-pairwise-disjoint collection.  Notice that, also, B_{n} \subseteq A_{n} for every n and hence \mu(B_{n})\leq \mu(A_{n}).  Now we may apply this and our countable union measure property:

\displaystyle\mu\left(\bigcup_{i=1}^{\infty}A_{i}\right) = \mu\left(\bigcup_{i=1}^{\infty}B_{i}\right) = \sum_{i=1}^{\infty}\mu(B_{i}) \leq \sum_{i=1}^{\infty}\mu(A_{i})

which completes the proof.  \Box

6 Responses to “Real Analysis Primer, part 1: Measures and Sigma-Algebras.”

  1. This is an interesting set of definitions–not the one’s I’m used to. Surely you are defining an outer measure opposed to an inner measure are you not going to consider inner measures in your posts? Also, your definition of measurable is interesting. So, you’re taking some topological space and considering the Borel algebra on the space and defining the measurable sets to be those in the Borel algebra? Doesn’t one usually define a measure first and then create a sigma algebra of measurable sets by considering the sets for which the Carathéodory condition holds?

    Out of curiosity, what book are you using a reference? I look forward to reading your posts on measure theory!

    Also, I sort of know what you mean about not liking analysis. The way I’d put it is that I like qualitative math instead of quantitative math (viz. soft vs. hard analysis). But, some really fascinating structure related math (like differential geometry and differential topology) can’t really avoid doing a fair amount of hard analysis implicitly in their work, right?

    • James said

      For the analysis classes that I’ve taken, usually outer measure is taken as “the big deal” and to be measurable means something with respect to this outer measure. I’m not sure why this is done. There may be some super-important reason, but it’s eluded me. Introducing the general measure first was just a pedagogical choice — in one of the next posts, we will define outer and inner measure using this general definition of a measure. Rudin’s complex book does some of this (in terms of defining measurable sets to be a sigma-algebra) and Kolmogorov’s book does something similar, but he uses semi-rings in a way which is not obvious to me right now.

      I don’t know how I feel about defining the sigma-algebra first verses defining the measure first. When I imagine this, I imagine the measure has a lot to do with the topology of the space, so the Borel construction is natural to do first and then just define some measure with some nice properties. Even if there is no topology, the way the set is broken up, for me, seems to be an important part of what the measure “should mean.” If we defined a (general) measure first, we’d need to have it satisfy all of the properties above, but these properties deal with subsets of our set — so which ones can we use? Or would we just throw away ones that didn’t make the theory work? Is this “throwing away” notion well-defined? Alternatively, we could build up the subsets that work, but, to me, this feels exactly like generating a sigma-algebra.

      For reference, I am doing the worst possible thing: I am attempting to do much of this from memory and from my terrible class notes, and then checking Rudin’s Complex Analysis to see if I’m anywhere near correct. I anticipate using Kolmogorov’s book for a lot of what comes next, just because I kind of remember liking his style better — and, really, I don’t really want to be doing measures in complete and total generality (yet).

      I always groan and then admit, “Yes, analysis is important.” And it is. Most of the theorems aren’t even so bad, intuitively. It’s just when I start reading a proof and the book starts using something like \frac{3\epsilon}{7} that my mind begins to wander.

    • James said

      Re-reading a bit of this, I think I misinterpreted what you said. And in this post I think I may have abused terminology a bit. It seems that (according to Kolmogorov) the notion of outer and inner measure coinciding for a set implies that set is Lebesgue Measurable. Rudin gives that the names of the elements of our sigma-algebra are Measurable sets. In fact, Rudin doesn’t even mention Outer Measure except in passing in the notes at the end of his book. I’m a bit concerned about these terms, but after this post I will mainly be doing Lebesgue measure with outer measures and such and so there shouldn’t be much confusion. Maybe. Hopefully.

  2. Right. What I always learned (this is in response to your concerns with defining measurable sets) is to define an outer measure on a set X to be a map \mu:2^X\to[0,\infty] which satisfies the axioms you listed and then to define the ‘measurable sets’ to be the ones which satisfy the Caratheodory condition (namely a set A\in 2^X is measurable if and only if for every set U\in 2^X one has \mu(A)=\mu(A\cap U)+\mu(U-A).

    I understand why this may seem like a stupid definition since intuitively ‘measurable’ for a measure \mu means A\in\text{Dom}\left(\mu\right) but historically this makes sense if you look at the progression of the usual Lebesgue measure on the reals.

    I used Royden for my learnings in measure theory, it’s the classic book I thought. I have Rudin and Kolmogorov–Rudin is Rudin nothing else can be said, good selection of material but the exposition is a little weak. Kolomogorov is fun because he’s so famous, but I’m not smitten with the book. Is there a reason you don’t use Royden?

    • James said

      Yeah, that’s usually the way things are done. The next post is just Lebesgue measure and it matches what you’ve said in your first paragraph there. I think I wanted to introduce it this way to emphasize that measure can be much more general than Lebesgue measure, since I may be going into other kinds of measures later.

      There’s an excellent reason that I don’t use Royden: I don’t have it. I ought to go to the library and page through it to see if there are any exciting insights that Kolmogorov and Rudin are missing out on.

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