Gracious Living

Measures and Non-Measurable Sets
November 23, 2010, 14:22
Filed under: Analysis, Math | Tags: , , , ,

This is very late, but don’t worry, I’ll get another one up tonight.

One of the big lessons learned from the Banach-Tarski paradox is that even in something as simple as a unit ball, we can find sets of impossible or undefinable volume. In the discussion preceding the proof, I also mentioned the paradoxes surrounding length of fractal curves in \mathbb{R}^2 and area in \mathbb{R}^3. Together, these present us with a crisis: how can we characterize length, area, and volume? The answer to this crisis was developed around the turn of the century by heroes like Borel and Lebesgue, and it’s called measure theory.

Rather than spewing definitions straight away, I’d first invite you to consider, as Lebesgue and Borel probably did, what properties something like “area” should hold in a set X. We’re going to define something called a measure, that encodes the properties of “area,” “volume,” and so on. The first guess is that it should be a real-valued function on \mathcal{P}(X); but we don’t want things to have negative measure, and it’s also possible to have sets with infinite measure. So we’re looking for a function \mu:\mathcal{P}(X)\rightarrow[0,\infty]. Such a function is called extended-valued; we avoid paradoxes with infinity by only using it in a limited way (here, we define anything plus infinity to be infinity, and we don’t perform any other operations on it).

What properties should this satisfy? Well, one would hope that you could add together measures: that \mu(E_1\cup E_2)=\mu(E_1)+\mu(E_2) if E_1 and E_2 are disjoint. Note that this also implies that if E\subset F, then \mu(E)\le\mu(F). In fact, the original definition went further: if E_1,E_2,\dotsc are a countable collection of disjoint sets, then \mu\left(\bigcup_{i=1}^\infty E_i\right)=\sum_{i=1}^\infty\mu(E_i). We call this property countable additivity. But we don’t want to allow this for uncountable unions, because any set in \mathbb{R}^n is an uncountable union of points, and we want points to have measure zero.

What values should this function take? There don’t need to be sets of infinite measure (X itself could have finite measure), but there do need to be sets of zero measure. Since \emptyset is disjoint from everything, it needs to have zero measure for the previous definition to work. No other specific values are mandatory — even in \mathbb{R}^2, the area of a set depends on what scale we’re using.

In \mathbb{R}^n, we also want this function to be preserved by isometries, but this is a special case — functions like this can be defined even on sets where there is no conception of “isometry.”

Now, the Banach-Tarski paradox proved that such a function, even one that’s only finitely additive, can’t be defined everywhere in \mathbb{R}^3. So we have to modify our domain from \mathcal{P}(X) to some subset \mathcal{M}. We’ll call its elements measurable sets. What properties must this set have? In order to make the above properties make sense, it needs to contain \emptyset and be closed under countable unions. We’re also going to want X itself to be measurable — since X=E\cup E^c for any set E, we want \mathcal{M} to be closed under complementation as well. Note that together, these properties make \mathcal{M} closed under countable intersections, since E\cap F=(E^c\cup F^c)^c. A set \mathcal{M} that’s nonempty and closed under complementation and countable union is called a \sigma-algebra (the sigma being topologist’s shorthand for “countable”, for some reason).

A measure space is a triple (X,\mathcal{M},\mu), where \mathcal{M} is a \sigma-algebra of subsets of X, called the measurable sets, and \mu:\mathcal{M}\rightarrow[0,\infty], a function called measure, satisfies \mu(\emptyset)=0,\mu\left(\bigcup_{i=1}^\infty E_i\right)=\sum_{i=1}^\infty\mu(E_i) for E_i disjoint.

Perhaps the most surprising thing is that you can perform integration with respect to a measure on any measure space! How you do this is a bit contrived, but it basically involves approximating your function as a sum of “characteristic functions,” which are 1 on a certain measurable set and 0 elsewhere.

Let’s look at some examples. The simplest example is the counting measure, which is defined everywhere and which sends each set to its cardinality if it’s finite, or to infinity if it’s infinite. Integration here is nothing more than taking a sum of the nonzero values of your function, and gives you infinity if there are infinitely many.

If \mu(X) is 1, then \mu is a probability measure. This is the theoretical basis for the study of probability. Integration of a function here gives you its expected value corresponding to the probability measure.

What Lebesgue did was define Lebesgue measure on \mathbb{R}^n, which is isometry-invariant, measures all open sets, and also measures every subset of a measure-zero set. This is the measure you implicitly use when you do a normal integral. The advantage of Lebesgue measure is that we can integrate functions “up to a set of measure zero”: that is, if we change the values of a function on a set of measure zero, the integral remains the same. For example, countable sets have measure zero, so if we start with a function which is, say, 1 at every rational and 0 at every irrational, it will be impossible to take Riemann sums, but the Lebesgue integral is zero because the function is zero outside a set of measure zero.

Now, when we talk about \sigma-algebras, we can often think of them in terms of a basis. The \sigma-algebra generated by a basis is the smallest \sigma-algebra containing all the sets in that basis. An important example is the Borel \sigma-algebra of a topological space. This is the \sigma-algebra generated by the open sets of that space: it thus contains all closed sets as well, and all countable intersections of open sets, countable unions of closed sets, et cetera. The Lebesgue \sigma-algebra is generated by the open sets and the subsets of sets of measure zero.

So what’s interesting is that it takes the axiom of choice to show that there’s a set not in this \sigma-algebra. The Banach-Tarski example is powerful, but here’s an example of a set in \mathbb{R} with no length, called the Vitali set. We form the quotient group $latex\mathbb{R}/\mathbb{Q}$; this is the set of cosets a+\mathbb{Q} in \mathbb{R}. Choose an element from each coset in [0,1], which we can do by subtracting integers if necessary, and call the set of these elements V. The rationals, and particularly the rationals in [-1,1], are countable, so we can label them q_1,q_2,\dotsc. Let V_i=V+q_i. Then [0,1]\subset\bigcup V_i\subset[-1,2] (every element of [0,1] differs from its representative in [0,1] by a rational in [-1,1]). Since all the V_i are disjoint, the measure of their union is equal to the sum of their measures. But since Lebesgue measure is translation-invariant, this is just \sum_{i=1}^\infty \mu(V), which is either 0 or \infty. This is certainly not between 1 and 3!

I hope this was an interesting overview. Pretty much everything I said deserves some expansion, which I’ll do in the future, starting with the construction of the Lebesgue measure. This should be an interesting segue into real analysis. I also plan to talk about solvable groups in the context of the Banach-Tarski and von Neumann paradoxes, and return to standard topology with metric and quotient topologies. Thanks for reading!

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[…] the interests of total confusion, let’s discuss metrics, with have almost nothing to do with measures, despite the similarity in name!  Measures have to do with the sizes of sets of points, and are […]

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[…] forms an (infinite) orthonormal basis for this vector space.  Regrettably, you have to use the Lebesgue integral to do this, and since S&S want their series to be self-contained, they only use the Riemann […]

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