Measure Theory, Sigma Algebra

Sigma Algebra

Before I define a sigma algebra, I want to emphasise that many of the notions that we will come across in measure theory have analogues in topology. For example, a sigma algebra, as we will see shortly, is similar to a topology on a set, i.e. a collection of subsets that obey certain properties. And measurable functions are analogous to continuous functions , and so on.

Let x be a set. A sigma algebra on X, sometimes denoted σ, is a collection of subsets of X such that:

  1. X is in σ.
  2. A is in σ iff the complement of A is in σ.
  3. The countable union of sets in σ also lies in σ.

(The term algebra is somewhat unfortunate, since this has nothing to do with rings.)

We call the pair (X,σ) a "measurable space", similar to a topological space. If A is in σ, we say A is "measurable", or a "measurable set", analogous to an open set.

From (1) and (2), it follows that the empty set is in σ.

Using Demorgan's law, one can show that the countable intersection of sets in σ also lies in σ.

Note that σ could be X and the empty set. This is the indiscrete algebra, similar to the indiscrete topology. At the other extreme, σ could contain every subset of X. This is the discrete algebra, similar to the discrete topology.

One sigma algebra may be stronger than another, if it includes more subsets of X. Thus algebras can be partially ordered, just as topologies on X are partially ordered. Note that the intersection of algebras on X produces another valid algebra. Thus we may talk about the weakest algebra on X that contains a given collection C of sets within X. It is the intersection of all algebras that contain C. There is at least one such algebra, namely the discrete algebra on X, thus the "generated algebra" makes sense, and is denoted σ[C]. We might think of C as a base for σ.

If C, in the above paragraph, is itself a topology on X, then σ[C] is a Borel sigma algebra. This will become clearer later on.

Union and Intersection

The arbitrary intersection of sigma algebras is another sigma algebra, but not so for unions. Let X be two points, and let Y be two more disjoint points. Let Z be X and Y taken together, consisting of 4 points. Let A be an algebra on Z with any combination of the points of X and none or both of the ppoints of Y. Let B be an algebra on Z with any combination of the points of Y and none or both of the ppoints of X. The union of these algebras A∪B consists of 12 sets out of 16. If this is sigma, then the union of any two of these 12 sets gives another of the 12 sets. Let one set have one point from X, and let another set have one point from Y. There union is not in A∪B, hence A∪B is not sigma.

Subset

Let Y be a measurable set within X. Restrict σ to the sets that are entirely contained in Y. This includes Y and the empty set, and the union of any countable collection of sets in Y. Finally, if A is measurable in X, and contained in Y, then Y intersect (A complement) is also in σ. This is of course the complement of A relative to Y. Therefore σ restricted to Y becomes a sigma algebra on Y.

Sigma Algebra on the Reals

Recall that the standard topology of the reals is based on the open intervals. Let this act as a base for σ. For any point z in R, a countable collection of open intervals closes in on z. Therefore every point belongs to σ. Join the open intervals with their endpoints to produce the half open intervals and the closed intervals.

Closed intervals include points, i.e. the degenerate closed intervals, and thus the endpoints of any closed interval lie in σ. Remove these to get the open intervals, as described above.

Half open intervals pointing up can be used to converge on any given point z. Thus the endpoints of these half open intervals are in σ, and can be removed to give open intervals. A similar result holds for half open intervals pointing down.

Closed intervals can be combined to produce closed rays pointing up and down, and two such rays intersect in a closed interval. Similarly, open intervals combine to produce open rays extending up and down, and open rays overlap in open intervals.

In summary, the following sets imply the same sigma algebra on the reals. Open intervals, closed intervals, half open intervals pointing up, half open intervals pointing down, closed rays, and open rays.

In contrast, start with the points of R, and produce a σ algebra. Note that σ must contain every countable collection of points, and the complements thereof. Show that these sets satisfy the criteria of a sigma algebra. Now, the countable union of points cannot create a closed interval, hence this is a weaker algebra on the reals.

Measurable Function

A function from X to Y is measurable if the preimage of every measurable set in Y is measurable in X. This assumes X and Y are measurable spaces, with sigma algebras. This definition should remind you of a continuous function from one topological space into another, where the preimage of every open set is open.

The composition of measurable functions, from X to Y to Z, is measurable from X to Z.

The identity map on X is measurable, for every σ on X.

Any function from a discrete space, or into an indiscrete space, is measurable.

Using the Base

Assume the measurable sets of Y are generated by a base C. A function f from X into Y is measurable iff the preimage of every base set in C is measurable. The reasoning is the same as that used in topology, where it is sufficient to show the preimage of every base open set is open. Arbitrary unions in the range pull back to arbitrary unions in the domain, and that proves continuity. Let's do the same thing here.

Assume each set in C has a measurable preimage under f. Each measurable set in Y is constructed by taking the union or complement of preexisting measurable sets. Proceed by induction on the number of steps in the construction of our measurable set. When n = 1, we have the assertion "belongs to C", and the preimage is measurable by assumption. If T is measurable in Y, with measurable preimage S in X, take complements, and the preimage of the complement of T is the complement of S. Both these sets are measurable, by the properties of a sigma algebra. Similarly, the countable union of measurable preimages produces a measurable set in X - and that completes the proof.

Linear Combination of Measurable Functions is Measurable

If σ on the reals is generated by closed intervals, and f is measurable, and b is a real constant, then b×f is measurable. (This clearly holds if b = 0.) By the above, we only need look at base sets. Given a closed interval, divide it by b, and the preimage of the scaled closed interval, under f, belongs to σ. This is the preimage of the original open interval under b×f. Therefore b×f is measurable.

As you move to the sum of two measurable functions f+g, beware of the topology trap. In topology, we considered an arbitrary point x in the preimage of a given base open set, then surrounded x with an open neighborhood that maps into our open set. The preimage is the union of these open neighborhoods, and is open. But that assumes the arbitrary union of open sets is open. In a sigma algebra, unions must be countable. We have a little more work to do.

Start with a closed interval [a,b], and graph, in the xy plane, a ≤ x+y ≤ b. This is a closed strip slanting down and to the right. Cover the strip with closed squares. This is not a partition; the squares overlap at their boundaries. The covering is fractal in nature - I'll leave the details to you. Since each square has a distinct rational point inside, the covering is countable. Now, each square implies a range of x and a range of y, i.e. two closed intervals. These pull back, through f and g respectively, to preimages that belong to #s'. Their intersection also belongs to #s'. These points in the domain put f in the proper range, and g in the proper range, as described by our square, and the square lies in our strip, so that f+g lies in [a,b]. The countable union of these preimages also lies in #s', and that covers the strip. The preimage of [a,b] is a measurable set, and f+g is measurable.

The proof for f×g works the same way. Graph a ≤ xy ≤ b, and cover the region between these two hyperbolas with squares. Similarly, a ≤ 1/x ≤ b is used to prove 1/g is measurable, provided g is nonzero. Combine this with multiplication, and f/g is measurable.

Concatenation

If A is any measurable set in X, and f is a measurable function on X, then f is measurable on A. The preimage of a measurable set is measurable in X, so intersect with A, and find another measurable set.

Conversely, let X be the countable union of disjoint measurable regions, and assume f is measurable on each region. The various preimages under f are measurable inside each region, and measurable in X. Take their union to find a measurable set, which is the preimage of f throughout X.

This is analogous to topology, where a function is continuous on a space iff it is continuous on the individual components.

Min Max

If the topology of R includes rays, then the absolute value of a measurable function is measurable. Partition X into two regions, that map to values ≥ 0, and values < 0. By the above, f is measurable on both regions. Negate f on the second region, then put the two pieces back together to find the measurable function |f|.

Max(f,g) is derived in the same way. Since f-g is measurable, we can separate X into regions where f ≥ g, and f < g. Apply f and g to these regions respectively, and max(f,g) is measurable. Similarly, min(f,g) is measurable. This generalizes to any finite set of measurable functions.