Measure Theory, Simple Function

Simple Function

A simple function has a finite range. In the case of a function from X into the reals, or the extended reals, finitely many real numbers are involved. Each of these real numbers has a preimage in X, and these preimages partition X into a finite collection of sets, or regions.

If u is a simple function, one can rewrite u as a linear combination of indicator functions, where an indicator function χ is 1 on the set A and 0 elsewhere. If u attains n values, then write u as the sum of cj × χj(X), as j runs from 1 to n. The values cj form the range of u, with Aj mapping to cj.

Let σ turn X into a measurable space. Assume the reals also form a measurable space, such that each point in R can be isolated in a measurable set. (This is not unusual.) If u is a measurable function from X into R, then each preimage Aj is a measurable set. Conversely, if each Aj is a measurable set, then the preimage of every set in R is measurable, and u is a measurable function.

The Integral of a Nonnegative Simple Function

Let u be a simple function on X, as described above, such that u maps X into the nonnegative extended reals, and write u as the sum of cjχj(X). The integral of u, with respect to a measure μ, over X, is the sum as j runs from 1 to n, of cj×μ(Aj). Since the nonnegative extended reals form a semiring, the terms can be added in any order. This will become important below, as we regroup terms to prove various theorems.

In this formula, X is the domain of integration, μ is the flavor of integration, and u is the integrand.

Refinement

What if we break up one of the regions upon which u is constant? If A7 was the last region, call it A8+A9. Take the integral of u, and instead of c7×μ(A7), we have c7×(μ(A8)+μ(A9)). (This assumes A8 and A9 live in σ.) Since A8 and A9 are disjoint, the sum of their measures equals the measure of their union. Multiply this relationship by c7, and we're done. Therefore the integral does not change when the regions of X are refined.

In fact, we can separate A7 into a countably infinite collection of regions. The proof is the same. We are distributing c7 across an infinite sum, but that's no problem. So countably infinite refinements are fair game.

Linear Combination

Let's verify the properties of linearity. Multiply u by a constant b, and the integral is multiplied by b. This is clear when b is nonzero, as the regions of X do not change. When b is 0, all of X maps to 0, and the regions of X that were constant under u combine into one region, namely X. Since the original regions are a refinement of X, the integral does not change. Therefore, integration and scaling commute.

Take the integral of u+v and find the integral of u plus the integral of v. The combined function u+v attains finitely many values, and is a simple function. For any two real numbers y and z, the preimage of y under u, and the preimage of z under v, are in σ, and so is their intersection. Partition X into regions that are constant with respect to u and to v. This is a finite refinement of the regions produced by u, and by v, and does not change the two integrals. Add the two integrals together, regroup terms, and find a formula for the integral of u+v, or perhaps a refinement thereof. Therefore, the integral of a linear combination of simple functions produces the linear combination of the individual integrals.

Concatenation

As with Riemann integration, adjacent regions can be concatenated, and the integrals are added together. Let u be a simple function on X, where X is the disjoint union of measurable sets X1 and X2. Refine the integral of u, so that each region is either in X1 or X2. This does not change the integral of u. Write the formula for the integral of u and regroup terms to find the integral of u over X1 plus the integral of u over X2. This generalizes to any finite partition of X, provided all subsets are measurable.

Countably infinite partitions of X are valid as well. Refine the integral of u into an infinite sum of terms. Because the terms are nonnegative, we can regroup terms to our heart's content - thus creating an infinite sum of integrals across the regions of X.

The Integral of a Nonnegative Function

Assume a curve in the xy plane never dips below the x axis. The area under the curve is the supremum of the lower Riemann sums. In other words, consider all possible step functions below the curve, take their integrals, and then find the least upper bound; that is the integral of the curve. The Lebesgue integral of a nonnegative function is defined in the same way, with simple functions playing the role of step functions.

First some notation. Let M+(X,σ) be the set of all nonnegative measurable functions from (X,σ) into the extended reals. This assumes a sigma algebra on the extended reals. I will postpone the definition of such an algebra; let's just say it exists. Then the integral of u ∈ M+(X,σ) is the supremum of the integrals of all simple functions in M+(X,σ) that are bounded by u.

The zero function, which is always measurable, has an integral of 0. This fits inside any nonnegative function u. Therefore the integral is well defined, and nonnegative, even if u is not measurable; but u is usually assumed to be a measurable function.

If v is bounded by u, every simple function that fits inside v also fits inside u, hence the integral of v is no larger than the integral of u. This is called monotonicity.

If u is a nonnegative simple function, the two definitions of integral coincide. Every simple function v bounded by u has an integral bounded by the integral of u, so the integral of u is, well, the integral of u.

Multiply u by a constant b, and each simple function below u can also be multiplied by b. This multiplies their integrals, and the supremum, by b. Therefore scaling and integration commute.

Two simple functions bounded by u and v respectively, sum to a simple function that is bounded by u+v. But the converse may not hold. Without more information, we can't prove linearity. For instance, let X be the unit interval, and let σ be X and the empty set. Let u be 0 on the rationals and 1 on the irrationals, and Let v be 1 on the rationals and 0 on the irrationals. Now u+v has an integral of 1, while u and v have integrals of 0. The problem here is that u and v aren't measurable functions. I'll return to this topic later.

If X can be partitioned into a finite number of disjoint measurable sets, concatenation still holds. A simple function bounded by u determines, and is determined by, simple functions on the regions of X. Take upper bounds with respect to u, and the integral of u over X is the sum of the integrals of u over the regions of X.

Another form of monotonicity, which follows from the above, is that the integral of u over X bounds the integral of u over any measurable subset of X.

Concatenation also holds across a countably infinite partition of X. One direction is easy. Each simple function on X, bounded by u, has an integral that is equal to the sum of the integrals of the piecewise simple functions on the individual regions Xi. This in turn is less than or equal to the sum of the individual integrals. Thus the integral over X ≤ the sum of the integrals over Xi. For the converse, suppose the integral over X is too large. This means the integral of each Xi is finite. Also, the partial sum of the integrals of X1 through Xn cannot increase without bound, else the sum of the integrals and the integral over X are both infinite. Therefore, the sum of the integrals has a finite upper bound b. Since the integral of X is larger, choose a simple function on X with integral beyond b. This pushes the sum of the integrals of Xi beyond b, and that is a contradiction.

Indefinite Integral

In real analysis, or in measure theory, the indefinite integral is the limit of a sequence of integrals of f, as the region of integration approaches the boundaries of a closed interval, or infinity (i.e. the real line). In the case of measure theory, the sequence must be countable, and measurable. Let A be the union over An. Let Gn be the new stuff brought in by An, that was not present in A1 through An-1. Now A becomes the disjoint unionn over Gn. Apply countable concatenation, as described above, to Gn, and the integral over A becomes the limit of the integrals over An.

If An is a decreasing sequence of regions, and A is their intersection, one can construct the reverse indefinite integral. The limit of the integrals over An is the integral over A. The proof is the same, but to regroup the terms properly, we need all the integrals to be finite. If the integral of any An is finite, start there, and the conditions of the theorem are met.