Measure Theory, Dominated Convergence Theorem

Almost Everywhere

Let (X,σ,μ) be a measure space, and let p be a property that depends on x. We say p is true "almost everywhere" if there exists a set A in σ such that the elements of X, where p is false, all live in A, and μ(A) = 0. Thus p fails inside a set of measure 0.

If f is a measurable function outside of A, set f(A) = 0, and f becomes measurable by concatenation. Also, the integral of any function on A is 0, so setting f(A) = 0 does not change the integral of f.

If f = g almost everywhere, failing inside a measurable set A, set f(A) = g(A) = 0. This does not change the respective integrals. Now f = g, whence ∫ f = ∫ g. In other words, functions that are equal almost everywhere have the same integral.

Let f lie in L(X,σ,μ), as described in the previous section. Let A be the preimage of ∞ under f. Assuming ∞ is measurable, A is measurable. If μ(A) is nonzero, the integral of the positive portion of f is infinite. This is a contradiction, hence μ(A) = 0. The same result holds for -∞. Therefore, f is finite almost everywhere.

Change the infinite values of f to anything else, and we are changing f across a region of measure 0. This does not affect the integral.

Dominated Convergence Theorem

This is another opportunity to interchange limits and integration. Let fn be a sequence of functions drawn from L, such that fn approaches f almost everywhere. Let g be a nonnegative function drawn from L, that bounds |fn| for all n. Assume the usual topology on R. First show f is in L.

This is somewhat problematic, since f is not defined in some areas. It is equal to the limit of fn almost everywhere, but not everywhere. We don't know what f is on A, or even if fn approaches a limit on A. Set fn(A) = 0 for each n. This does not affect the measurability of fn, nor its integral. Now the limit of fn exists, and it is 0 on the set A. Let f(A) = 0, so that f is the limit of fn. This does not change the integral of f. Being the limit of measurable functions, f is now measurable.

The positive portion of f is bounded by g, which has a finite integral, thus ∫ f+ is finite. Similarly, f- is finite, hence f lies in L.

Now I am ready to state the theorem, and it's what you would expect. The integral of f equals the limit of the integrals of fn.

Recall our earlier equation.

∫ f = ∫ f+ - ∫ f-

Extend this result to each fn in the sequence.

∫ fn = ∫ fn+ - ∫ fn-

The limit of the integral of fn is now the limit of the sum, which is the sum of the limits.

limit(∫ fn+) - limit(∫ fn-)

Consider the first limit. The integrands are measurable, nonnegative functions that are bounded by g. The nonnegative dominated convergence theorem allows us to interchange limit and integration. The result is the integral of the limit of fn+. From this we subtract the integral of the limit of fn-. If the limit of fn+ is f+, and similarly for f-, then we have the formula for the integral of f, and we're done. The only trick is showing the limits converge to the respective pieces of f.

Fix an x, and assume f(x) = y, where y is positive. Move out in the sequence, so that fn(x) is always within ε of y. Now fn(x) is positive, and belongs to fn+, hence fn+(x) approaches y, which belongs to f+. On the other hand, for large n, fn-(x) = 0, which approaches 0, which is f-(x). So there is no trouble here.

A similar result holds when f(x) is negative.

Finally, let f(x) = 0. In the sequence fn(x), values can be held arbitrarily close to 0. Thus values in fn+ and fn- can be held arbitrarily close to 0. Both sequences converge to 0 at x, and f+(x) = f-(x) = 0. That completes the proof.