Measure Theory, Linear Combination of Integrals

From Below

Let u be a measurable function from X into the nonnegative extended reals. A sequence of functions s_n approaches u. Let s₀ = 0. Let s₁ = 0 where u is finite, and ∞ when u = ∞. Since the positive rationals are countable, arrange them in an ordered list. Let z be the first one on the list. Let s₂ = 0 when u is less than z, and z when u is greater than z and finite, and ∞ when u = ∞. Continue down the list in the same manner. If the n^th rational is y, and w is the largest rational on the list prior to y, let s_n+2 = s_n+1, or y, if u is ≥ y and < w.

Each s_n is a simple function.
The sequence s_n is monotonically increasing.
Each s_n is bounded by u.
Since half open intervals pull back through u to measurable sets, each s_n is measurable.
Fix an x, and a rational z below u(x), arbitrarily close to u(x). Go far enough out on the list, until you find z. Each s_n beyond this point is at least z, and bounded by u(x). The sequence s_n approaches u.
By the previous theorem, the integrals of s_n approach the integral of u.

Linear Combination of Integrals

Let s_n approach u, as described above, and let t_n approach v. By adding limits pointwise, (s_n+t_n) approaches u+v. Apply the monotone convergence theorem, and the integral of u+v is the limit of the integrals of s_n+t_n. Being a simple function, the integral of s_n+t_n is the integral of s_n plus the integral of t_n. Let n approach infinity and find the integral of u plus the integral of v. I left this theorem unfinished earlier; now it is complete. With modest assumptions about the topology of R, (to make the result measurable), the integral of a linear combination of functions is the linear combination of their integrals.

Reverse Monotone Convergence Theorem

Now that we know the integral of the sum is the sum of the integrals, we can prove the reverse monotone convergence theorem. This is the same as the monotone convergence theorem, but the functions f_n are monotonically decreasing. Still, the limit of the integrals of f_n is the integral of the limit function f.

However, there is an additional constraint. The functions f_n must be bounded by another measurable function g, having a finite integral. This makes the proof easy. Let h_n = g-f_n. This sequence is measurable, (with modest assumptions on the topology of R), and increasing, so apply the monotone convergence theorem, and the limit of the integrals of h_n equals the integral of g-f. Since all integrals are finite, we can regroup terms to our heart's content. For instance, ∫ h_n = ∫ g - ∫ f_n. Take the limit, and cancel ∫ g from both sides, and the limit of ∫ f_n becomes the integral of f.

If any f_n has a finite integral, let g = f_n, and start the sequence there. We only get in trouble if each f_n has an infinite integral. Let's look at a traditional example from calculus. (Lebesgue integrals behave like riemann integrals for continuous functions.) Let the domain be the open interval (0,1), and let f_n = 1 over nx. This is a sequence of descending hyperbolas. Each f_n has an infinite area below it, hence an infinite integral. Yet the limit function f is identically 0, and has an integral of 0.