Measure Theory, Fatou's Lemma

Fatou's Lemma

Let f_n be a sequence of functions on X. The liminf of f is the limit, as m approaches infinity, of the infimum of f_n for n ≥ m. When m = 1, we're talking about the infimum of all the values of f_n(x). As m marches along, more and more values are removed from consideration. The infimum can only rise. As the infimum moves up the real line, it may approach a limit, or it may approach or reach infinity. Either way, the result is liminf(f_n).

Limsup is defined similarly. The limsup is the limit, as m approaches infinity, of the supremum of f_n for n ≥ m. As m marches along, the supremum can only decrease.

Now for Fatou's lemma. (biography) The integral of liminf of f_n ≤ liminf of ∫ f_n. This assumes each f_n is nonnegative and measurable.

Let t_n be the infimum of f_k, for all k ≥ n. Thus liminf of f_n is the limit of t_n. We already said t_n is monotonically increasing. Clearly t_n is nonnegative. Let's prove t_n is measurable.

Start with f_n, which is measurable. Note that min(f_n,f_n+1) is measurable. By induction, the minimum of f_n through f_k is measurable. These functions are monotonically decreasing, and t_n is the limit as k approaches infinity. Since the limit is measurable, t_n is measurable.

Again, liminf(f_n) is the limit of t_n. By the monotone convergence theorem, the integral of liminf(f_n) is the limit of the integrals of t_n. Since these integrals are increasing, the limit of the integrals of t_n equals the liminf of the integrals of t_n. Across all of X, f_n is at least as large as t_n. Thus ∫ t_n ≤ ∫ f_n. Apply the liminf operator, and each infimum is at least as large, whence the resulting limit is at least as large. Therefore, ∫ liminf(f_n) ≤ liminf of ∫ f_n.

If the functions can be bounded by a measurable function with a finite integral, a similar result holds in the opposite direction. The proof is the same as above, but we use the supremum instead of the infimum, and then invoke the reverse monotone convergence theorem.

∫ limsup(f_n) ≥ limsup ∫ f_n

Limit is Measurable

This is a good time to prove the limit of measurable functions is measurable - assuming the usual topology on R. Let f_n be a sequence of measurable functions that approaches f. Build the sequence of funtions t, where t_n is the infimum of f_n and beyond, and s, where s_n is the supremum of f_n and beyond. We already showed these are measurable functions. Remember that t_n is increasing, and s_n is decreasing. Also, s_n is everywhere ≥ t_n.

Fix an x, and assume f(x) = ∞. For any large positive value b, there is some n such that f_n, and values beyond f_n, are above b. This puts s_n and t_n above b. Thus s and t approach infinity, just like f. A similar result holds for -∞.

Assume f(x) = b, where b is finite. With t increasing, and s decreasing, the limits exist. With s ≥ t, the limit of s is at least the limit of t. Suppose s remains above b+ε infinitely often. Move out in the sequence, so that f_n and beyond are all within ε of b. This places s_n within ε of b. This is a contradiction, hence the limit of s is at least b. Similarly, the limit of t is at most b. Put this all together and the limit of s_n = the limit of t_n = the limit of f_n = b = f(x).

Concentrate on the sequence t_n, which is measurable and increasing. We already showed that the limit is measurable, hence f is a measurable function.

Nonnegative Dominated Convergence Theorem

Continue the above, where f_n approaches f, and let each f_n be nonnegative. Derive s_n and t_n as before, and they are all nonnegative. Remember that t_n ≤ f_n ≤ s_n, and they all approach the same function f. Apply the monotone convergence theorem, and the integrals of t_n approach the integral of f. If the functions f_n are bounded by a function g, having a finite integral, then the reverse monotone convergence theorem applies. The sequence s_n is descending, and the integrals of s_n approach the integral of f. Since the integral of f_n is trapped between the integral of t_n and the integral of s_n, the integrals of f_n approach the integral of f.

This may remind you of trapping a riemann sum between a lower sum and an upper sum.

Let's see what goes wrong when g does not exist. Let X be the open interval (0,1), and let lebesgue integration reflect riemann integration. Let f_n be the step function that is n from 0 to 1/n, and 0 elsewhere. Each f_n has integral 1, and the limit is of course 1. Yet the limit function f is identically 0, hence its integral is 0.