Could you help me with measure theory (uniform convergence)I'm?

Question

I'm stil at the beginning and have some difficulties

Let (X, M, u) be a mesure space such that u(X) < oo. Let f_n be a sequence of integrable complex valued functions defined on X that converges uniformly to a function f. Show that f is integrable and that Int f du = lim Int f_n du.

Show that the condition u(X) < oo is really essencial for this theorem.

Thank you for any help. Does this problem get easier if at first we suppose the functions f_n are real valued and then extend the conclusion to the complex case?

Answer 1

No, I don't think this gets any easier if we first suppose the f_n are real valued. The proof - which is not that difficult - would be exactly the same.

Let eps >0 be given. Since f_n converges uniformly on X, the Cauchy criterion implies the existence of a positive integer k such that n >=k implies

|f_n| <= |f_k| + eps (|f_n| means the function x --> |f_n(x)|)

which shows the k tail of f_n is dominated by the function |f_k| + eps Since f_n --> f, so does its k tail.

Since f_k is integrable, so is |f_k|, so that Int |f_k| du < oo. And since u(X) < oo, it follows that

Int (|f_k| + eps) du = Int |f_k| du + eps Int du = Int |f_k| du + eps u(X) < oo, which shows |f_k| + eps is integrable. It follows the k tail of f_n converges to f and is dominated by the integrable function |f_k| + eps. By Lebesgue Dominated Convergence Theorem, it follows f is integrable and

Int f du = lim (n >=k) Int f_n du, which shows the k tail of Int f_n du converges to Int f du. So, the whole sequence converges to the same limit, and we have

Int f du = lim Int f_n du.

To see the condition u(X) < oo is really essential, put X = [0, oo), M = Lebesgue sigma-algebra, u = Lebesgue measure, so that u(X) = oo. Define

f_n(x) = c_n(x)/n, where c_n is the characteristic function of [0, n]. Given eps >0, for n >1/eps we have

If x is in [0, n], then 0 < f_n(x) = 1/n < eps
If x >n, then f_n(x) = 0 < eps

which shows f_n converges uniformly on [0, oo) to the identically zero function f =0. So, Inf f du =0. But, for every n, we have

Int f_n du = 1/n * u([0,n]) =1/n * n =1, so that

lim Int f_n du = 1 > 0 = Int fdu (In virtue of Fatou's Lemma, the left hand side could never be less then the right hand side).

So, although f_n converges uniformly to 0 and, in addition, the sequence f_n is uniformly bounded by 1, we don't have Int f du = lim Int f_n du. The condition u(X) < oo is really essential.

In the case of this sequence f_n, it's easy to see that, if a function g dominates f_n, then we must have g(x) >=1/1 for x in [0,1], g(x) >= 1/2 for x in [1,2], g(x) >= 1/3 for x in [2,3]........so that Int g du >= 1 + 1/2 + 1/3.....Since this is the harmonic series, which diverges, we see no function that dominates f_n is integrable, and the Dominated Convergence Theorem can't be applied.

Answer 2

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