Could someone help with this proof on measure theory?

Question

Let (X, M u) be a measure space, where X is a set, M is a sigma-algebra on X and u is a measure defined on M. Let f_n be a sequence of functions defined on X and with values on [0, oo] such that lim f_n = f. Suppose that lim Integral f_n du = Integral f du < oo (the integrals taken over X). Show that, for every set E of M, Integral_E f_n du = Integral_E f du, where Integral_E means integral over E. Also, show that this conclusion may fail if we have lim Integral f_n du = Integral f du = oo
Thank you for any help.

Answer 1

I'll assume you know the basics of measure theory and Lebesgue integral. Well, since you asked this question, you sure know the basics.

The given conditions imply that, for sufficiently large n, f_n is integrable over X and, therefore, over each set E of M. So, we can assume, without loss of generality, that such conditions hold for every n (if necessary, discard a finite number of values of n). Since f_n is a sequence in L+ (the space of measurable functions on X with values on [0, oo]), for every n we have 0 <= Int_E f_n du <= Int f_n du. Since the sequence (Int f_n du) is bounded (for it converges on R), these inequalities show (Int_E f_n du) is bounded and, therefore, contains convergent subsequences (Bolzano Weierstrass Theorem). Let (Int_E f_n_k du) be one of such subsequences. Then, (f_n_k) is subsequence of (f_n), so that lim f_n_k = f.

According to the properties of the integral, for every k we have

Int_E f_n_k du + Int_E' f_n_k du = Int f_n_k du , where E' is the complement of E with respect to X.

Since (Int f_n_k du) is a subsequence of (Int f_n du), which converges to Int f du, the same is true of (Int f_n_k du). And since Int_E f_n_k du converges, the equality above shows that Int_E' f_n_k du converges and that

lim Int_E f_n_k du + lim Int_E' f_n_k du = lim Int f_n_k du = Int fdu. In addition, since Int fdu = Int_E f du + Int_E' fdu, it follows that

lim Int_E f_n_k du + lim Int_E' f_n_k du = Int_E f du + Int_E' fdu (1)

Since (f_n_k) is a sequence in L+ whose limit is f (on X and on every set of M), it follows from Fatou's Lemma that

lim Int_E f_n_k du >= Int_E f du (2) and, similarly, lim Int_E' f_n_k du >= Int_E' f du (3).

In order for (1), (2) and (3) to be simultaneously satisfied, (2) and (3) must be equalities, which implies that

lim Int_E f_n_k du = Int_E f du (4). Since (Int_E f_n du) is bounded and (4) holds for all of its convergent subsequences, it follows (Int_E f_n du) itself is convergent and

lim Int_E f_n du = Int_E f du, valid for every set of M, that is, for every measurable subset of X. This proves the theorem.


To see this conclusion may fail if lim Int f_n du = Int f du = oo, we can take X = (0, oo), M = Lebesgue sigma-algebra on X , u = Lebesgue measure and

f_n(x) = 1/(nx) if x is in (0, 1]
f_n(x) = 1 if x is in (1, oo) n=1,2,3.... ....


Then, f_n conveges to the function f given by

f(x) = 0, if x is in (0, 1] and f(x) = 1 if x is in (1, oo). Let E = (0, 1]. Then, recalling in this case the Lebesgue and Riemann integrals yields the same value, we have

For every n, Int_E f_n du = Int (0 to 1)1/nx dx = 1/n ln(x) (0 to 1) = oo, Int_(1,oo) f_n du = 1 * oo = oo and, therefore, Int f_n du = oo. It follows that lim Int_E f_n du = oo and lim Int f_n du = oo.

On the other hand, Int_E f du = 0, Int_(0, 1) f du = oo and Int f du = oo

So, we have lim Int f_n du = Int f du = oo, but lim_E f_n du = oo > 0 = Int_E f du. This shows the condtion Int f du < oo cannot be dropped.

Answer 2

It's easy if you know "Fatou's Lemma": apply it to Integral_E f_n. If the inequality is strict that is "lim Integral_E f_n < Integral_E f" then the equality Integral f_n = Integral_E f_n + Integral_E' f_n when taking limits would give us "lim Integral_E' fn > integral_E' f contradicting Fatou.

For the contradiction, just do what the other guy did

Answer 3

i could use your reasoning to tutor the case f=0, then the consequence trivially generalizes to any uniform minimize of integrable complicated valued applications. This in the present day simplifies the evidence.