Can 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

The given conditions imply that f is integrable over every E of M and that f_n converges to f on E. Let E' be the complement of E, so that E' is in M. Then, for every natural n,

Int_E f_n du + Int_E' f_n du = Int f_n du

Since lim Int f_n du = Int f du = Int_E f du + Int_E' f du, it follows that

lim (Int_E f_n du + Int_E' f_n du) = Int_E f du + Int_E' f du (1).

By Fatou's Lemma,

lim inf Int_E f_n du >= Int_E lim inf f_n du = Int_E lim f_n du = Int_E f du (2)

and , similarly (skipping the details),

lim inf Int_E' f_n du >= Int_E' f du (3)

The righ hand sides of (2) and (3) are finite, since f is integrable over E and over E'.

We know that if a_n and b_n are real sequences such that (a_n + b_n) --> a + b, a and b in R, and

lim inf a_n >=a and lim inf b_n >= b, then

a_n --> a and b_n --> b. (The proof of this result is given at the end of this post, anyway)

So, applying this result with a_n = Int_E f_n du , b_n = Int_E' f_n du, a = Int_E f du and b = Int_E' f du and considering (1), (2) and (3), it follows that

lim Int_E f_n du = Int_E f du, proving 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.

In this proof, we used the following result:

If a_n and b_n are real sequences such that (a_n + b_n) --> a + b, a and b in R, and

lim inf a_n >=a and lim inf b_n >= b, then

a_n --> a and b_n --> b

Proof:

a_n contains a subsequence a_n_k such that lim a_n_k = lim sup a_n. Since (a_n_k + b_n_k) is subsequence of (a_n + b_n), it follows that lim (a_n_k + b_n_k) = a + b. So, lim b_n_k = lim (a_n_k + b_n_k) - lim a_n_k = a + b - lim sup a_n >= lim inf b_n >= b. Hence, lim sup a_n <= a, so that

lim sup a_n <= a <= lim inf a_n, which implies lim a_n = a and therefore, lim b_n = b.