Measurew theory - can you help me, please?

Question

Could you please help me prove this?

Let (X, M, m) be a mesure space and let E_n be a sequence of sets in M such that Sum (n=1, oo) m(E_n) < oo. Show that almost every x of X belongs to finitely many sets E_n.

Answer 1

Let E be the set of the elements x in X that belong to infinitely many sets E_n (usually called lim sum E_n). Then, E = Intersection (n =1, oo) (Union (m = n, oo) E_n = Intersection (n =1, oo) F_n, where F_n = Union (m = n, oo) E_m. We have to prove u(E) = 0.

The sequence F_n is decreasing (F_1 contains F2 contains F_3....) and, by the subadditivity of the measure, m(F1) <= Sum (n=1, oo) u(E_n) < oo. According to the continuity from above of the measure, we have

u(E) = lim u(F_n)

For every n, 0 <= u(F_n) < = Sum (m= n, oo) u(E_m). Since Sum(n =1, oo) u(E_n) < oo, lim ( n --> oo) Sum (m= n, oo) u(F_m) = 0, which implies lim u(F_n) = 0.

Hence, u(E) = 0, proving the assertion.

EDIT: Oh, I used the letter u, instead of m, to represent the measure. Well, no problem

EDIT2: There's another interesting, though not so natural, proof for this theorem.

For each n, put f_n(x) = c_1(E_1) ...+ c_n(E_n), where c_k is the characteristic function of E_k. Then, f_n is monotone increasing and has as limit the function f such that

f(x) = number of sets E_n to which x belongs, if x belongs to finitely many E_n s
f(x) = oo, if x belongs to infinitely many E_n s

Hence, E = {x in X | x belongs to infinitely many E_n} = { x in X | f(x) = oo}.

For each n, Integral f_n du = u(E_1)...+ u(E_n), so that

lim Integral f_n du = Sum (n=1, oo) u(E_n) < oo. By the Monotone Convergence Theorem,

Integral f du = lim Integral f_n du < oo. Since f is non-negative and has finite integral, the set where it assumes the value oo is null. And since this set is just the set E, it follows u(E) = 0.

Answer 2

EDIT
Let A_n=union{i=n..infinity}E_i. Let A=intersection{n=1..infinity}A_n. Then A={x in X;x is in infinitely many sets E_n}. Note that for every j, m(A)<=m(A_j)<=Sum (n=j..infinity) m(E_n)infinity, we get m(A)<=0, hence m(A)=0.

Answer 3

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Answer 4

Let Y be the set of x in X that belongs to infinitely many sets E_n.
Let E_n_k be a subsequence of E_n.
Y is a subset of intersection of all E_n_k so
0 ≤m(Y) ≤ lim(k->infinity) m(E_n_k) = 0
m(Y) = 0
QED