(UPDATED) Let $(X,S,\mu)$ a measure space, $(A_n)_{n\in \mathbb{N}}$ a sequence in S.

Question

Let $(X,S,\mu)$ a measure space, $(A_n)_{n\in \mathbb{N}}$ a sequence in S. $A=\bigcup\limits_{n=1}^{\infty} A_n$ , where $A_n\in S$ , $\mu(A)< \infty$ and $\sum_{n=1}^{\infty} \mu(A_n) \leq \mu(A)+\epsilon $ for some $\epsilon>0$.

1. Show that, $\sum_{n=1}^{\infty} \mu(E \bigcap A_n) \leq \mu(E \bigcap A)+\epsilon$ for any $E \in S$. [Solved]

Now:

2. Let $C=\{ x\in A | x \in A_n$ for 2 o more $A_n \}$. Show that $C \in S$ and $\mu(C) \leq \epsilon$.

Answer 1

I got the following:

Let $E\in S$. We can express A, An as union of disjoint sets.

$A=(A \bigcap E^c) \bigcup (A \bigcap E)$.

$A_n=(A_n \bigcap E^c) \bigcup (A_n \bigcap E)$.

Then

$\mu(A)= \mu(A \bigcap E^c) + \mu(A \bigcap E)$...(1)

$\mu(A_n)= \mu(A_n \bigcap E^c) + \mu(A_n \bigcap E)$...(2)

substituting (1),(2) in the original inequality:

$ \sum_{n=1}^{\infty} [\mu(A_n \bigcap E^c) + \mu(A_n \bigcap E)] \leq \mu(A \bigcap E^c) + \mu(A \bigcap E) + \epsilon $.

However:

There exists $(F_n)_{n \in \mathbb{N}}$ a sequence of disjoint elements in S, such that, $F_n \subset A_n$ for all n, and $\bigcup\limits_{n=1}^{\infty} F_n = \bigcup\limits_{n=1}^{\infty} A_n$.

Also:

3... $\mu(A \bigcap E^c) = \mu((\bigcup\limits_{n=1}^{\infty} A_n) \bigcap E^c) = \mu((\bigcup\limits_{n=1}^{\infty} F_n) \bigcap E^c) = \mu(\bigcup\limits_{n=1}^{\infty} (F_n \bigcap E^c))= \sum_{n=1}^{\infty} \mu(F_n \bigcap E^c) $ (because $F_n \bigcap E^c$ are disjoint).

And:

4... $\mu(\bigcup\limits_{n=1}^{\infty} (A_n \cap E^c))= \mu((\bigcup\limits_{n=1}^{\infty} A_n) \cap E^c) = \mu((\bigcup\limits_{n=1}^{\infty} F_n) \cap E^c)=\mu(\bigcup\limits_{n=1}^{\infty} (F_n \cap E^c))=\sum_{n=1}^{\infty} \mu(F_n \bigcap E^c)$

And:

$\sum_{n=1}^{\infty} \mu(F_n \bigcap E^c)= \mu(\bigcup\limits_{n=1}^{\infty} (A_n \cap E^c)) \leq \sum_{n=1}^{\infty} \mu(A_n \bigcap E^c)$ (by subadditivity).

Then

$ \sum_{n=1}^{\infty} [\mu(A_n \bigcap E) + \mu(F_n \bigcap E)^c] \leq \sum_{n=1}^{\infty} [\mu(A_n \bigcap E) + \mu(A_n \bigcap E^c)] $

Thus:

$\sum_{n=1}^{\infty} [\mu(A_n \bigcap E) + \mu(F_n \bigcap E^c)] \leq \mu(A \bigcap E) + \mu(A \bigcap E^c) + \epsilon $

by (3)

$\sum_{n=1}^{\infty} [\mu(A_n \bigcap E) + \mu(F_n \bigcap E^c)] \leq \mu(A \bigcap E) + \sum_{n=1}^{\infty} \mu(F_n \bigcap E^c) + \epsilon $

Finally:

$\sum_{n=1}^{\infty} \mu(A_n \bigcap E) \leq \mu(A \bigcap E)+ \epsilon $

Answer 2

Using some MathJax/Latex/markdown stuff we get more readable math:

Solution for (1):

$(DC)$: Let $F_n$ a disjoint (measurable) cover of $A$ such that $F_n\subset A_n$ $$ \sum \mu(A_n)=\boxed{\sum_n \mu(A_n\cap E)+\sum_n \mu(A_n\cap E^c)}\stackrel{\text{assumption}}{\le}\\ \varepsilon+\mu(A)=\varepsilon+\mu(A\cap E)+\mu(A\cap E^c)\stackrel{\text(DC)}{=}\\ \varepsilon+\mu(\cup_n F_n\cap E)+\mu(\cup_n F_n\cap E^c)\stackrel{\text(DC)}{=}\\ \varepsilon+\sum_n \mu(F_n\cap E)+\sum_n \mu(F_n\cap E^c)\stackrel{\text(DC)}{\le}\\ \varepsilon+\sum_n \mu(F_n\cap E)+\sum_n \mu(A_n\cap E^c)\stackrel{\text(DC)}{=}\\ \boxed{\varepsilon+\mu(A\cap E)+\sum_n \mu(A_n\cap E^c)} $$ Subtract the finite $\sum_n \mu(A_n\cap E^c)$ from both sides of the inequality to get the result.

Hint for (2): Under the given constraints one can prove that: $$ \mu(\cup_n A_n)+\mu(C) \le \sum_n \mu(A_n) $$ The derivation is based on easy part of Borel-Cantelli lemma. The claim of the problem (2) is a direct consequence of this inequality.