Suppose that $\sum_{n=1}^{\infty}\mu(A_n)\le \mu (A)+\epsilon$ Let $(X,S,\mu)$ a measure space. Let $A=\displaystyle\bigcup_{n=1}^{\infty}A_n$ where each $A_n\in S$. Suppose $\mu (A)<\infty$ and  $\displaystyle\sum_{n=1}^{\infty}\mu(A_n)\le \mu (A)+\epsilon$ for some $\epsilon >0$. Show that:
a) $\displaystyle\sum_{n=1}^{\infty}\mu(E\cap A_n)\le \mu (E\cap A)+\epsilon$ for every $E\in S$.
b) Let $C$ the set of all $x\in A$ such that $x\in A_n$ for at least two $n$. Then $C\in S$ and $\mu (C)\le \epsilon$.
I tried this:
a) It is easy to see that there exists $(B_n)$ a sequence of disjoint measurable sets such that $B_n\subseteq A_n$ and $\displaystyle\bigcup_{n=1}^{\infty}B_n=\displaystyle\bigcup_{n=1}^{\infty}A_n=A$.
Thus $\mu(E\cap A)=\displaystyle\sum_{n=1}^{\infty}\mu(E\cap B_n)$, but I don't know if this will work...
b) I think that:
$C=\displaystyle\bigcup\left\{{\displaystyle\bigcap_{n\in J}A_n:|J|\ge 2, J\subseteq\mathbb{N}}\right\}$
Hence measurable. But how can we show that $\mu (C)\le \epsilon$?
Thanks.
 A: First I should note that we are probably using $\mu(A)<\infty$ for countability from below somewhere in here, but I haven't checked where.

*

*This approach will work. $\bigsqcup_{n\in\mathbb{N}}B_{n}=A$ and measures satisfy countable disjoint additivity, so we have
$$
\sum_{n\in\mathbb{N}}\mu(A_{n})\leq\mu(A)+\epsilon=\sum_{n\in\mathbb{N}}\mu(B_{n})+\epsilon
$$
Subtracting, we find
$$
\epsilon\geq\sum_{n\in\mathbb{N}}\mu(A_{n})-\mu(B_{n})=\sum_{n\in\mathbb{N}}\mu(A_{n}\setminus B_{n})
$$
again using disjoint additivity (since $B_{n}\sqcup(A_{n}\setminus B_{n})=A_{n}$). Now we can similarly split $E$ into the part intersecting with $B_{n}$ and the part intersecting with $A_{n}\setminus B_{n}$:
$$
\sum_{n\in\mathbb{N}}\mu(E\cap A_{n})=\sum_{n\in\mathbb{N}}\mu(E\cap B_{n})+\mu(E\cap(A_{n}\setminus B_{n}))\leq\mu(E\cap A)+\sum_{n\in\mathbb{N}}\mu(A_{n}\cap B_{n})
$$
which is in turn bounded above by $\mu(E\cap A)+\epsilon$, using disjoint additivity to get $\mu(E\cap A)$ and monotonicity ($E\cap(A_{n}\setminus B_{n})\subseteq A_{n}\setminus B_{n}$) along with our previous inequality to get $\epsilon$.

*I believe your proof that $C\in S$ is technically incorrect since it gives $C$ as an uncountable union (there are uncountably many such $J$s) of measurable sets, and this is not guaranteed to be measurable. However, you may rewrite your expression for $C$ as
$$
C=\bigcup_{\substack{i,j\in\mathbb{N}\\i\neq j}}A_{i}\cap A_{j}
$$
(it suffices to take intersections of just two sets), which is a countable union of measurable sets, and so $C\in S$.
To show that $\mu(C)\leq\epsilon$, use what we just proved. We have
$$
\sum_{n\in\mathbb{N}}\mu(C\cap A_{n})\leq\mu(C\cap A)+\epsilon=\mu(C)+\epsilon
$$
Intuitively, the sum of the lefthand side is at least double-counting every bit of $C$, since every element of $C$ appears in at least two $A_{n}$s.
Let's show that this double-counts every bit of $C$. We build $C$ by iterating over the $A_{n}$s and keeping the new repeats. Construct a sequence $\{C_{n}\}_{n\in\mathbb{N}}$ by $C_{0}=\emptyset$ and
$$
C_{n}=\left(\bigsqcup_{m<n}A_{n}\cap B_{m}\right)\setminus\left(\bigsqcup_{m<n}C_{m}\right)
$$
The left disjoint union grabs all the repeats (everything in $A_{n}$ and was also $A_{m}$ for some $m<n$); the right disjoint union discards the 'repeated repeats.'
Using countable disjount additivity we have
$$
\mu(C)=\sum_{n\in\mathbb{N}}C_{n}
$$
But now we have both $C_{n}\subseteq C\cap A_{n}$ and $C_{n}\subseteq\bigcup_{m<n}C\cap A_{m}$ and so, summing,
$$
\sum_{n\in\mathbb{N}}\mu(C\cap A_{n})\geq 2\sum_{n\in\mathbb{N}}\mu(C\cap C_{n})=2\mu(C)
$$
This last bit feels shaky, but you get the idea and it looks like someone else has already answered more concisely. Please let me know if there are any questions.

A: I'll use integral, but just for non-negative step functions (you can use $\int\sum a_n\,\chi_{C_n}\,d\mu=\sum a_n\,\mu(C_n)$, $C_n$'s disjoint, as a definition, and check that $\int\sum a_n\,\chi_{C_n}\,d\mu=\sum a_n\,\mu(C_n)$ ($a_n\geq0$) even if $C_n$'s are not disjoint, by refining $C_n$'s). 
The inequality says that $\int\sum\chi_{A_n}\,d\mu\leq\mu(A)+\epsilon$, i.e. $\int(-1+\sum\chi_{A_n})\,d\mu\leq\epsilon$, but then $\int(-1+\sum\chi_{A_n})\chi_E\,d\mu\leq\epsilon$ as $0\leq(-1+\sum\chi_{A_n})\chi_E\leq-1+\sum\chi_{A_n}$. This solves a).
For b), let $f(x)=1$ for $x$'s which are in a single $A_n$, and $f(x)=2$ for those which are in at least two. As $f\leq \sum\chi_{A_n}$, you get $\int(f-1)d\mu\leq\int(-1+\sum\chi_{A_n})\,d\mu\leq\epsilon$, as you wanted.
