Upper bound for some measurable sets given the inequality $\sum_{n=1}^{\infty} \mu (A_{n}) \leq \mu (\bigcup_{n=1}^{\infty} A_{n}) + \epsilon$

Question

I've thinking about this problem for several days now and it didn't seem that difficult, but I haven't been able to find a satisfying solution.

Let $(X, \Sigma , \mu)$ be a measure space, $(A_{n})_{n=1}^{\infty}$ a sequence of measurable sets and $A = \bigcup_{n=1}^{\infty} A_{n}$. Suppose that $\mu (A) < \infty$ and that $\epsilon \geq 0$ is such that $\sum_{n=1}^{\infty} \mu (A_{n}) \leq \mu(A) + \epsilon$. For every $k \in \mathbb{N}$, let $C_{k} = \left\{ x \in A : x \ \text{belongs to at least k of the sets} \ A_{n} \right\}$ Show that:

$a) \ \sum_{n=1}^{\infty} \mu(A_{n} \cap E) \leq \mu (A \cap E) + \epsilon$ for any $E \in \Sigma$.

$b) \ \mu(C_{k}) \leq \frac{\epsilon}{k-1}$ for any $k \geq 2$.

There are hints for both $a)$ and $b)$. The hint for $a)$ says to consider $A$ as a union of pairwise disjoint measurable sets (in the usual way, namely, by defining $B_{1} = A_{1}$ and $B_{n} = A_{n} \setminus \bigcup_{j=1}^{n-1} A_{j}$ for $n \geq 2$ and noticing that $A = \bigcup_{n=1}^{\infty} B_{n}$) and the hint for $b)$ says to consider the set $A \setminus C_{k} = \left\{ x \in A : \sum_{n=1}^{\infty} \chi_{A_{n}} \leq k-1 \right\}$.

So my attempts included expressing $\mu (A \cap E)$ as the series $\sum_{n=1}^{\infty} \mu(B_{n} \cap E)$ which is lower or equal than $\sum_{n=1}^{\infty} \mu(A_{n} \cap E)$ but I didn't get anywhere since i wanna establish and upper bound for $\sum_{n=1}^{\infty} \mu(A_{n} \cap E)$ instead of a lower one. I also tried to prove that $\sum_{n=1}^{\infty} \chi_{A_{n}} \leq \chi_{A} + \epsilon \chi_{E}$ for $x \in E$ so that I could use monotonocity of the integral and the Monotone Convergence Theorem, but I wasn't able to prove that inequality either.

Haven't tried many attempts for $b)$, but I think it could be closely related to $a)$ by substituting $E$ for $A \setminus C_{k}$ or some trick along those lines. The hint for $b)$ leads me to believe that this exercise could be attacked via integration but I could be wrong and maybe the hint points towards something else that I'm missing.

Any hints or ideas (even those ones that don't use the hints of course) would be greatly appreciated and welcome. Thanks in advance.

Answer 1

First note that $B_n$'s are disjoint and their union is $A$. Hence $\sum \mu(A_n) <\sum \mu(B_n)+\epsilon$ which implies $\sum \mu(A_n\setminus B_n) <\epsilon$. Now $\mu(A\cap E)+\epsilon =\sum \mu(B_n \cap E) +\epsilon > \sum \mu(A_n \cap E) $ (because $\sum \mu(E \cap [A_n\setminus B_n]) <\epsilon$). This proves the first part.

For the second part take $E=C_k$ in the first part. Note that $\int \sum_n I_{A_n} I_{C_k} \geq k \int I_{C _k}=k \mu(C_k)$. Hence $k \mu(C_k) \leq \sum_n \mu(A_n\cap C_k) \leq \sum_n \mu(B_n\cap C_k)+\epsilon = \mu (A\cap C_k) +\epsilon \leq \mu(C_k)+\epsilon$. This give part 2).

Answer 2

a)

Let $D_{n}=\bigcup_{i=1}^{n-1}A_{i}$ and $B_{n}=A_{n}-D_{n}$.

The sets $B_{n}$ are disjoint with $A=\bigcup_{n=1}^{\infty}B_{n}$ so that $\mu A=\sum_{n=1}^{\infty}\mu B_{n}$.

Also $A_{n}=B_{n}\cup D_{n}$ where $B_{n}$ and $D_{n}$ are disjoint, so that $\mu A_{n}=\mu B_{n}+\mu D_{n}$.

Then: $$\sum_{n=1}^{\infty}\mu A_{n}=\sum_{n=1}^{\infty}\mu B_{n}+\sum_{n=1}^{\infty}\mu D_{n}=\mu A+\sum_{n=1}^{\infty}\mu D_{n}$$ so that: $$\sum_{n=1}^{\infty}\mu A_{n}\leq\mu A+\epsilon\iff\sum_{n=1}^{\infty}\mu D_{n}\leq\epsilon$$

Based on that we find for $E\in\Sigma$: $$\sum_{n=1}^{\infty}\mu\left(A_{n}\cap E\right)=\sum_{n=1}^{\infty}\mu\left(B_{n}\cap E\right)+\sum_{n=1}^{\infty}\mu\left(D_{n}\cap E\right)=$$$$\mu\left(A\cap E\right)+\sum_{n=1}^{\infty}\mu\left(D_{n}\cap E\right)\leq\mu\left(A\cap E\right)+\sum_{n=1}^{\infty}\mu D_{n}\leq\mu\left(A\cap E\right)+\epsilon$$

b)

Substituting $E=C_k$ we find:$$\sum_{n=1}^\infty\mu(A_n\cap C_k)\leq\mu(C_k)+\epsilon\tag1$$

But we also have:$$\sum_{n=1}^\infty\mathbf1_{A_n\cap C_k}\geq k\mathbf1_{C_k}$$and consequently (just integrate on both sides):$$\sum_{n=1}^\infty\mu(A_n\cap C_k)\geq k\mu(C_k)\tag2$$

Combining $(1)$ and $(2)$ we come to the result.