Prove $\sum_{n=1}^{\infty} n \mu(A_n) = \sum_{n=1}^{\infty}\mu(B_n) = \sum_{n=1}^{\infty} \mu(E_n)$

Here is the full question:

Let $(X, \mathfrak{U},\mu)$ be a measure space and let $E_k$ be measurable sets s.t. $\sum_{k=1}^{\infty}\mu(E_k) \lt \infty$. Define:

$$A_n = \{x \in X : x \in E_k \text{ for exactly $n$ values of $k$}\}$$ $$B_n = \{x \in X : x \in E_k \text{ for at least $n$ values of $k$}\}$$

Show that the sets $A_n, B_n$ are measurable and

$$\sum_{n=1}^{\infty} n \mu(A_n) = \sum_{n=1}^{\infty}\mu(B_n) = \sum_{n=1}^{\infty} \mu(E_n)$$

Partial solution:

The measurability part was no problem. (Both $A_n$ and $B_n$ can be written as countable unions of countable intersections).

For the first equality $\sum_{n=1}^{\infty} n \mu(A_n) = \sum_{n=1}^{\infty}\mu(B_n)$, I used the fact $B_n=B_{n+1} \cup A_n$ and that it is a disjoint union (take measures of both sides then multiply by $n$...).

I can't figure out how to approach the second equality, there is no obvious identity i can use.

ADDED: The proof should use nothing more than simple theorems and definitions of $\sigma$-algebras algebras semi rings and measures. (In particular no integrals).

Solution 1:

We define

$$f = \sum_{k=1}^\infty \chi_{E_k}.$$

Since the $E_k$ are measurable, their characteristic functions are measurable, and hence so is $f$. Since $\sum_{k=1}^\infty \mu(E_k) < \infty$, $f$ is integrable, and hence $f^{-1}(\infty)$ is a null set.

Then we have

$$A_n = f^{-1}(\{n\}),\quad B_n = f^{-1}([n,+\infty))$$

which shows these sets are measurable, and

$$\sum_{k=1}^\infty \mu(E_k) = \int_X f\,d\mu = \sum_{n=1}^\infty \int_{A_n} f\,d\mu = \sum_{n=1}^\infty n\mu(A_n)$$

by - for example - the monotone convergence theorem.

Further, we have

$$f = \sum_{n=1}^\infty \chi_{B_n}$$

which implies

$$\sum_{k=1}^\infty \mu(E_k) = \int_X f\,d\mu = \int_X \sum_{n=1}^\infty \chi_{B_n}\,d\mu = \sum_{n=1}^\infty \mu(B_n).$$

Note that if $\bigcap\limits_{n=1}^\infty B_n$, the set of points lying in infinitely many $E_k$, has positive measure, then we still have the equality $\sum\limits_{k=1}^\infty \mu(E_k) = \sum\limits_{n=1}^\infty \mu(B_n)$, both are $+\infty$ then, but it can be that $\sum_{n=1}^\infty n\mu(A_n) < +\infty$ in that case.

Without integrals, it is not as straightforward.

I hoped to avoid it, but I don't see a nice direct way without some construction of the following sort.

First we note that the set of points lying in infinitely many $E_k$ is a null set. If available, that's an application of the Borel-Cantelli lemma, otherwise we prove it here.

$$N := \bigcap_{n=1}^\infty \left(\bigcup_{k=n}^\infty E_k\right)$$

is the set of points lying in infinitely many $E_k$. And since for all $n\in \mathbb{N}\setminus \{0\}$ we have

$$\mu(N) \leqslant \mu \left(\bigcup_{k=n}^\infty E_k\right) \leqslant \sum_{k=n}^\infty \mu(E_k)$$

by the montonicity and $\sigma$-subadditivity of $\mu$, the finiteness of $\sum \mu(E_k)$ implies $\mu(N) = 0$. By replacing all $E_k$ with $E_k\setminus N$, all the measures remain unchanged, hence we may assume $N = \varnothing$.

Let $\mathcal{P}$ denote the set of nonempty finite subsets of $\mathbb{N}\setminus \{0\}$, and for $K \in\mathcal{P}$, we define

$$F_K = \bigcap_{k\in K} E_k$$

and

$$G_K = F_K \setminus \bigcup_{k\notin K} E_k.$$

Since $\mathcal{P}$ is countable, the $F_K$ and $G_K$ are measurable. Also $G_{K} \cap G_M = \varnothing$ for $K\neq M$, and

$$\bigcup_{k=1}^\infty E_k = \bigcup_{K\in \mathcal{P}} G_K.$$

Further we have

$$A_n = \bigcup_{\substack{K\in\mathcal{P}\\ \operatorname{card} K = n}} G_K,\qquad B_n = \bigcup_{\substack{K\in\mathcal{P}\\ \operatorname{card} K \geqslant n}} G_K,$$

which shows that $A_n$ and $B_n$ are countable unions of measurable sets, hence measurable.

Next, for every $k$ we have

$$E_k = \bigcup_{\substack{K\in\mathcal{P}\\ k\in K}} G_K$$

and therefore

\begin{align} \sum_{k=1}^\infty \mu(E_k) &= \sum_{k=1}^\infty \mu\left(\bigcup_{\substack{K\in\mathcal{P}\\ k\in K}} G_K\right)\\ &= \sum_{k=1}^\infty \sum_{\substack{K\in\mathcal{P}\\ k\in K}}\mu(G_K)\\ &= \sum_{K\in\mathcal{P}}\sum_{k\in K} \mu(G_K)\\ &= \sum_{K\in\mathcal{P}} \operatorname{card} K\cdot \mu(G_K)\\ &= \sum_{n=1}^\infty \sum_{\substack{K\in\mathcal{P}\\\operatorname{card} K = n}} n\cdot \mu(G_K)\\ &= \sum_{n=1}^\infty n\left(\sum_{\substack{K\in\mathcal{P}\\\operatorname{card} K = n}} \cdot \mu(G_K)\right)\\ &= \sum_{n=1}^\infty n \mu\left(\bigcup_{\substack{K\in\mathcal{P}\\ \operatorname{card} K = n}} G_K\right)\\ &= \sum_{n=1}^\infty n\mu(A_n). \end{align}

And finally

\begin{align} \sum_{n=1}^\infty \mu(B_n) &= \sum_{n=1}^\infty \mu\left(\bigcup_{\substack{K\in\mathcal{P}\\ \operatorname{card} K \geqslant n}} G_K\right) \\ &= \sum_{n=1}^\infty \sum_{\substack{K\in\mathcal{P}\\ \operatorname{card} K \geqslant n}}\mu(G_K)\\ &= \sum_{n=1}^\infty \sum_{k=n}^\infty \sum_{\substack{K\in\mathcal{P}\\\operatorname{card} K = k}}\mu(G_K)\\ &= \sum_{k=1}^\infty \sum_{n=1}^k \sum_{\substack{K\in\mathcal{P}\\\operatorname{card} K = k}}\mu(G_K)\\ &= \sum_{k=1}^\infty k\cdot \sum_{\substack{K\in\mathcal{P}\\\operatorname{card} K = k}}\mu(G_K)\\ &= \sum_{k=1}^\infty k\cdot \mu(A_k). \end{align}

Unless I'm overlooking something very simple, this exercise can serve as a good illustration of the power of the Lebesgue integral.