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Let $(M, F,\mu)$ be a finite measure space, $f:M\to M$ be a $\mu$-measurable function such that $\forall A\in F:\mu(A)=\mu(f^{-1}(A))$ and $E\in F:\mu(E)>0$. Define

$$\rho_E(x) = \min\{n\geq 1:f^n(x)\in E\}$$

$$E_0 = \{x\in E:f^n(x)\not\in E \text{ for every $n\geq 1$}\}$$

$$E_0^* = \{x\in M:f^n(x)\not\in E \text{ for every $n\geq 0$}\}$$

$$E_n = \{x\in E:f^k(x)\not\in E \text{ for every $1\leq k\leq n-1$, but $f^n(x)\in E$}\}$$

$$E_n^* = \{x\in M:f^k(x)\not\in E \text{ for every $0\leq k\leq n-1$, but $f^n(x)\in E$}\}$$

I am trying to understand why $\sum_{i=n}^\infty \mu(E_i) = n\mu(E_n)$. The following is known:

  1. $E_n, E_n^*$s form a disjoint partition of $\mu$-measurable sets for $M

  2. $f^{-1}(E_n^*) = E_{n+1}^*\cup E_{n+1}$

  3. $\mu(E_n^*) = \mu(E_m^*) + \sum_{i=n+1}^m\mu(E_i) \text{ for every $m>n$}$

  4. The prior parts imply that $\mu(E_n^*) = \sum_{i=n+1}^\infty \mu(E_i)$

  5. Thus $\mu(M) - \mu(E_0^*) = \sum_{n=1}^\infty\sum_{i=n}^\infty\mu(E_i)$

How can we then say that $\sum_{n=1}^\infty\sum_{i=n}^\infty\mu(E_i) = \sum_{n=1}^\infty n\mu(E_n)$?

Cartesian Bear
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2 Answers2

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I am trying to understand why $\sum_{i=n}^\infty \mu(E_i) = n\mu(E_n)$.

This is not necessarily true; but see below.

How can we then say that $\sum_{n=1}^\infty\sum_{i=n}^\infty\mu(E_i) = \sum_{n=1}^\infty n\mu(E_n)$?

One can write the infinite double sum on the LHS as a sum of terms that are typographically arranged so that on row $1$ the terms of the inner sum with $n=1$ are seen, on row $2$ the leftmost position is skipped and the terms of the inner sum with $n=2$ are seen etc:

\begin{align*} \sum_{i=1}^\infty\mu(E_i) &= \mu(E_1) + \mu(E_2) + \mu(E_3)+\cdots\\ \sum_{i=2}^\infty\mu(E_i) &= \phantom{\mu(E_1) +}\, \mu(E_2) + \mu(E_3)+\cdots\\ \sum_{i=3}^\infty\mu(E_i) &= \phantom{\mu(E_1) +\mu(E_2) +} \, \mu(E_3)+\cdots\\ \vdots \end{align*}

Reading the terms column by column gives the RHS. (This reordering is admissible; see Why can we interchange summations?)

Alp Uzman
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I think some of the facts you wrote are irrelevant to your question of why $\sum_{i=n}^\infty \mu(E_i) = n\mu(E_n) $?

I think that the solution to your question comes from the layer cake representation\formula. Define $\tau(x):= \rho_E(x)\cdot \mathbf{1}_E(x)$. Your question amounts to showing $\int_{E_n}\tau d\mu = \mu \big( \infty> \tau(x)\geq n \big)$.

If you know that $\tau$ is finite almost surely, this follows immediately from the following version of the layer cake formula.

$$ n\mu(E_n) =\int_{E_n} \tau d\mu = \int_{0}^\infty \mu \Big( \tau(x)>t \Big)dt = \int_{n}^\infty \mu \Big( \tau(x)>t \Big)dt= \sum_{k=n}^\infty \mu(E_k) +\mu(E_0) . $$

Poincaré recurrence theorem says that $\mu(E_0)=0$.

Keen-ameteur
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