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Let $(X,\mathcal{A},\mu)$ such that $\mathcal{A}$ an algebra of subsets of $X$ and $\mu:\mathcal{A}\rightarrow [0,1]$ a function such that

i) $\mu(\emptyset)=0$

ii) $\mu(\bigcup_{k=1}^{n}A_k)=\sum_{k=1}^{n}\mu(A_k)$ where the sets $A_k\in\mathcal{A}$ for $1\leq k\leq n$ are disjoints.

Prove that the following ítems are equivalents

iii) $\mu(\bigcup_{k=1}^{\infty}A_k)=\sum_{k=1}^{\infty}\mu(A_k)$ where the sets $A_k\in\mathcal{A}$ for $k\in\mathbb{N}$ are disjoints

iv) $\lim_{n\rightarrow\infty}\mu(A_n)=0$ where $(A_n)_{n\in\mathbb{N}}$ is a sequence of elements of $\mathcal{A}$ such that $A_{n+1}\subset A_n$ for all $n\in\mathbb{N}$ and $\bigcap_{n\in\mathbb{N}}A_n=\emptyset$

I'm able to prove the part iii)$\rightarrow $ iv), using the fact that if $\mu(\bigcap_{n\in\mathbb{N}}A_n)=\lim_{n\rightarrow\mathbb{N}}\mu(A_n) $ following the same proof showed on "The elements of integration and lebesgue measure, Bartle" chapter 3, lemma 3.4.

But I'm not being able to do iv)$\rightarrow$ iii). I think that I have to build a sequence of elements in the algebra $\mathcal{A}$ that is decreasing using a sequence that is disjoint to use the hypothesis, but I can't do it.

I'll appreciate any suggestions, thanks.

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1 Answer 1

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First of all, i) and ii) imply that if $A\subseteq B$ are sets in $\mathcal{A}$ then $\mu(B\setminus A)=\mu(B)-\mu(A)$.

Let's prove iv)$\Rightarrow$iii):

Suppose the sets $A_k\in\mathcal{A}$ for $k\in\mathbb{N}$ are disjoint. For every $k\in\mathbb{N}$, let $B_k=\bigcup_{i=k+1}^\infty A_i$. Notice that $B_{k+1}\subseteq B_k$ and $\bigcap_{k=1}^\infty B_k=\varnothing$. By iv), $\lim\mu(B_k)=0$.

Now, notice that $B_k=(\bigcup_{i=1}^\infty A_i)\setminus(\bigcup_{i=1}^kA_i)$, so i) and ii) imply that $\mu(B_k)=\mu(\bigcup_{i=1}^\infty A_i)-\mu(\bigcup_{i=1}^k A_i)$ (see the first paragraph above). Thus $$0=\lim_{k\to\infty} \mu(B_k)=\lim_{k\to\infty}\left(\mu\left(\bigcup_{i=1}^\infty A_i\right)-\mu\left(\bigcup_{i=1}^k A_i\right)\right)=\mu\left(\bigcup_{i=1}^\infty A_i\right)-\lim_{k\to\infty}\mu\left(\bigcup_{i=1}^k A_i\right),$$ hence $$\mu\left(\bigcup_{i=1}^\infty A_i\right)=\lim_{k\to\infty}\mu\left(\bigcup_{i=1}^k A_i\right)\overset{ii}{=}\lim_{k\to\infty}\sum_{i=1}^k\mu(A_i)=\sum_{i=1}^\infty \mu(A_i)$$

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  • $\begingroup$ I did something similar before but the problem in your proof is the same that I found on mine: to use iii) it's necessary that each $B_k\in\mathcal{A}$, but enumerable union of elements os an algebra isn't on an Algebra necessarily. Is there a way to prove that each $B_k$ belongs to $\mathcal{A}$? $\endgroup$
    – Irene
    Jun 7, 2014 at 10:21
  • $\begingroup$ I was thinking that the fact that I need to prove iii) implies that I can apply $\mu$ to $\bigcup_{n\in\mathbb{N}}A_n$ when $A_n\in\mathcal{A}$ disjoint, right? I just have to prove that $\mu (\bigcup_{n\in\mathbb{N}}A_n)$ is $\sum_{n\in\mathbb{N}}A_n$, assuming that $\bigcup_{n\in\mathbb{N}}A_n$ is already on $\mathcal{A}$ so I can apply $\mu$ to it. Am I right? $\endgroup$
    – Irene
    Jun 7, 2014 at 11:11
  • $\begingroup$ @Irene Your second comment is correct. Notice that for iii) to make sense, you have to assume that $\bigcup_{n=1}^\infty A_n\in\mathcal{A}$, otherwise you cannot evaluate $\mu$ at it. In other words, you should read iii) as "$\mu(\bigcup_{n=1}^\infty A_n)=\sum_{n=1}^\infty \mu(A_n)$ whenever $A_n\in\mathcal{A}$ and $\bigcup_{n=1}^\infty A_n\in\mathcal{A}$." $\endgroup$ Jun 7, 2014 at 16:55

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