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I have been told in university that:

The three following conditions are equivalent for an additive function defined over a $\sigma$-algebra (in fact, over a ring, but let's make life easier by consenting countable unions) $\mathcal{E}\subseteq\mathcal{P}(X)$, where $X$ is any set:

  1. $\mu$ is $\sigma$-additive over a $\sigma$-algebra $\mathcal{E}$, i.e. for all countable disjoint sequences of sets $A_i\in\mathcal{E}$ we have that: $$\mu\left(\bigcup_{i=1}^\infty A_i\right)=\sum_{i=1}^\infty\mu(A_i);$$
  2. For all sequences of sets $A_i\in\mathcal{E}$ such that $A_i\subseteq A_{i+1}$ for all $i$ we have that: $$\mu\left(\bigcup_{i=1}^\infty A_i\right)=\lim_{i\to+\infty}\mu(A_i),$$ which in turn is $\sup\limits_{i}\mu(A_i)$;
  3. For all sequences of sets $A_i\in\mathcal{E}$ satisfying $A_i\supseteq A_{i+1}$ for all $i$ we have that: $$\mu\left(\bigcap_{i=1}^\infty A_i\right)=\lim_{i\to+\infty}\mu(A_i),$$ which in turn is $\inf\limits_{i}\mu(A_i)$.

That 1 implies 2 has been shown to me in the following fashion:

Suppose 1 holds. Let $A_i$ be a sequence of the kind described in 2. Set: $$A_1'=A_1,$$ $$A_2'=A_2\smallsetminus A_1,$$ $$A_3'=A_3\smallsetminus A_2,$$ $$\vdots$$ $$A_n'=A_n\smallsetminus A_{n-1}.$$ These sets are disjoint since for all $i$ we have $\bigcup\limits_{j=1}^{i-1}A_j\subseteq A_i$ due to the form of the sequence, and therefore removing $A_i$ is just like removing that union plus $A_i$, and writing the primed sequence with the unions explicitly written makes the disjointness clear. Furthermore, it is also clear that the union of these sets is the same as that of the $A_i$. By this equality, $\sigma$-additivity and another thing I will explicit after stating the equality, we have that: $$\mu\left(\bigcup_{i=1}^\infty A_i\right)=\mu\left(\bigcup_{i=1}^\infty A_i'\right)=\sum_{i=1}^\infty\mu(A_i')=\sum_{i=1}^\infty(\mu(A_i)-\mu(A_{i-1})).$$ Indeed, $\mu(A_i')=\mu(A_i\smallsetminus A_{i-1})$, and since $A_{i-1}\subseteq A_i$ we have precisely $\mu(A_i\smallsetminus A_{i-1})=\mu(A_i)-\mu(A_{i-1})$. The sum we have written is telescopic, and going to infinity one sees that what "is left", if we add $A_0=\varnothing$ to cancel the $\mu(A_1)$ left uncanceled by the rest of the terms, is precisely the limit of the measures, as we wanted to prove.

The rest was left to the student, with the following remark:

To prove that 3 implies one you play this trick in reverse, and to prove that 2 implies 3 and vice versa you have to build opportune sets.

I tried to complete the proof.

  1. To prove 2 implies 1, I took a sequence of disjoint sets $A_i\in\mathcal{E}$, and considered: $$A_i'=\bigcup_{j=1}^iA_j.$$ They form a nondecreasing sequence, thus by point 2, which is our hypothesis, we have that: $$\mu\left(\bigcup_{i=1}^\infty A_i\right)=\mu\left(\bigcup_{i=1}^\infty A_i'\right)=\sup_i\mu(A_i'),$$ since the unions clearly coincide. But now what is the measure of the primed sets? By additivity, since $A_i'$ is the union of $A_j$ for $j=1,\dots,i$, we get that: $$\mu(A_i')=\sum_{j=1}^i\mu(A_j),$$ and the supremum of that is the infinite sum, proving $\sigma$-additivity.
  2. Then I actually tried to prove 2 implies 3 and vice versa. For 2 implies 3, suppose 2 holds, and take a sequence $A_i$ as in 3. Set $A_i'=A_1\smallsetminus A_i$. This is clearly a sequence as in 2, so by 2 we have: $$\mu\left(\bigcup_{i=1}^\infty A_i'\right)=\sup\limits_i\mu(A_i').$$ But what is the measure of these primed sets? $$\mu(A_i')=\mu(A_1\smallsetminus A_i)=\mu(A_1)-\mu(A_i),$$ since $A_i\subseteq A_1$ for all $i$, so what is the supremum? $$\sup\limits_i(\mu(A_i'))=\sup\limits_i(\mu(A_1)-\mu(A_i))=\mu(A_1)-\inf\limits_i\mu(A_i).$$ But then it is clear that: $$\bigcup_{i=1}^\infty A_i'=A_1\smallsetminus\bigcap_{i=1}^\infty A_i.$$ That intersection is contained in $A_1$, so we have, combining this statement with the passages above, that: $$\mu(A_1)-\mu\left(\bigcap_{i=1}^\infty A_i\right)=\mu\left(A_1\smallsetminus\bigcap_{i=1}^\infty A_i\right)=\mu\left(\bigcup_{i=1}^\infty A_i'\right)=\mu(A_1)-\inf_i\mu(A_i),$$ Q.E.D.
  3. For 3 implies 2, I managed it only with $\mu$ finite. Take $A_i$ a sequence of sets in $\mathcal{E}$ as in 2. Set $A_i'=A_i^c=X\smallsetminus A_i$. By the De Morgan law we have: $$\bigcup_{i=1}^\infty A_i=\left(\bigcap_{i=1}^\infty A_i'\right).$$ So if $\mu(X)=C<+\infty$, we have: $$\mu\left(\bigcup_{i=1}^\infty A_i\right)=C-\mu\left(\bigcap_{i=1}^\infty A_i'\right).$$ Now evidently $A_i'$ is a sequence as in 3, so by 3, which we supposed to hold, the right side above is $C-\inf\limits_i\mu(A_i')$, but that is $\sup\limits_i(C-\mu(A_i'))=\sup\limits_i\mu(A_i)$, Q.E.D.

So what do you think? Does this proof work or are there problems? And how do I extend the last point to non-finite measures?

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  • $\begingroup$ Quick comment: It seems you are doing too much work, since equivalence of all 3 things can be found by just proving "1 implies 2" and "2 implies 3" and "3 implies 1." You have the first, so you just need "2 implies 3" and "3 implies 1." $\endgroup$ – Michael Jun 22 '14 at 19:07
  • $\begingroup$ Yes I know, but with what I have done proving 3 implies 1 or 3 implies 2 concludes the proof anyway, supposing what I have done has no problems. I don't know why I tried 2 implies 1, maybe because it seemed the easiest to do and was suggested by my professor. $\endgroup$ – MickG Jun 22 '14 at 19:10
  • $\begingroup$ Indeed it took me quite a while to prove 2 implies 3, while it took me a second to prove 2 implies 1, which justifies my impression. Of course, if you prove 3 implies 1 I'll accept the answer as concluding the proof :). $\endgroup$ – MickG Jun 22 '14 at 19:12
  • $\begingroup$ Oh, and 2 implies 3 is in the question. $\endgroup$ – MickG Jun 22 '14 at 19:13
  • $\begingroup$ What is an "additive function"? Clearly we can define $\mu(A)=1$ for all sets $A$, in which case 3 holds but 1 does not. Does "additive" mean that $\mu(A \cup B) = \mu(A) + \mu(B)$ whenever $A$ and $B$ are disjoint? $\endgroup$ – Michael Jun 22 '14 at 19:18
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As Michael pointed out in comments, my proofs seem to be correct, and I am doing more work than needed. I have $1\implies2$ from class, I proved $2\implies3$, I only need $3\implies1$.

Let $A_i$ be disjoint measurable sets. The following equality, which holds if the sets are disjoint, will be of use in our proof:

$$\bigcup_{i=N+1}^\infty A_i=\bigcup_{i=1}^\infty A_i\smallsetminus\bigcup_{i=1}^N A_i.$$

For $N\to\infty$, the RHS tends to $\bigcup_1^\infty A_i\smallsetminus\bigcup_1^\infty A_i=\emptyset$, so the LHS does too, it is evidently decreasing, so by our hypothesis of 3 we have:

$$\lim_{N\to\infty}\mu\left(\bigcup_{i=N+1}^\infty A_i\right)=\mu(\emptyset)=0.$$

But by finite additivity and that equality, we have:

$$\mu\left(\bigcup_{i=N+1}^\infty A_i\right)=\mu\left(\bigcup_{i=1}^\infty A_i\right)-\mu\left(\bigcup_{i=1}^N\right),$$

so the LHS goes to zero, thus so does the RHS, giving 1.

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