Nonatomic finite space, arbitrary small partition.

Question

I am trying to understand the proof that any finite non-atomic measure space can be always be finitely partitioned by sets of arbitrary small size. The proof essentially goes as follows:

For $\epsilon>0$ We denote $\epsilon_1:=\sup\{\mu(A): A\subset X, \mu(A)\le\epsilon\}$. Obviously $0<\epsilon_1<\epsilon$ (if it was 0, then were done). Since it is positive, by non-atomic of our space we can find a set $A_1$ such that $\epsilon_1/2<\mu(A_1)<\epsilon_1<\epsilon$. Now we can consider $A_1^c$ and find $\epsilon_2=\sup\{\mu(A):A\subset A_1^c, \mu(A)\le \epsilon\}$. Now this $\epsilon_2>0$. We can now find an $A_2$ such that $\epsilon/2<\mu(A_2)<\epsilon_2$. We can keep up doing this until we reach a $\mu(A_n^c)=0$ set in which case we are done. So assuming this never happens then we have a disjoint collection $A_n$, $n=1,2,3,..$ such that $\epsilon_n/2<\mu(A_n)<\epsilon_n<\epsilon$. I do not see how this arrives at a contradiction? If you try to sum over all the $n$, the $\epsilon_n$ can be so small that the summation is still smaller than $\mu(X)$? In this case we have a valid infinite partition, and I am sure where the contradiction is.

Answer 1

We are not trying to arrive at a contradiction at all! We are trying to prove that there exists a partition $(A_n)$ such that $\mu (A_n) \leq \epsilon$ for all $n$. Since $A_n$'s are disjoint it follows that $\epsilon_n \to 0$. Now note that if $A$ is disjoint from all the $A_n$'s then we get $\epsilon_n \geq \mu(A)$ for each $n$ by definition of $\epsilon_n$ so $\mu (A)=0$. Thus $A_n$'s do exhaust $X$ up to a null set and we are done.

EDIT:

To get a finite partition note that $\sum \mu (A_n) <\infty$. There exists $N$ such that $\sum\limits_{k=N+1}^{\infty} \mu (A_n) <\epsilon$. Now $A_1,A_2,...,A_N,B$ is your desired partition where $B =X\setminus \bigcup_{k\leq N} A_k$

Answer 2

Here is another approach based on Zorn's Lemma which avoids the tricky business of the $\varepsilon_n$'s.

Given $\varepsilon>0$, let us say that an $\varepsilon$-pre-partition of $X$ is any collection $\mathcal A$ formed by pairwise disjoint measurable subsets $E\subseteq X$, each of which sasisfies $$ 0<\mu(E)<\varepsilon. $$

Besides the above requirement in terms of the measure, the difference between this notion and the usual notion of partition is that we are not requiring the members of $\mathcal A$ to cover $X$.

Since the measure of $X$ is finite, it is easy to see that every $\varepsilon$-pre-partition must be at most countable (this is because we have required the members of a $\varepsilon$-pre-partition to have strictly positive measure).

Next consider the family $\mathcal F$ formed by all $\varepsilon$-pre-partitions, equipped with the order relation given $$ \mathcal A_1 \leq \mathcal A_2 \Leftrightarrow \mathcal A_1 \subseteq \mathcal A_2. $$ Clearly $\mathcal F$ is an inductive ordered set, that is, it satisfies the hypothesis of Zorn's Lemma, so there exists a maximal element, say $\mathcal A_m$.

Arguments like this, including transfinite induction, tend to produce really big things but fortuntely there is no room for monsters in this game. What I mean is that $\mathcal A_m$ must be separable, just like any other element of $\mathcal F$!

Seting $Y=\bigcup \mathcal A_m$ (union of the members of $\mathcal A_m$) is easy to see that $Y$ has full measure, by the maximality of $\mathcal A_m$.

If we choose any member $E\in\mathcal A_m$ and replace it by $E\cup (X\setminus Y)$ then $\mathcal A_m$ will become a true partition by sets of measure $<\varepsilon$.

If $\mathcal A_m$ happens to be finite we are done and otherwise we may proceed as in @KaviRamaMurthy's answer.