Non-atomic finite measure space $(X,\mathcal{A})$ allows a partition $\left( X_i\right)_{i=1}^m$

Question

Let $(X,\mathcal{A},\mu)$ be a finite and non-atomic measure space. I want to show that for every $\varepsilon >0$ there is a finite partition of $X$ into subsets $X_1,...,X_m$ with $\mu(X_i)<\varepsilon$.

I think that I have found a simple proof, but have doubts because it is so simple.

Proof:

By the property of non-atomic I can find a sequence $X\supset A_1^1\supset A_2^1\supset ... \supset A_n^1\supset ... $ with $\mu(X)> \mu(A_1^1)> \mu(A_2^1)> ... > \mu(A_n^1)> ...>0$ where $\mu(A_n^1)<\varepsilon$

Now we do the same for $X\setminus A_n^1$:

$X\setminus A_n^1\supset A_1^2\supset A_2^2\supset ... \supset A_n^2\supset ... $ with $\mu(X\setminus A_n^1)> \mu(A_1^2)> \mu(A_2^2)> ... > \mu(A_n^2)> ...>0$ where $\mu(A_n^2)<\varepsilon$

Then we just repeat with $X\setminus (A_n^1\cup A_n^2)$. Since the measure space is finite, there must be (use monotonicity of the measure) an $m\in\mathbb{N}$ with $X=\bigcup_{i=1}^m A_n^i$ where the $A_n^i$ are disjoint by construction.

You're pretty close; you just haven't justified why your decreasing sequence $A^1_i$ eventually has a term with measure $< \epsilon$ (they could all be greater than $\epsilon$ and still be decreasing). — Mees de Vries, Dec 19 '19 at 09:45
@KaboMurphy I just used the definition and forgot that the sequence can be decreasing but still be greater $\varepsilon$. — EpsilonDelta, Dec 19 '19 at 09:51
As the other comments say, you have to justify that $\mu(A_1^n)$. Maybe you can find inspiration here, i.e. the theorem due to Wacław Sierpiński. With that theorem, you don't even need the sequence. — Norse, Dec 19 '19 at 09:51
@KaboMurphy For a fairly small value of "highly non-trivial"... — David C. Ullrich, Dec 19 '19 at 10:50
This old answer to a question of mine is very much related. (From 2011). — Giuseppe Negro, Dec 19 '19 at 11:00

David C. Ullrich · Accepted Answer · 2019-12-19T10:54:42.703

This is immediate from

Theorem (Seirpinski). If $\mu$ is a non-atomic measure on $X$ and $0<\alpha<\mu(X)$ then there exists $E\subset X$ with $\mu(E)=\alpha$.

Incidentally, it seems to me that the comments regarding what's wrong with your proof of the existence of that partition are close to backwards. First, although you didn't explain why, which you need to do, getting $\mu(A)<\epsilon$ is easy; perhaps "trivial", certainly not "highly non-trivial":

Easy Lemma. If $\mu$ is a non-trivial finite non-atomic measure on $X$ then there exists $E\subset X$ with $0<\mu(E)\le\mu(X)/2$.

Proof: Since $\mu$ is non-atomic there exists $E_1\subset X$ with $0<\mu(E_1)<\mu(X)$. Now either $E=E_1$ or $E=X\setminus E_1$ works.

Applying that finitely many times gives $0<\mu(A)<\epsilon$, no problem. The actual problem with your construction is this:

We start by finding $E_1\subset X$ with $0<\mu(E_1)<\epsilon$, then $E_2\subset X\setminus E_1$ with $0<\mu(E_2)<\epsilon$, etc. It does not follow that there exists $n$ with $X=E_1\cup\dots\cup E_n$ (modulo null sets).

The problem is that $\mu(E_k)$ may be too small; for example if $\mu(E_k)=\mu(X)/3^k$ then $$\mu\left(X\setminus\bigcup_{k=1}^\infty E_k\right)>0.$$

(Indeed, this is exactly why Seirpinski's Theorem is not a trivial consequence of the Easy Lemma; note that the theorem does follow in a very straightforward way from an improved version:

Somehwhat Harder Lemma. If $\mu$ is a non-trivial finite non-atomic measure on $X$ then there exists $E\subset X$ with $\frac13\mu(X)\le\mu(E)\le\frac23\mu(X)$.)

Great answer! So the theorem follows from the "somewhat harder theorem" because we now have a lower bound on the $E_k$ we create in this iteration process as described in my post? Could you elaborate on why it would immediately follow and how to prove the "somewhat harder lemma"? — EpsilonDelta, Dec 19 '19 at 13:01
@EpsilonDelta See for example https://math.stackexchange.com/questions/3322079/sequence-that-converges-in-l2-on-the-real-line-but-nowhere-pointwise/3322108#3322108 — David C. Ullrich, Dec 19 '19 at 13:15

Non-atomic finite measure space $(X,\mathcal{A})$ allows a partition $\left( X_i\right)_{i=1}^m$

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