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Talking about

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

The proofs I see all use Zorn's lemma or transfinite recursion. Don't get me wrong, I think the proof by transfinite recursion is simple, natural and elegant: We try to construct $E$ as a union of sets with small measure. After we have $\omega$ of them it may be that the measure of the union is still less than $t$, so we just go ahead and "define" $E_\alpha$ for countable ordinals $\alpha$, and for some countable $\alpha$ we must have $\mu\left(\bigcup_{\beta<\alpha}E_\beta\right)=t$. (One could easily convert that to a proof by Zorn's lemma that seems simpler and more natural than the Zorn proofs I've seen: Consider a maximal collection $(E_\alpha)$ of disjoint sets of positive measure with $\sum\mu(E_\alpha)\le t$...)

So that's nice and "intuitive", but it seems like there "should" be a more elementary argument. There is a proof using nothing but epsilons and deltas, although it's a little bit complicated. (I kinda like the proof of the Main Lemma there; it's clear that the ML is something tending in the direction of the theorem, and indeed the proof of the theorem from ML is straightforward.)

Anyway, my question is whether anyone sees how to make the following proof work:

Totally Cool Proof. Let $A$ be the "measure algebra", that is, the space of equivalence classes of measurable sets modulo null sets. Define a metric on $A$ by $d(E,F)=\mu(E\Delta F)=||\chi_E-\chi_F||_1$ (say we're in the case $\mu(X)<\infty$.) Then $(A,d)$ is connected (because why?), hence $\mu(A)$ is connected.

Note I tend to believe that $(A,d)$ is in fact path-connected, because it seems to me that a limit of successive refinements of the chain obtained in proof of the $\frac12$theorem at that link should give a path. But no points for working that out; what I'm wondering about is a totally trivial proof that there are no non--trivial clopen sets...

David C. Ullrich
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1 Answers1

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It is true that $(A,d)$ is path-connected, but proving it involves the same ideas that you need to prove Sierpinski's theorem directly.

In fact, even though it is not actually phrased this way, this is essentially how Sierpinski's theorem is proven on Wikipedia. More precisely, they give a slick proof (using Zorn's lemma) of the following strong form of Sierpinski's theorem:

Proposition: For any $B\in \Sigma$ with $\mu(B)=c$, then there exist a monotone function $s: [0,c]\to \Sigma$ with $s(c)=B$ which is a right inverse to $\mu$ (namely $\mu(s(t))=t$).

Then $s$ is a continuous path between $\emptyset$ and $B$ (in fact, it is an isometric embedding), which shows the path-connectedness of your metric space.

Antoine Labelle
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