To what extent will Radon-Nikodym's theorem hold if we do not admit axiom of choice?

Question

In the proof of the Radon-Nikodym theorem, the function $f = \frac{\mathrm{d} \nu}{\mathrm{d} \mu}$ is constructed as the supremum of measurable functions satisfying $\int_A f \mathrm{d} \mu \leq \int_A \mathrm{d}\nu$, and the existence of this supremum is by Zorn's lemma, an equivalent statement of the axiom of choice (AC), which is not admitted by many mathematicians.

So, does Radon-Nikodym's theorem still hold if we don't admit AC? And if it doesn't, then can we get a similar statement like Radon-Nikodym's theorem (probably weaker)?

Answer 1

You can actually prove versions of Radon-Nikodym without AC or LEM! Indeed, there are constructive (in the sense of Bishop) versions of the Radon-Nikodym theorem, though I haven't spent any time thinking about them, so I'm not sure exactly how much weaker they are, or which ~bonus assumptions~ you might need to make them work.

For instance, Bridges proves in The Constructive Radon-Nikodym Theorem that

Let $I$, $J$ be the complete extensions of two integrals defined on the same initial integration set $L$, such that $J$ is $I$-absolutely continuous and $X$ is $J$-finite. Suppose that there exists a sequence $X_1 < X_2 < \cdots$ of $I$-integrable sets such that $\chi_{X_n}$ is converges in $I$- and $J$-measure to $1$, and $J_{X_n}$ is normable with resepct to $I_{X_n}$ for each $n \geq 1$. Then there exists an essentially unique $I$-integrable function $f_0$ such that, for each $f \in L_1(I) \cap L_1(J)$, $ff_0$ is $I$-integrable and $I(ff_0) = Jf$.

According to the introduction of this paper, there is a different constructive version of Radon-Nikodym which can be found in Bishop's Foundations of Constructive Analysis (Ch 7. Thm 2), which applies only to measures on a locally compact space. I haven't looked at this reference, so it's possible that the restriction to locally compact spaces makes some other aspect of it more nice, which will work better for your purposes.

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I hope this helps ^_^