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Let $\mu$ and $\mu^\prime$ be probability measures over the sigma algebra $\Sigma$ consisting of the Lebesgue measurable subsets of $[0,1]$. Suppose also that $\mu$ and $\mu^\prime$ assign measure $0$ to all and only null sets. (Notation: if $X$ is measurable, let $\mu_X$ denote the renormalized measure over measurable subsets of $X$: $\mu_X(Y)=\mu(Y)/\mu(X)$ (and similarly for $\mu^\prime$).)

Now suppose that we have a collection of functions:

$$\{f_X:X\rightarrow \overline{X} \mid X\in \Sigma, 0<\lambda(X)<1\}$$

such that for every way of partitioning $[0,1]$ into two sets, $X$ and $\overline{X}$, $f_X$ preserves the measure both between $\mu_X$ and $\mu_{\overline{X}}$ and between $\mu^\prime_X$ and $\mu^\prime_{\overline{X}}$. (In other words $\mu_X(f^{-1}(Y)) = \mu_{\overline{X}}(Y)$, and similarly for $\mu^\prime$.)

Does it follow that $\mu=\mu^\prime$?

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    $\begingroup$ Firstly, μ, μ′ and λ are absolutely continuous wrt one another. Its therefore okay to assume μ′=λ. From there, you've got some real problems. You could assume μ([0,1/2])=a<1/2, for instance, and then chase what that means about the measures of other sets, until you have enough information to determine μ. Then, telling if μ∼λ is actually tricky. The best result is $$\int\sqrt{ \frac{d\mu}{dm} \frac{d\lambda}{dm} } dm >0$$ iff μ∼λ where m=μ+λ. There's also a Haar measure thing going on maybe, related to $\langle f_X \ |\ \lambda(X)=1/2 \rangle$ but that's an ugly group. Great problem! $\endgroup$ – user24142 May 12 '15 at 20:14
  • $\begingroup$ I agree with the other commenter that this is a "great problem", but could you provide some motivation or context for your question? It might provide some insight into a solution. $\endgroup$ – Matt Rosenzweig May 14 '15 at 16:34
  • $\begingroup$ I can try! I'm a philosopher, not a mathematician so my motivations may not be that helpful for solving the problem... $\endgroup$ – Andrew Bacon May 14 '15 at 16:57
  • $\begingroup$ If you define a operation on propositions, written $A\rightarrow B$, by $AB \cup f_{\bar{A}}^{-1}(B)$ then this operation has the property that the probability of $A\rightarrow B$ is the conditional probability of $B$ on $A$. Here the probability is measured by $\mu$ or $\mu^\prime$, as given in the question. I suspect that for certain choices of families of functions there's an interesting class of measures that makes the probability of $A\rightarrow B$ the same as conditional probability. Connectives like $A\rightarrow B$ are potentially useful for modelling natural language conditionals. $\endgroup$ – Andrew Bacon May 14 '15 at 16:58
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I assume that all $f_X$ are bijections (so that the inverse is measurable) - or at least I need the sets $f_X^{-1}(Y)$ to generate the $\sigma$-algebra on $X$. I suspect that a similar argument could somehow work in the arbitrary case, but just suspect at this point.

In the notations of the question, denote $p=d\mu/d\mu'$, then for a measurable $Y\subset \overline{X}$ \begin{multline*} \int_{f_X^{-1}(Y)}p(x)\mu'_X(dx)= \frac{\mu(X)}{\mu'(X)}\int_{f_X^{-1}(Y)}\mu_X(dx)= \frac{\mu(X)}{\mu'(X)}\int_Y \mu_\overline{X}(dx)=\\= \frac{\mu(X)\mu'(\overline{X})}{\mu'(X)\mu(\overline{X})} \int_Yp(x)\mu'_\overline{X}(dx) =\frac{\mu(X)\mu'(\overline{X})}{\mu'(X)\mu(\overline{X})} \int_{f_X^{-1}(Y)}p(f_X(x))\mu'_X(dx). \end{multline*}

Since $Y$ is arbitrary, we get that $\mu'$-a.s. on $X$ $$ p(x)=\frac{\mu(X)\mu'(\overline{X})}{\mu'(X)\mu(\overline{X})} p(f_X(x))=C_Xp(f_X(x)). $$ Suppose that $\mu'(p>1+\delta)>0$ for some $\delta$, then $\mu'(p<1)>0$, as $p$ is the Radon-Nikodim density between probability measures. Choose $1+\delta<a<b$ such that $\mu'(a<p<b)>0$ and $b-a<\delta$ (by dividing $(\delta,\infty)$ on the equal intervals of length less than $\delta$). Take $\overline{X}\subset \{a<p<b\}$ such that $\mu'(\overline{X})>0$ and $\mu'(X\cap \{a<p<b\})>0$ (it can be done since $\mu'$ is equivalent to Lebesgue measure).

Depending on the constant $C_X$, we have that $p$ has values in $(C_Xa,C_Xb)$ a.s. on $X$. Since $X$ contains $\{p<1\}$, $C_X<1/a$ and hence $p(X)\subset (-\infty, 1+\delta/a)$. But $X$ also contains a positive subset of $\{a<p<b\}$, thus, since $1+\delta/a<a$, we get a contradiction and it follows that $p\equiv 1$, hence $\mu=\mu'$.

EDIT (idea):

My idea is that if \begin{array}{ll} \nu_X=\nu_{\overline X}f^{-1},\\ \mu_X=\mu_{\overline X}f^{-1},\\ \nu=p\mu, \end{array} then, applying heuristically $f^{-1}$ two times with and without the density $p$, we get (up to normalizing constants): \begin{array}{ll} \nu_{\overline X}f^{-1}=(p\mu_{\overline X}) f^{-1}=(pf^{-1})(\mu_{\overline X}f^{-1}),\\ \nu_{\overline X}f^{-1}=\nu_X=p\mu_X=p(\mu_{\overline X}f^{-1}). \end{array} It means that $p=pf^{-1}$, up to a constant, which seems unlikely for a non-trivial $p$. Rigorously, after the corresponding sequence of integral equations above, we get $(p\circ f)\mu_{\overline X}= p\mu_{\overline X}$ only on the sets of the form $f^{-1}(Y)$, so I need the family of such sets to be rich enough to distinguish values of $p$ to get $p=p\circ f$ a.s. In particular, when $f$ is bijection, this family is the whole Borel $\sigma$-algebra and the idea works.

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  • $\begingroup$ Thanks, this was the most helpful answer. I'm still a little bit unsure about the last of those integral identities. Is this where the bijectivity assumption is coming in? Do you know if it's possible to do without it? $\endgroup$ – Andrew Bacon May 20 '15 at 17:20
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    $\begingroup$ I added the general idea and explanation on where the bijectivity is used. $\endgroup$ – Jorkug May 20 '15 at 18:20
  • $\begingroup$ Thanks -- that explanation was v. helpful! $\endgroup$ – Andrew Bacon May 20 '15 at 19:06
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Your assumption about having a collection of functions satisfying ... holds for any pair of mutually absolutely continuous atomless probability measures on the unit interval. So every pair $\mu \neq \mu'$ is a counterexample. To show this it is enough to prove the following.

Claim 1: Let $X, Y$ be complete separable metric spaces. Let $\mu_1, \mu_2$ be two atomless Borel measures on $X$ and $\nu_1, \nu_2$ be atomless Borel measures on $Y$. Then there exists a mod-null bijection $f: X \to Y$ such that $f$ is measure preserving w.r.t. $\mu_i, \nu_i$ for $i = 1, 2$.

To show Claim 1 you should first show something like the following:

Claim 2: Let $f:[0, 1] \to \mathbb{R}$ be such that $\int_{[0, 1]} f = 0$ (lebesgue intergral). Then for every $\delta \in [0, 1]$ there is a subset $A \subseteq [0, 1]$ of lebesgue measure $\delta$ such that $\int_{A} f = 0$.

Using Claim 2 you can construct countably branching trees of compact subsets of $X, Y$ respectively such that the nodes agree in their respective measures. This helps you prove claim 1.

This is the rough idea. I think it should work.

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  • $\begingroup$ Would you please flesh out your argument a little more? Is the idea to use the first claim to give the existence of the measure-preserving maps $f:X\rightarrow\overline{X}$, for any partition $X\cup\overline{X}=[0,1]$? Regarding the second claim, I don't see how to obtain a probability measure on $[0,1]$ that gives a counterexample. $\endgroup$ – Matt Rosenzweig May 16 '15 at 0:41
  • $\begingroup$ I added some more comments. I hope this helps. $\endgroup$ – hot_queen May 16 '15 at 1:05
  • $\begingroup$ I agree that for any measure of a certain sort on $[0,1]$, you can find a family of functions meeting the conditions stated in the question. But how does this show the possibility of two distinct measures, $\mu$ and $\mu^\prime$, satisfying those conditions relative to a single family of functions? As far as I can see, the family of functions you construct for $\mu$ might be different from the family constructed for $\mu^\prime$. Unless I'm missing something. $\endgroup$ – Andrew Bacon May 16 '15 at 2:07
  • $\begingroup$ That is the content of claim 1 which should follow from claim 2. The idea is to construct countably branching trees of compact subsets of $X, Y$ such that each node in the tree for $X$ has same $\mu_i$-measure and similarly on the $Y$ side. $\endgroup$ – hot_queen May 18 '15 at 19:58
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    $\begingroup$ I think I do get bijections (modulo null) in claim 1 - so my answer is incompatible with Jorkug's :P. I will try to check the details and post them if they work out. Sorry for the sktechiness. $\endgroup$ – hot_queen May 20 '15 at 17:30

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