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Let $\mathcal{B}$ be the Borel $\sigma$-algebra of $[0,1]$, and let $\mathcal{M}$ be the set of probability measures on $([0,1],\mathcal{B})$, equipped with the evaluation $\sigma$-algebra $\ \sigma(\ …

It is known that a measurable bijection $f \colon [0,1] \to [0,1]$ has a measurable inverse. (Here, all measurability is simply with respect to the Borel $\sigma$-algebra of $[0,1]$.) Now fix an arbi …

Let $F$ be the set of bijective Borel-measurable functions $f \colon [0,1] \to [0,1]$ that preserve the Lebesgue measure. Is it the case that for every non-Lebesgue-measurable set $A \subset …

Let $(X,\Sigma)$ be a standard measurable space, let $\rho$ be a probability measure on $(X,\Sigma)$, and let $\mathcal{E}$ be a sub-$\sigma$-algebra of $\Sigma$. We will say that a stochastic kernel …

I'm trying to pinpoint the "intuitive argument" for Freiling's Axiom of Symmetry. It's meant to be a "probabilistic" argument, so thinking about what seems to me to be the probabilistic intuition, it …

[This question is an extension of my question Does a positive-measure subset of the unit interval almost surely intersect a random translation of some countable subgroup of $\mathbb{R}$?. I'm asking i …

Let $X$ be a compact metric space, $P_X$ the set of Borel probability measures on $X$, and $K_X$ the set of non-empty closed subsets of $X$. I will define the "topology of weak+Hausdorff convergence" …

Okay, I think I've worked out that the answer is no, i.e. there exists a sub-$\sigma$-algebra $\mathcal{G}$ of $\mathcal{B}$ such that $\mathbb{E}_\mathcal{G}$ is not universally measurable. (We will …

Following the nice answer to Do the Lebesgue-null sets cover "all the sets can naturally be regarded as sort-of-null sets"?, the particular situation that I am especially interested in (which is a kin …

I've found the answer - it's NO! The paper I found addressing the question is the following: http://projecteuclid.org/euclid.aop/1175287757 ("0-1 Laws for Regular Conditional Probabilities") A simple …

The Hausdorff paradox is an incredibly counter-intuitive consequence of the axiom of choice; it is also important for demonstrating the non-existence, under AC, of a rotation-invariant measure on the …

I'm going to assume that your space is locally compact (as well as $\sigma$-compact), so that $X$ is the union of a sequence of compact sets where each lies in the interior of the next. In this case, …

Let $(X,\Sigma)$ be a measurable space [which we can assume to be a standard Borel space if we wish]. Let $\mathcal{S}$ be a set of probability measures on $(X,\Sigma)$. [If we wish, we can assume th …

Please forgive me if this is a very easy question. Let $A \subset [0,1]$ be a Borel-measurable set with strictly positive Lebesgue measure. Does there necessarily exist a countable subgroup $G$ of $\ …

I'm pretty sure I can prove the "Theorem" given further below (without very much difficulty), but it seems way too basic not to have been noticed before. So I'm wondering if there are any papers/text …