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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 $\ …

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 …

Well, the question was asked a long time ago, so my answer might not be of much help to the asker any more; but perhaps for the sake of future readers I'll write an answer anyway. Since densities are …

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 …

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(\ …

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 …

[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 …

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 …

Following the suggestion in the first comment below my question (and with the help of the second comment), I can give an example of a scenario that is "even worse" than what I requested, where $A \cup …

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 …

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 $X$ be a Banach space, and for any compactly supported Borel probability measure $\mathbb{P}$ on $X$, define the mean $\mu_\mathbb{P}$ by $\mu_\mathbb{P}=\int_X x \, \mathbb{P}(dx)$. I want to say …

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 …

All the notions of convergence of measures that I know of are either in the purely measure-theoretic category (e.g. strong convergence, total variation), or in the topological category (e.g. weak conv …