By way of introduction: As expressed in some of the comments, I find the "locally compact" assumption possibly a bit too strong. A weaker assumption than having a locally compact second-countable ...

The claim is true. (As proved here, it is quite unique to sinusoidal vector fields. The same reference also mentions, in its introduction, existing applications of equation (1) with $g$ taking the ...

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 ...

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") ...

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, ...

I have the answers to my two questions. (I've actually had them for a while; apologies for the delay in posting.) I will give them in reverse order: The answer to Q2 is yes; the structure of 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 $A \cup ... View answer Accepted answer 1 votes Obviously in the case that$n=1$, we have$s(x,x)=0$and so if$f'(x) \neq 0$then$\frac{\partial h}{\partial y_1}$doesn't exist at$(x,x)$. So I will assume that$n \geq 2$. Answer to Question 1. ... View answer 1 votes Let$\Omega=\{-1,1\} \times \{-1,1\}$with the discrete$\sigma$-algebra and uniform measure, let$X(\omega_1,\omega_2)=\omega_2$, and let$$\mathcal{A}_n \ = \ \left\{ \begin{array}{l l} \sigma(\{\... View answer Accepted answer 1 votes Yes. As in the comments: take$X=\mathbb{S}^1$; and let$\nu$be the law of the random set constructed by taking a positive-Lebesgue-measure Cantor set$K \subset \mathbb{S}^1$and rotating$K$... View answer 0 votes Having read Mateusz Kwaśnicki's answer, I will now write it in my own way: Lemma. Let$S_\infty$and$T$be separable metric spaces, and let$(S_j)_{j \in \mathbb{N}}$be a sequence of Borel subsets ... View answer 0 votes I think I can now prove the following (which covers the case requested in the bounty): Theorem. Let$g=P/Q$for polynomials$P$and$Q$where$\mathrm{order}(P) \leq \mathrm{order}(Q)$and$Q$has no ... View answer 0 votes I have found an answer; it is based on Proposition 3.10 of here. Claim:$\mathbb{P}_W \otimes \lambda$is the only$\Theta$-invariant probability measure whose projection onto$\Omega$is$\mathbb{P}...