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It is true that there is a universal constant $c$ such that $\mathbb E[f(Z)]\le c\mathbb E[f(Y)]$. The best (smallest) value for $c$ that I can prove now is $c=2.187...$, but expect that it holds for …

No. Suppose that $\Omega$ consists of four points arranged in a square, where adjacent points have distance 1 between them and opposite points have distance 2. Specifically, if the points are labeled …

No, it is not possible for $\mu$ to be complete. There exists a closed subset $K$ of $X$ with $\mu(K)=0$ and a continuous onto map $f\colon K\to2^\omega$. With $K,f$ as above, if $A\subseteq …

There always exists a dominating measure. First, given two finite measures $\mu,\nu$ on $(\Omega,\mathcal{F})$, the Lebesgue decomposition theorem says that there is an $A\in\mathcal{F}$ such that $1 …

You could alternatively try defining Gaussian measures as $2$-stable distributions. This does remove any reliance on finite dimensional projections, and even removes reference to topology. Let $V$ be …

Showing that the two definitions agree almost everywhere is easy! Using the truncated transform $$ \mathcal{H}\_\epsilon\mu(x)=\frac1\pi\int_{\lvert y-x\rvert > \epsilon}\frac{d\mu(y)}{x-y} $$ then, b …

As suggested in my comment, here's a simple fact which applies to any probability measure $\mu$ on (the Borel σ-algebra of) a second countable topological space $X$. There is a unique minimal closed s …

The set $Z_{\mathcal{P^\ast}}$ is never compact except in the case where it is a singleton (or empty). This is for the general case with $S=(S^1,S^2,\ldots,S^d)$ being an $\mathbb{R}^d$-valued semimar …

As suggested in the question, $P_t$ need not be a well defined operator on $L^\infty(W,\mu)$. That is, $F$ can be zero $\mu$-almost everywhere, but $P_tF$ is nonzero on a set of positive $\mu$-measure …

Yes, it is true that every such operator factors through a canonical map. Theorem: Let $(\Omega,\mathcal{F},\mathbb{P})$ be a probability space and $A\colon H\to L^0(\mathbb{P})$ be a continuous l …

No, there does not exist any such universal constant C. I'll build up a counterexample inductively. First, suppose that we have the following. (i) Let $K_1,K_2$ be compact and absolutely convex s …

Here's a proof that $\exp(z)$ is not a characteristic function using the product expansion for the determinant, which is essentially equivalent to Lidskii's theorem stating that the trace of a trace c …

No, $S$ does not have to span $C(X)$. Taking the case with $X=[0,1]$, let $\mu$ be any atomless finite signed measure whose positive and negative parts $\mu^+$,$\mu^-$ have full support, so that $\mu …

Yes, it is possible to extend the state space with respect to which $Y$ is a Feller process. Then, $X$ will be a dense open subset of the extension $\hat X$. Furthermore, for any initial distribution …

No. In fact, every Lebesgue measurable function $f\colon I\to E$ is equal almost everywhere to a limit of simple Lebesgue measurable functions. As you hint at in the question, this is easy to show in …