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$\newcommand\Om\Omega\newcommand\A{\mathcal A}\newcommand\I{\mathbb I}$Your original statement is trivial: Take $B_1^i=\Omega_1$ and $B_2^i=\Omega_2$ for all $i$. If you additionally require that $B_1 …

$\newcommand{\R}{\mathbb R}\newcommand{\ep}{\varepsilon}\newcommand{\tX}{\tilde X}\newcommand{\tY}{\tilde Y}\newcommand{\tpi}{\tilde\pi} $Here is how this can be done in an explicit way, at least when …

$\newcommand{\ep}{\varepsilon}\newcommand{\R}{\mathbb R}\newcommand{\N}{\mathbb N}$The answer is yes, the $f_n$'s are uniformly integrable wrt to $\mu$. Indeed, let us follow the proof of Theorem 4.5. …

The condition $$\int_X f\,d\rho=\int_Y\int_X f\,d\rho_y\,\mu(dy) \tag{1}$$ holds, by the definition of $\rho$, in the case when $f$ is the indicator of any $\Gamma\in\mathcal X$. By the linearity in $ …

$\newcommand{\N}{\mathbb N} \newcommand{\R}{\mathbb R} \newcommand{\B}{\mathcal B} \newcommand{\F}{\mathcal F} \newcommand{\X}{\mathcal X} \newcommand{\ep}{\epsilon} \newcommand{\la}{\lambda} \newcomm …

The statement Let $\phi:X\to\mathbb{R}$. When $\phi$ depends only on the coordinates $x_k$ for $k\geq n$, this is, $$\phi(x)=\phi(x_n,x_{n+1},x_{n+2},\ldots)\quad\mbox{for }x\in X$$ we hav …

$\newcommand{\F}{\mathcal{F}} \newcommand{\G}{\mathcal G} \newcommand{\HH}{\mathcal H}$ The answer to both your questions is yes, even without assuming the existence of regular versions of conditional …

$\newcommand{\si}{\sigma} \newcommand{\om}{\omega} \newcommand{\Om}{\Omega} \newcommand{\F}{\mathcal{F}}$ The answer to your first question is negative. E.g., let $\Omega:=2^S=\mathcal P(S)$, with $\m …

Let $\nu:=\sigma$. This answer, based mainly on comments by Anthony Quas, provides a necessary and sufficient condition for $\mu\ll\nu$ (the absolute continuity of $\mu$ with respect to $\nu$) in term …

Assuming that $L^1_{\mathbb R}$ denotes the space of all classes of $\mu$-equivalent functions $f\colon E\to\mathbb R$ with $\|f\|_1=\int_E|f|\,d\mu<\infty$, the answer is yes. Indeed, we have $$\s …

Disintegration is associative, in the following sense: Suppose that we have a disintegration $$\mu(dx)=\int_Y(\mu h_1^{-1})(dy)\mu_{h_1}(y,dx)$$ of a probability measure $\mu$ with a kernel $\mu_{h_1 …

Answer to Question 3: Yes, there is such a sequence. E.g., for all $n=0,1,\dots$ and all Borel $B\subseteq[0,1]$, let $$\mu_n(B):=\frac12\int_B(2+\sin2\pi nx)\,dx.$$ Then $\mu_n$ converges to $\mu_0$ …

$\newcommand\de{\delta}$ $\newcommand\ep{\epsilon}$ $\newcommand\al{\alpha}$ Fix any natural $k$. Let $$\de_k:=\inf_{n\ge k}\|f_n 1_{\{|f_n|\le k\}}\|_2$$ and $$\eta_k:=\ep_{k-1}/(4k).$$ We have $$ …

Your desired conclusion does hold without the assumption of a probability density of $\bf x$ and $\bf z$, provided that the functions $f$ and $g$ are assumed to be Borel measurable. Indeed, let $X:= …

The answer is yes. Indeed, a probability measure $\mu$ over $\mathbb R$ has a density bounded by a real $K>0$ iff the cdf of $\mu$ is $K$-Lipschitz, that is, Lipschitz with the Lipschitz constant $K$. …