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Without more assumptions, it's not true. Take $d=1$ and let $$f(x,y) = \begin{cases} x, & \text{if } y=0 \\ 0, & \text{otherwise} \end{cases}$$ Let $X$ be any nontrivial random variable with mean zero …

No. Take any sequence $r_n \to \infty$ and let $X_{n,k}$ be iid $N(0,1)$ for $k \ne r_n$, and $X_{n,r_n} = r_n$. The hypothesis is satisfied because $(X_{n,1}, \dots, X_{n,k})$ are iid $N(0,1)$ as …

I suppose here that $E$ is separable. Let $\epsilon > 0$ and fix a sequence $\epsilon_i > 0$ such that $\sum_{i=1}^\infty \epsilon_i < \epsilon$. Let $B_r$ denote the closed ball of radius $r$. Set $ …

Let $\phi \in C_c$ be nonzero (say for simplicity that $\phi$ is supported in $(0,1)$) and let $f_n(x) = \phi(x-n)$ be its translation to the left by $n$. This sequence converges to 0 uniformly on co …

Here's a partial answer: For $p$ an even integer, it's true. Maybe someone else can see how to handle the other cases. In this case, the desired integral can be written as $$\left(-\sum_{k=0}^{p-1} …

Obviously $f(B) \subseteq f(A)$. And since $B$ is dense in $A$, we have $A \subseteq \overline{B}$, and so $f(A) \subseteq f(\overline{B})$. But by the continuity of $f$, we have $f(\overline{B}) \s …

The topology generated by $d^{PH}$ is, in general, neither coarser nor finer than $d^{TV}$. The Hausdorff distance between the supports of two measures, even when combined with the Prokhorov distance …

Let $\mu_n$ be the measure defined by $$\mu_n(A) = \frac{1}{\tau_n} \int_0^{\tau_n} \mu(A-t)\,dt.$$ In general, this sequence doesn't converge weakly. Suppose for instance that $\mu$ is a point mass …

No. Your first hypothesis is true for any Lebesgue measurable function $f$; there is a sequence of continuous functions $f_n$ such that $f_n \to f$ almost everywhere. This is standard. But there do …

Posting the answer already given in comments: All you really need is that $X_n(t) \to X(t)$ in $L^1$ for each $t$. Conditional expectation with respect to any $\sigma$-field is continuous with respe …

No, this is not true. Let $X \sim N(0,1)$ and define $Y_n$ to be independent of $X$, such that $Y_n = \sqrt{n}$ with probability $1/n$ and 0 otherwise. Set $X_n = X+ Y_n$. Since $Y_n \to 0$ in $L^1 …