A nonidentical Bernoulli weak law

Chebyshev weak law for uncorrelated arrays: For each $n\ge 1$, let $X_{n,1},\dots ,X_{n,r_n}$ be square-integrable real random variables on one probability space, pairwise uncorrelated within the row, where $r_n\ge 0$ is finite. Set $S_n={\sum }_{k=1}^{r_n}{X}_{n,k}$ and let $b_n>0$ be deterministic. If $v_n:=b_n^{-2}{\sum }_{k=1}^{r_n}\mathrm{Var}(X_{n,k})⟶0,$ then $(S_n-\mathbb{E}{S}_n)\mathrm{/}{b}_n\to 0$ in $L^2$ and in probability. More precisely, its second moment is $v_n$, and its probability of absolute value at least $\epsilon >0$ is at most $v_n\mathrm{/}{\epsilon }^2$. No independence between rows is required.

Coordinate random elements of a countable product are independent: Under the measure of thm-countable-product-of-probability-spaces, the coordinate maps $X_n(x)=x_n$ have laws ${\mu }_n$ and are independent.

Assuming countable and dependent choice, countable products of arbitrary probability spaces: Assume countable choice and dependent choice. For probability spaces $(E_n,{\mathcal{E}}_n,{\mu }_n{)}_{n\in \mathbb{N}}$ there is a unique probability measure $\mu$ on ${\mathcal{C}}_{\mathbb{N}}$ such that, for every finite $F$, its $F$-coordinate marginal is ${⨂}_{n\in F}{\mu }_n$.

A Bernoulli$(p)$ variable has mean $p$ and variance $p(1-p)$; a binomial$(n,p)$ variable has mean $np$ and variance $np(1-p)$: If $X$ is Bernoulli$(p)$, then $\mathbb{E}[X]=p$ and $\mathrm{Var}(X)=p(1-p)$. If $S$ is binomial$(n,p)$, then $\mathbb{E}[S]=np,\mathrm{Var}(S)=np(1-p)\mathrm{.}$ These formulas include $p=0$, $p=1$, and $n=0$.

Expectations factor over finite products of independent random variables: Let $n\ge 1$, let $X_0,\dots ,X_{n-1}$ be independent real random variables on a common probability space, and let $g_i:\mathbb{R}\to \mathbb{R}$ be Borel measurable for each $i<n$. 1. If every $g_i$ is nonnegative, then $\mathbb{E}[{\prod }_{i<n}{g}_i(X_i)]={\prod }_{i<n}\mathbb{E}[g_i(X_i)]$ in $[0,+\mathrm{\infty }]$. 2. If every $g_i(X_i)$ is integrable, then ${\prod }_{i<n}{g}_i(X_i)$ is integrable and the same factorization holds in $\mathbb{R}$.