Kolmogorov three-series theorem

Zero truncation at a positive level: For a real random variable $X$ and a deterministic level $A>0$, its zero truncation is $X^{(A)}=X{\mathbf{1}}_{\{\mathrm{\mid }X\mathrm{\mid }\le A\}}\mathrm{.}$ The threshold event is measurable because $X$ is measurable and $[-A,A]$ is Borel; its indicator and the product are measurable by thm-arithmetic-and-lattice-operations-preserve-measurability. Thus $X^{(A)}$ is a real random variable as in def-random-element-and-real-random-variable. It equals $X$ at both cutoff endpoints and is zero outside the interval. Since $\mathrm{\mid }{X}^{(A)}\mathrm{\mid }\le A$, for every $0<p<\mathrm{\infty }$ its absolute $p$th moment is at most $A^p\mathbb{P}(\mathrm{\Omega })=A^p$. This is not clipping to the endpoints.

Necessity of the truncated mean and variance conditions: Let independent real random variables $(X_n{)}_{n\ge 1}$ have ${\sum }_n{X}_n$ convergent almost surely. For every fixed $A>0$, set $Y_n=X_n^{(A)}=X_n{\mathbf{1}}_{\{\mathrm{\mid }{X}_n\mathrm{\mid }\le A\}}$. Then ${\sum }_n\mathbb{P}(\mathrm{\mid }{X}_n\mathrm{\mid }>A)<\mathrm{\infty },{\sum }_n\mathrm{Var}(Y_n)<\mathrm{\infty },$ and the real numerical series ${\sum }_n\mathbb{E}{Y}_n$ converges.

Kolmogorov two-series sufficiency: Let $(X_n{)}_{n\ge 1}$ be independent square-integrable real random variables. If ${\sum }_{n\ge 1}\mathbb{E}{X}_n$ converges in $\mathbb{R}$ and ${\sum }_{n\ge 1}\mathrm{Var}(X_n)<\mathrm{\infty }$, then ${\sum }_{n\ge 1}{X}_n$ converges almost surely and in $L^2$.

First Borel-Cantelli lemma for events: Let $(A_n{)}_{n\in \mathbb{N}}$ be events in a probability space. If ${\sum }_{n=0}^{\mathrm{\infty }}\mathbb{P}(A_n)<+\mathrm{\infty },$ then $\mathbb{P}(A_n\mathrm{i}\mathrm{.}\mathrm{o}\mathrm{.})=0.$ No independence hypothesis is needed.

Measurable coordinatewise functions preserve independence: Let $(X_i{)}_{i\in I}$ be an independent family of random elements $X_i:(\mathrm{\Omega },\mathcal{F},\mathbb{P})\to (S_i,{\mathrm{\Sigma }}_i)$. For each $i$, let $g_i:(S_i,{\mathrm{\Sigma }}_i)\to (T_i,{\mathcal{T}}_i)$ be measurable. Then the family $(g_i\circ X_i{)}_{i\in I}$ is independent.

A series converges iff each of its tail series converges, and the sum splits as $s_N$ plus the $N$-th tail: Let $(a_k)$ be a sequence of reals with partial sums $s_n={\sum }_{k<n}{a}_k$, let $N\in \mathbb{N}$, and let $t_j:={\sum }_{i<j}{a}_{N+i}$ be the partial sums of the $N$-th tail series ${\sum }_{k\ge N}{a}_k$ (def-series). Then: 1. $t_j=s_{j+N}-s_N$ for every $j\in \mathbb{N}$; 2. $\sum a_k$ converges if and only if its $N$-th tail series converges, and in that case ${\sum }_{k=0}^{\mathrm{\infty }}{a}_k=s_N+{\sum }_{k=N}^{\mathrm{\infty }}{a}_k;$ 3. hence the following are equivalent: $\sum a_k$ converges; every tail series of $\sum a_k$ converges; some tail series of $\sum a_k$ converges. In words: convergence of a series is a property of its terms from any index on, and changing finitely many terms changes the sum but not the fact of convergence.