Cesaro limit of truncated means

Change of variables for expectation: Let $X:(\mathrm{\Omega },\mathcal{F},\mathbb{P})\to (S,\mathrm{\Sigma })$ be a random element, let ${\mathbb{P}}_X$ be its law, and let $g:(S,\mathrm{\Sigma })\to \mathbb{R}$ or $g:(S,\mathrm{\Sigma })\to \mathbb{C}$ be measurable.

1. If $g\ge 0$, then $\mathbb{E}[g(X)]={\int }_{S}gd{\mathbb{P}}_X\mathrm{.}$
2. If $g(X)$ is integrable, then $g$ is integrable with respect to ${\mathbb{P}}_X$ and the same formula holds: $\mathbb{E}[g(X)]={\int }_{S}gd{\mathbb{P}}_X\mathrm{.}$

Dominated convergence: Let $f$ and $(f_n)$ be measurable complex-valued functions such that $f_n\to f$ almost everywhere and $\mathrm{\mid }{f}_n\mathrm{\mid }\le g$ almost everywhere for a single nonnegative measurable function $g$ with $\int gd\mu <+\mathrm{\infty }$. Then $f\in L^1(\mu )$, $\int \mathrm{\mid }{f}_n-f\mathrm{\mid }d\mu ⟶0,$ and hence $\int f_{n}d\mu ⟶\int fd\mu \mathrm{.}$

If $x_k\to L$ then ${\sigma }_n\to L$: convergence implies $(C,1)$-summability to the same value: Let $(x_k)$ be a sequence of reals that converges (def-sequence, def-real-limit), and let $({\sigma }_n)$ be its sequence of Cesaro means (def-cesaro-mean). Then $({\sigma }_n)$ converges as well, and

${\lim }_n{\sigma }_n={\lim }_k{x}_k\mathrm{.}$

Both limits are asserted to exist: the right-hand one by hypothesis, the left-hand one as part of the conclusion. Equivalently: a convergent sequence is $(C,1)$-summable, to its own limit. The notation is licensed by uniqueness of limits of real sequences (lem-limit-unique).

The converse is false (fs-cesaro-converse).