Statistical convergence

Remarks

Every convergent sequence with limit $L$ converges statistically to $L$, since each of the sets above is finite and finite sets have density $0$. The statistical limit is unique when it exists.

Example: let $x_n = 1$ if $n$ is a perfect square and $x_n = 0$ otherwise. For $0 < \varepsilon \le 1$, the set of $n$ with $\abs{x_n} \ge \varepsilon$ is the set of squares, which has at most $\sqrt{N}$ elements up to $N$, so its density is $0$. Hence $(x_n)$ converges statistically to $0$, although it does not converge.

Non-example: $x = (1, 0, 1, 0, \dots)$ does not converge statistically. For any $L$, one of $\abs{1 - L}$ and $\abs{L}$ is at least $1/2$. So for $\varepsilon = 1/2$ the set above contains all odd or all even $n$, and its density is not $0$.

A sequence converges statistically to $L$ if and only if there is a set $K$ of density $1$ such that $x_n \to L$ as $n \to \infty$ through $K$. A bounded sequence converges statistically to $L$ if and only if $\frac{1}{n} \sum_{k=1}^{n} \abs{x_k - L} \to 0$, which is the condition of strong Cesàro summability with exponent $1$ applied to the sequence $(x_n)$.