# Convergence of a stochastic sequence?

I am reading this paper related to an algorithm for nonsmooth optimization problems. After many simplifications, I was able to formalize the method as follows: let $$\Bbb B$$ denote the unit ball in $$\Bbb R^n$$ and consider a set $$D\subseteq \Bbb B.$$ We have an operator $$T: \Bbb R^n \times D \to \Bbb R^n,$$ and the probability space $$(\Bbb B, \mathcal{B}|_{\Bbb B}, \mu),$$ where

• $$\mathcal{B}|_{\Bbb B}$$ denotes the Borel sigma algebra in $$\Bbb R^n$$ restricted to the unit ball

• $$\mu$$ is the Lebesgue measure restricted to $$\Bbb B$$ scaled so that it is a probability measure, i.e, for $$B\in \mathcal{B}|_{\Bbb B},$$ $$\mu(B):=\frac{\mu'(B)}{\mu'(\Bbb B)},$$ and $$\mu'$$ is the Lebesgue measure in $$\Bbb R^n.$$ Furthermore, we have that $$\mu(D)=\mu (\Bbb B).$$

The algorithm is then as follows:

• Step 0: Take $$x^0\in \Bbb R^n,\; k=0.$$

• Step 1: Sample $$u^k$$ from $$\Bbb B$$ according to the probability space described. If $$u^k \notin D:$$ Stop, return $$x^k.$$ Else, go to Step 2

• Step 2: Take $$x^{k+1}:= T(x^k, u^k),\; k=k+1.$$ Go to Step 1.

The main result of the paper states that:

• With probability one, the method above never stops and every accumulation point of the generated sequence belongs to a set $$S$$ (of stationary points).

I am not able to formalize this statement in order to attempt a proof by myself due to the following questions:

1. According to such statement, the set

$$A:= \{\{x^k\}: \{x^k\} \textrm{ is generated by the method(that does not stop) and every accumulation point belongs to } S\}$$ is an event in some probability space $$(X,\mathcal{A},\nu)$$ with $$\nu(A)=1.$$

Who is $$(X,\mathcal{A},\nu)$$? In the paper, this is not explained because it seems not to be needed.

1. If you don't need the definition of $$(X,\mathcal{A},\nu),$$ is there a branch of probability theory that is devoted to the study of these types of statements? To my little knowledge, this seems to be related to stochastic processes or random walks.

$$\newcommand{\F}{\mathcal{F}}$$
It appears that by $$\mu(D)=\mu(B)$$ you meant $$\mu(D)=\mu(\Bbb B)$$; otherwise, this condition would not make sense.
It also appears that the sentence "Sample $$u^k$$ from $$\Bbb B$$ according to the probability space described", which you quoted, means that $$u^0,u^1,\dots$$ are independent. More generally, we may assume that the joint distributions, say $$\mu_k$$, of the finite sequences $$(u^0,\dots,u^k)$$ on the corresponding product sigma-algebras $$\F^{\otimes (k+1)}$$ (where $$\F:=\mathcal{B}|_{\Bbb B}$$) are defined consistently for all $$k=0,1,\dots$$, in the sense that for each such $$k$$ the push-forward measure $$\mu_{k+1}\pi_{k+1,k}^{-1}$$ of the probability measure $$\mu_{k+1}$$ under the projection $$\pi_{k+1,k}$$ from $$\Bbb B^{k+2}$$ onto $$\Bbb B^{k+1}$$ coincides with $$\mu_k$$. (If the $$u_j$$'s are independent, then $$\mu_k$$ is the product measure $$\mu^{\otimes (k+1)}$$.) Let $$\F^{\otimes\infty}$$ be the smallest sigma-algebra over $$\Bbb B^\infty$$ containing the set of all cylindrical sets, that is, of all sets of the form $$\pi_{\infty,k}^{-1}(E)$$, where $$k=0,1,\dots$$, $$E\in\F^{\otimes (k+1)}$$, and $$\pi_{\infty,k}$$ is the projection from $$\Bbb B^\infty$$ onto $$\Bbb B^{k+1}$$.
Then, by the Kolmogorov extension theorem, there is a unique probability measure $$\mu_\infty$$ on $$\F^{\otimes\infty}$$ such that for all $$k=0,1,\dots$$ and all $$E\in\F^{\otimes (k+1)}$$ we have $$\mu_\infty(\pi_{\infty,k}^{-1}(E))=\mu_k(E)$$. So, for your probability space $$(X,\mathcal{A},\nu)$$ you can take $$(\Bbb B^\infty,\F^{\otimes\infty},\mu_\infty)$$.