# Proof that $(x,\omega)\mapsto X^n(x,\omega)$ measurable and $X^n(x,\cdot)$ convergence in probabiliity gives a uniform limit in $(x,\omega)$

This is part of the proof of Lemma 16.12 from Schilling's Brownian Motion.

Let $$(E,\mathscr{E})$$ be a measurable space and $$(\Omega, \mathscr{A},P)$$ be a probability space. Assume that $$(X^n(x,\cdot))_{n\ge 1}$$ is a sequence of random variables depending on a parameter $$x$$. Also let $$(x,\omega)\mapsto X^n(x,\omega)$$ be $$\mathscr{E}\otimes \mathscr{A}$$ measurable and for every $$x\in E$$, the sequence $$X^n(x,\omega)$$ converges in probability.

The goal is then to show that there is a version of the limit $$X(x,\omega)$$ such that $$(x,\omega)\mapsto X(x,\omega)$$ is $$\mathscr{E}\otimes \mathscr{A}$$ measurable.

The proof proceeds as follows.

By assumption, $$\lim_{m,n\to \infty}P(|X^n(x,\omega)-X^m(x,\omega)|>2^{-k})=0$$ for all $$k\ge 0$$ and $$x\in E$$. Next we define $$n_k(x):=\inf\{m>n_{k-1}(x):\sup_{i,j\ge m} P(|X^i(x,\cdot)-X^j(x,\cdot)|>2^{-k})\le 2^{-k}\}.$$

The text then says obviously, $$x\mapsto n_k(x)$$ is $$\mathscr{E}$$-measurable and we see that $$P(\sup_{x\in E} |X^{n_k(x)}(x,\cdot)-X^{n_l(x)}(x,\cdot)|>2^{-k})\le 2^{-k}$$ for all $$l>k$$. And then using the completeness of the convergence in probability, this shows that $$X^{n_k(x)}(x,\omega)$$ converges in probability, uniformly in $$x\in E$$ and we can set this limit as $$X(x,\omega)$$.

My questions are $$1$$. Why is $$x\mapsto n_k(x)$$ $$\mathscr{E}$$ measurable? I can't see why this is obvious.

$$2$$. From the definition of $$n_k$$, how can we get the boundedness in probability, uniformly in $$x$$ of the form $$P(\sup_{x\in E} |X^{n_k(x)}(x,\cdot)-X^{n_l(x)}(x,\cdot)|>2^{-k})\le 2^{-k}$$ for all $$l>k$$?

$$3$$.Finally, how can we show that this implies convergence in probability for $$X^{n_k(x)}(x,\omega)$$ uniformly in $$x\in E$$?

I would greatly appreciate if anyone could provide the details here.

• Have I answered your question? Commented Feb 6, 2023 at 8:24
• @AdityaDhawan No it only shows that $\sup_{i,j\ge m} P(|X^i(x)-X^j(x)|>2^{-k}$ is $\mathscr{E}$ measurable. It doesn't show why we get the ucp convergence. Commented Feb 8, 2023 at 11:13
• Like I mentioned, I do not believe that UCP convergence is guaranteed. That being said, as I have shown, it is not necessary to have UCP convergence to show that the limit $X(x,\omega)$ has a measurable version. Feel free to let me know if you don't understand the argument, or see an error. Commented Feb 8, 2023 at 12:15

To show $$x \to n_k(x)$$ is $$\mathscr{E}$$-measurable, it suffices to show that the map $$x \to P(|X^i(x,\cdot)-X^j(x,\cdot)|>2^{-k})$$ is $$\mathscr{E}$$ measurable $$\forall i,j,k \in \mathbb{N}$$, since then the map $$Y_m(x) = \sup_{i,j\ge m} P(|X^i(x,\cdot)-X^j(x,\cdot)|>2^{-k})$$ is $$\mathscr{E}$$ measurable $$\forall \ m$$, so that the event $$\{n_k = n\} = \{n_{k-1} \lt n\} \cap \{Y_m \gt 2^{-k} \ \forall m \lt n \} \cap \{Y_n \leq 2^{-k}\}$$, which is of course in $$\mathscr{E}$$ under the inductive hypothesis $$x \to n_{k-1}$$ is $$\mathscr{E}$$-measurable, so that by induction $$x \to n_k(x)$$ is measurable.

Note that since $$(x,\omega) \to X^{n}(x,\omega)$$ is measurable in the product $$\sigma$$-algebra $$\forall \ n$$, it is true that $$x \to |X^i(x,w)-X^j(x,w)|$$ is also measurable. One property of the product $$\sigma$$-algebra is that if $$(x,\omega) \to Z(x,\omega)$$ is bounded-measurable in the product sigma-algebra, then $$x \to \int Z(x,\omega) dP(\omega)$$ is $$\mathscr{E}$$-measurable, and so taking $$Z = 1_{A}(|X^{i}-X^j)|$$, where $$A$$ is a Borel set, we get that the map $$x \to P(|X^{i}(x,\cdot)-X^{j}(x,\cdot)| \in A)$$ is $$\mathscr{E}$$ measurable, and in particular this holds for $$A = (2^{-k},\infty)$$.

The definition of $$n_k$$ tells us that $$\sup_{x\in E} P(|X^{n_k(x)}(x,\cdot)-X^{n_l(x)}(x,\cdot)|>2^{-k})\le 2^{-k}$$, and it is not clear to me how this can be extended to $$P(\sup_{x\in E} |X^{n_k(x)}(x,\cdot)-X^{n_l(x)}(x,\cdot)|>2^{-k})\le 2^{-k}$$ without further regularity conditions on $$\mathscr{E}$$ and $$X$$. In fact it is not even clear why $$\omega \to \sup_{x\in E} |X^{n_k(x)}(x,\omega)-X^{n_l(x)}(x,\omega)|$$ is measurable!

However, it is enough to have $$\sup_{x\in E} P(|X^{n_k(x)}(x,\cdot)-X^{n_l(x)}(x,\cdot)|>2^{-k})\le 2^{-k}$$ : For each $$x$$, Borel Cantelli shows that almost surely, $$|X^{n_k(x)}(x,\cdot)-X^{n_l(x)}(x,\cdot)| \leq 2^{-k}$$ eventually, which shows that $$\forall x, X^{n_k(x)}(x,\cdot)$$ is almost surely Cauchy, and therefore we deduce that $$\forall x, X^{n_k(x)}(x,\omega) \to X(x,\omega)$$ almost surely.

Thus $$X(x,\omega)$$ is almost surely $$\limsup_{k} X^{n_k(x)}(x,\omega)$$, a sequence of $$\mathscr{E} \times \mathscr{A}$$ measurable random variables, and is therefore itself measurable.

• Why does almost surely Cauchy imply almost surely convergence? Also, the limit $X(x,\omega)$ is obtained by fixing $x$ and taking the limit in $\omega$, so how do we guarantee that we get a convergence of the mapping $(x,\omega)\mapsto \lim sup_k X^{n_k(x)}(x,\omega)$? Commented Aug 26, 2023 at 16:05
• In fact, where is the measurability of $n_k$ used here? Commented Aug 26, 2023 at 16:58
• $n_k$ depends on $x$ so how do we ensure $X^{n_k(x)}(x,\omega)$ is measurable in $\mathscr{E}\times \mathscr{\Omega}$? Commented Aug 26, 2023 at 17:05

I can give you some details about point 1.

I assume we start with something like $$n_0(x)=0$$. First we note that measurability of $$n_1$$ follows if we show that $$\{x\in E\mid n_1(x)>t\}$$ is measurable for each $$t\in\mathbb{N}$$. For such a $$t$$ we see that $$n_1(x)>t$$ if and only if $$\sup_{i,j\geq s}P(\lvert X^i(x,\cdot)-X^j(x,\cdot)\rvert>2^{-1})> 2^{-1}$$ for each $$s\leq t$$. Hence, we may write $$\{x\in E\mid n_1(x)>t\}=\bigcap_{s\leq t}\{x\in E\mid\sup_{i,j\geq s}P(\lvert X^i(x,\cdot)-X^j(x,\cdot)\rvert>2^{-1})> 2^{-1}\}.$$ From the measurability of $$(x,\omega)\mapsto X^n(x,\omega)$$ you can show that for each $$s\in\mathbb{N}$$ the map $$x\mapsto\sup_{i,j\geq s}P(\lvert X^i(x,\cdot)-X^j(x,\cdot)\rvert>2^{-1})$$ is measurable. This gives you measurability of the right-hand side above.

For $$k>1$$ it should be similar, and you will likely need measurability of $$n_{k-1}$$ to prove measurability of $$n_k$$.