# proof of completeness under ucp convergence

Let $(\Omega,\mathcal{F},(\mathcal{F}_{t})_{t\in\mathbb{R}_{+}},\mathbb{P})$ be a filtered probability space, $\mathcal{M}$ be a sub-$\sigma$-algebra of $\mathcal{B}(\mathbb{R}_{+})\otimes\mathcal{F}$, and $S$ be a set of real valued functions on $\mathbb{R}_{+}$ which is closed (http://planetmath.org/Closed) under uniform convergence on compacts. We show that both the set of $\mathcal{M}$-measurable processes and the set of jointly measurable processes with sample paths almost surely in $S$ are complete (http://planetmath.org/Complete) under ucp convergence. The method used will be to show that we can pass to a subsequence which almost surely converges uniformly on compacts.

We start by writing out the metric generating the topology of uniform convergence on compacts (compact-open topology) for functions $\mathbb{R}_{+}\rightarrow\mathbb{R}$. This is the same as uniform convergence on each of the bounded intervals $[0,n)$ for positive integers $n$,

 $d(X)\equiv\sum_{n=1}^{\infty}2^{-n}\min\left(1,\sup_{t

Then, the metric is $(X,Y)\mapsto d(X-Y)$. Convergence under the ucp topology is given by

 $D^{\rm ucp}(X)=\mathbb{E}[d(X)]$

for any jointly measurable stochastic process $X$, with the (pseudo)metric being $(X,Y)\mapsto D^{\rm ucp}(X-Y)$.

Now, suppose that $X^{n}$ is a sequence of jointly measurable processes such that $X^{n}-X^{m}\xrightarrow{\rm ucp}0$ as $m,n\rightarrow\infty$. Then, $D^{\rm ucp}(X^{n}-X^{m})\rightarrow 0$ and we may pass to a subsequence $X^{n_{k}}$ satisfying $D^{\rm ucp}(X^{n_{j}}-X^{n_{k}})\leq 2^{-j}$ whenever $k>j$. So,

 $\mathbb{E}\left[\sum_{k}d(X^{n_{k}}-X^{n_{k+1}})\right]=\sum_{k}D^{\rm ucp}(X^% {n_{k}}-X^{n_{k+1}})\leq\sum_{k}2^{-k}=1.$

In particular, this shows that $\sum_{k}d(X^{n_{k}}-X^{n_{k+1}})$ is almost surely finite and, therefore,

 $d(X^{n_{j}}-X^{n_{k}})\leq\sum_{i=j}^{k-1}d(X^{n_{i}}-X^{n_{i+1}})\rightarrow 0$

as $k>j\rightarrow\infty$, with probability one.

So, the sequence $X^{n_{k}}$ is almost surely Cauchy (http://planetmath.org/CauchySequence), under the topology of uniform convergence on compacts. We set

 $X_{t}(\omega)\equiv\left\{\begin{array}[]{ll}\lim_{k\rightarrow\infty}X^{n_{k}% }_{t}(\omega),&\textrm{if the limit exists},\\ 0,&\textrm{otherwise}.\end{array}\right.$

As measurability of real valued functions is preserved under pointwise convergence, it follows that if $X^{n}$ are $\mathcal{M}$-measurable, then so is $X$. In particular, $X$ is a jointly measurable process. Furthermore, since convergence is almost surely uniform on compacts, if $X^{n}$ have sample paths in $S$ with probability one then so does $X$.

It only remains to show that $X^{n}\xrightarrow{\rm ucp}X$. However, we have already shown that $d(X^{n_{k}}-X)\rightarrow 0$ with probability one, hence $D^{\rm ucp}(X^{n_{k}}-X)\rightarrow 0$.

 $D^{\rm ucp}(X^{n}-X)\leq D^{\rm ucp}(X^{n}-X^{n_{k}})+D^{\rm ucp}(X^{n_{k}}-X).$

Letting $k$ go to infinity, this is bounded by $\sup_{m>n}D^{\rm ucp}(X^{n}-X^{m})$, which goes to zero as $n\rightarrow\infty$.

Title proof of completeness under ucp convergence ProofOfCompletenessUnderUcpConvergence 2013-03-22 18:40:35 2013-03-22 18:40:35 gel (22282) gel (22282) 5 gel (22282) Proof msc 60G07 msc 60G05