# proof of completeness under ucp convergence

Let $(\mathrm{\Omega},\mathcal{F},{({\mathcal{F}}_{t})}_{t\in {\mathbb{R}}_{+}},\mathbb{P})$ 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 ${\mathbb{R}}_{+}\to \mathbb{R}$. This is the same as uniform convergence on each of the bounded intervals $[0,n)$ for positive integers $n$,

$$ |

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

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

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

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

$$\mathbb{E}\left[\sum _{k}d({X}^{{n}_{k}}-{X}^{{n}_{k+1}})\right]=\sum _{k}{D}^{\mathrm{ucp}}({X}^{{n}_{k}}-{X}^{{n}_{k+1}})\le \sum _{k}{2}^{-k}=1.$$ |

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

$$d({X}^{{n}_{j}}-{X}^{{n}_{k}})\le \sum _{i=j}^{k-1}d({X}^{{n}_{i}}-{X}^{{n}_{i+1}})\to 0$$ |

as $k>j\to \mathrm{\infty}$, 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

$${X}_{t}(\omega )\equiv \{\begin{array}{cc}\underset{k\to \mathrm{\infty}}{lim}{X}_{t}^{{n}_{k}}(\omega ),\hfill & \text{if the limit exists},\hfill \\ 0,\hfill & \text{otherwise}.\hfill \end{array}$$ |

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 ${X}^{n}\stackrel{ucp}{\to}\mathrm{X}$. However, we have already shown that $d({X}^{{n}_{k}}-X)\to 0$ with probability one, hence ${D}^{\mathrm{ucp}}({X}^{{n}_{k}}-X)\to 0$.

$${D}^{\mathrm{ucp}}({X}^{n}-X)\le {D}^{\mathrm{ucp}}({X}^{n}-{X}^{{n}_{k}})+{D}^{\mathrm{ucp}}({X}^{{n}_{k}}-X).$$ |

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

Title | proof of completeness under ucp convergence |
---|---|

Canonical name | ProofOfCompletenessUnderUcpConvergence |

Date of creation | 2013-03-22 18:40:35 |

Last modified on | 2013-03-22 18:40:35 |

Owner | gel (22282) |

Last modified by | gel (22282) |

Numerical id | 5 |

Author | gel (22282) |

Entry type | Proof |

Classification | msc 60G07 |

Classification | msc 60G05 |