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Can anyone suggest how to prove a right continuous stochastic process is measurable?

Thanks Indrajit

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1 Answer 1

Let $(X_t)_{t\geq 0}$ be a right-continuous stochastic process. For every $n\in\mathbb{N}$ we define a process $(X_t^n)_{t\geq 0}$ given by $$ X_t^n(\omega)=X_{i/2^n}(\omega)\quad \text{if } \tfrac{i-1}{2^n}\leq t<\tfrac{i}{2^n}\; \text{and } i\geq 1. $$

Show that for each $n\in\mathbb{N}$ the process $(X_t^n)_{t\geq 0}$ is indeed a measurable process.

Convince yourself that the right-continuity ensures that $X_t(\omega)=\lim_{n\to\infty} X_t^n(\omega)$ for every $(t,\omega)\in [0,\infty)\times \Omega$. Conclude.


To show that $(X_t^n)_{t\geq 0}$ is a measurable process for every $n\in\mathbb{N}$ we note that $$ \{(t,\omega)\mid X_t^n(\omega)\in B\}=\bigcup_{i=1}^\infty \left[ \tfrac{i-1}{2^n},\tfrac{i}{2^n}\right[\times \{X_{i/2^n}\in B\} $$ for every $B\in\mathcal{B}(\mathbb{R})$.

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Since i just started to learn stochastic process, could you please elaborate a bit on how show that $X^n(\omega)$ is measurable. I was able to convince the limit part but could establish how it ensure measurability of $(X_t)$ –  Indrajit Sep 25 '12 at 16:13
    
I have added something to the measurability part. Hope it helps. –  Stefan Hansen Sep 25 '12 at 16:59
    
Thanks Stefan, this makes much easier. One more issue given we establish the measurability of $X_t^n(\omega)$ and showing $X_t^n()$ converges to $X_t$, which argument are we using to show $X_t$ is measurable? –  Indrajit Sep 25 '12 at 17:38
    
We are using that taking limit preserves measurability. In general if $(X,\mathcal{E})$ is a measurable space and $(f_n)_{n\in \mathbb{N}}$ is a sequence of measurable functions converging pointwise to a function $f$, i.e. $f_n(x)\to f(x)$ for all $x \in X$, then $f$ is also measurable. Here we have $X=[0,\infty)\times \Omega$ and $\mathcal{E}=\mathcal{B}(\mathbb{R}_+)\otimes \mathcal{F}$. –  Stefan Hansen Sep 25 '12 at 18:14

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