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I was reading the Stochastic Calculus notes by George on this website http://almostsure.wordpress.com/2009/11/03/stochastic-processes-indistinguishability-and-modifications/ and I cannot understand why he wrote that any right/left continuous function can be approximated as the limit of the following sequence of processes expressed below

$X_t^n(\omega)=\sum_{1}^{\infty} \mathbb{1}_{\{(k-1)/n \leq t \leq k/n} X_{k/n}(\omega)$

Why is this true? I mean I see that this sequence approximates $X_t^n$ but I could someone give me a hint as to how could I rigorously show that?

The author also mentions that the following is jointly measurable

${(t,\omega)\mapsto 1_{\{(k-1)/n\le t <k/n\}}X_{k/n}(\omega)}$ and then uses this to imply that $X$ is jointly measurable as it is the limit of sequence of these jointly measurable random variables

Why is that?

Thanks you

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1 Answer 1

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First of all, the author of the linked website does not claim that any right- and left-continuous process can be approximated by

$$X_t^n(\omega) := \sum_{k=1}^{\infty} 1_{\left\{\frac{k-1}{n} \leq t \leq \frac{k}{n} \right\}} X_{\frac{k}{n}}(\omega).$$

Instead, it should read

$$X_t^n(\omega) := \sum_{k=1}^{\infty} 1_{\left\{\frac{k-1}{n} \leq t <\frac{k}{n} \right\}} X_{\frac{k}{n}}(\omega) \tag{1}$$

for any right-continuous process $(X_t)_{t \geq 0}$ and

$$X_t^n(\omega) := \sum_{k=1}^{\infty} 1_{\left\{\frac{k-1}{n}< t \leq \frac{k}{n} \right\}} X_{\frac{k-1}{n}}(\omega) \tag{2}$$

for any left-continuous process $(X_t)_{t \geq 0}$.


Turning to your original question: Suppose that $(X_t)_{t \geq 0}$ is right-continuous, i.e. the approximation $(X_t^n)_{t \geq 0}$ is given by $(1)$ for each $n \in \mathbb{N}$. By assumption, $t \mapsto X_t(\omega)$ is right-continuous for fixed $\omega$. In particular,

$$X_t(\omega) = \lim_{n \to \infty} X_{t_n}(\omega) \tag{3}$$

for any sequence $t_n \downarrow t$. Now for any $t \geq 0$, we choose for each $n \in \mathbb{N}$ the (unique) $k=k(n)$ such that $$\frac{k(n)-1}{n} \leq t < \frac{k(n)}{n}.$$ It follows from the definition of $(X_t^n)_{t \geq 0}$ and $(3)$ (with $t_n := \frac{k(n)}{n}$) that

$$X_t^n(\omega) = X_{\frac{k(n)}{n}}(\omega) \to X_t(\omega).$$

This finishes the proof. The argumentation for left-continuous processes is very similar, I leave it to you.


Concerning jointly measurability: Note that the product of two measurable functions is again measurable. Using the very definition of measurability, it is not difficult to see that both mappings

$$(t,\omega) \mapsto 1_{\frac{k-1}{n} \leq t < \frac{k}{n}} \quad \text{and} \quad (t,\omega) \mapsto X_{\frac{k}{n}}(\omega)$$

are jointly measurable.

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  • $\begingroup$ Thank you. We say that the mapping $(t,\omega) \mapsto \mathbb{1}_{\frac{k-1}{n} \leq t < \frac{k}{n}$ because $\{\mathbb{1}_{\frac{k-1}{n} \leq t < \frac{k}{n} <a\} \in \mathcal{B}([o,t]) \otimes \mathcal{F} \forall a \in \mathcal{R}$ $\endgroup$ Oct 28, 2014 at 22:04
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    $\begingroup$ Note that $t$ is a variable, so $\mathcal{B}[0,t]$ doesn't make sense. But I guess you mean the right thing. $\endgroup$
    – saz
    Oct 29, 2014 at 6:38
  • $\begingroup$ Nice answer. +1. But you have missed the last question "...uses this to imply that $X$ is jointly measurable as it is the limit of sequence of these jointly measurable random variables. Why is that?" You have shown the $X^n_t(\omega)$ is measurable and $X^n_t\rightarrow X_t(\omega)$ as $n\rightarrow\infty$ pointwise for every given $t$ and $\omega$ but have not proven $X_t(\omega)$ is jointly measurable with respect to the product $\sigma-$ meaure. $\endgroup$
    – Hans
    Nov 19, 2018 at 0:51
  • $\begingroup$ @Hans Sorry but I don't really understand your question/problem. If $f_n$ is a sequence of $\mathcal{A}/\mathcal{B}(\mathbb{R}^d)$-measurable mappings which converges pointwise to a mapping $f$, then $f$ is $\mathcal{A}/\mathcal{B}(\mathbb{R}^d)$-measurable. Since this holds for any $\sigma$-algebra $\mathcal{A}$, we can use this fact to deduce the (joint) measurability of $X$. $\endgroup$
    – saz
    Nov 19, 2018 at 6:17
  • $\begingroup$ I am saying that the proof of the proposition that pointwise limit of a sequence of joint measurable function is still a joint measurable function needs to be presented explicitly to answer the OP's question. $\endgroup$
    – Hans
    Nov 19, 2018 at 7:00

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