# All right or left continuous processes are jointly measurable

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

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

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

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

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.

• Thank you. We say that the mapping $(t,\omega) \mapsto \mathbb{1}_{\frac{k-1}{n} \leq t < \frac{k}{n}$ because $\{\mathbb{1}_{\frac{k-1}{n} \leq t < \frac{k}{n} <a\} \in \mathcal{B}([o,t]) \otimes \mathcal{F} \forall a \in \mathcal{R}$ – user3503589 Oct 28 '14 at 22:04
• Note that $t$ is a variable, so $\mathcal{B}[0,t]$ doesn't make sense. But I guess you mean the right thing. – saz Oct 29 '14 at 6:38
• 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. – Hans Nov 19 '18 at 0:51
• @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 $X$. – saz Nov 19 '18 at 6:17
• 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. – Hans Nov 19 '18 at 7:00