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
 A: 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.
