Measurability question for càglàd process

Question

I have encounter the following question, which is probably naive for probabilists, but let me still ask it:

Let $\{X_t(\omega): t \in \mathbb{R}\}$ be a real valued càglàd process (that is ''continue à gauche et il existe limite à droite'' instead of the usual càdlàg assumption ), is the following function

$$ \limsup_{t \downarrow 0 } X_t(\omega)$$

measurable? If this is the case, would you please tell me a reference for it, if this is not the case, then what kinds of conditions on the process will such a limit function measurable ?

Thanks a lot.

Answer 1

If $(X_t)_{t \geq 0}$ is a càdlàg process, then it follows from the very definition of "càdlàg" that

$$\lim_{t \downarrow 0} X_t(\omega) = \limsup_{t \downarrow 0} X_t(\omega) = X_0(\omega).$$

Consequently, the $\limsup$ is measurable since $\omega \mapsto X_0(\omega)$ is measurable (as $(X_t)_{t \geq 0}$ is a stochastic process).

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Edit (càglàd case) If a function $f$ is right-continuous at some point $x \in \mathbb{R}$, then we have

$$\limsup_{y \downarrow x} f(y) = \limsup_{n \to \infty} f(y_n)$$

for any sequence $y_n \downarrow x$. Since the (pointwise) $\limsup$ of a sequence of measurable functions is measurable, we conclude that

$$\limsup_{t \downarrow 0} X_t(\omega) = \limsup_{n \to \infty} X_{t_n}(\omega)$$

is measurable whenever $(X_t)_{t \geq 0}$ is a càglàd process. Here $t_n$ denotes an arbitrary sequence satisfying $t_n \downarrow 0$.