Is the Ito process always measurable?

Question

I recently read the literature where there is a sentence that says, "Since the process $X(t), t\in [0,T]$: $$X(t) = X(0) + \mu(t,X(t)) ~ dt + \sigma(t,X(t)) ~ dW(t)$$ is an Ito process, then $X(t)$ is $\mathcal{F}_T$ measurable." My question is whether this statement is true, and what is the explanation behind it?

Answer 1

Another result which is commonly quoted is a similar statement -- every Ito integral process $X_t = \int_0^t f(s,B_s) dB_s$ is an $\mathcal{F}_t$-martingale, which implies that $X_t$ must be $\mathcal{F}_t$-measurable.

If you want to extend that to any Ito process in a form you defined, i.e. $$ dX_t = \mu(t, X(t)) dt + \sigma(t, X(t)) dB_t, $$ you can claim that $X_t$ is a martingale if $\mu \equiv 0$ and $$ \mathbb{E}\left[ \sqrt{\int \sigma(t)^2 \mathrm{d}t} \right] < \infty. $$

Indeed, as @Andrew Zhang posted in a comment, the idea of a process being measurable with respect to a filtration (a time-sequence of $\sigma$-algebras) is that the process does not rely on the information from the "future" -- so $X_t$ is not allowed to depend on what happens after a time $t$.

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For example, if $B_t$ is standard Brownian motion at time $t$, the processes $B_t, \sin(B_t)$ and $B_{t/2}$ are $\mathcal{F}_t$-measurable, while $B_{2t}$ is not.

Another useful example is that Stratonovich integral is not necessarily $\mathcal{F}_t$-measurable, since the Stratonovich integral is defined using the middle point of the interval (as opposed to Ito integral which is defined using the left-hand endpoint only), so when you are at a time point, corresponding to the left-hand point, the value for Ito integral is known but Stratonovich is not yet fixed...