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Suppose I have a sequence of stochastic processes, $X_n(t)$, adapted to a common filtration, which I know converge to $X(t)$ at all $t$, in the mean square sense. Is it possible to infer that $X(t)$ is also adapted? Why/why not?

EDIT: Assume the filtration satisfies the usual conditions (complete and right continuous).

saz
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user2379888
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  • You will need to assume that the filtration is complete (otherwise the assertion fails to hold). If the filtration is complete, then the measurability follows from the fact that $X_n(t) \to X(t)$ in $L^2$ implies that $X_{n_k}(t) \to X(t)$ almost everywhere for a suitable subsequence. – saz May 04 '19 at 19:16
  • Yes, I'm working with a complete filtration. Do I need to redefine the limiting process to be $\tilde{X}(t)$ to get everywhere convergence along the subsequence to make the conclusion? – user2379888 May 04 '19 at 19:27

1 Answers1

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Convergence in $L^2$ implies almost sure convergence of a subsequence. Therefore we are done if we can show the following statement:

Lemma: Let $(\Omega,\mathcal{A},\mathbb{P})$ be a probability space and let $(Y_n)_{n \in \mathbb{N}},Y$ be $\mathcal{A}$-measurable random variables. Assume that each $Y_n$ is $\mathcal{F}$-measurable for a complete sub-sigma algebra $\mathcal{F} \subseteq \mathcal{A}$. If $Y_n \to Y$ almost surely then $Y$ is $\mathcal{F}$-measurable.

Proof: Set

$$C := \{\omega; \lim_{n \to \infty} Y_n(\omega) \, \, \text{exists and is finite}\}.$$

Since $Y_n$ is $\mathcal{F}$-measurable for all $n \geq 1$, it follows that $C \in \mathcal{F}$ (see e.g. this question for a proof). Clearly,

$$Y_{\infty} := 1_C(\omega) \lim_{n \to \infty} Y_n(\omega),\qquad \omega \in \Omega$$

is well-defined. As $Y_n 1_C$ is $\mathcal{F}$-measurable, we find that $$Y_{\infty}(\omega)= \lim_{n \to \infty} (1_C Y_n)(\omega), \qquad \omega \in \Omega$$ is $\mathcal{F}$-measurable as pointwise limit of $\mathcal{F}$-measurable functions. On the other hand, we have $Y=Y_{\infty}$ almost surely, and since $\mathcal{F}$ is complete, this implies that $Y$ is $\mathcal{F}$-measurable (see this question).

saz
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