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I wonder if the following general fact is true: If $f_n$ is a sequence of $\mathcal {F} _n$-measurable random variables converging to $f$ and $\mathcal{F}_n \downarrow \mathcal{F}$ [meaning $\mathcal{F}_{n+1 } \subset \mathcal{F}_n$ and $\bigcap_{n=1 } ^{\infty } = \mathcal {F } $] does it follow that $f$ is $\mathcal{F}$-measurable and how would we show this?


The general context of my question, which I think may be reduced to the above, is the following:

Let $X $ be a stochastic process where for every omega and any point $s$, the limit from the right at $s$, $\lim_{t \downarrow s } X(\omega,t)$, exists. Assume further that $T$ is a finite stopping time and define

$$T_n = \frac{\lfloor T \rfloor + 1}{2^n}$$

so that $T_n \downarrow T$. Further take it as known that for every $n$, $T_n$ is measurable w.r.t. the stopped sigma algebra $\mathcal{F}_{T_{n } + }:=\{ \Lambda \in \mathcal{F} : \ \forall t: \ \Lambda \cap \{T_n < t \} \in \mathcal {F} _t \} $, and if $T = \lim T_n$ then $ \bigcap_{n=1 } ^{\infty } \mathcal{F}_{T_{n } + } \downarrow \mathcal{F}_ {T+}$.

Assume we have shown that for every $n, \ X(T_n)$ is $\mathcal{F}_{T_{n } + }$-measurable. Does it follow that $X(T)$ is $\mathcal{F}_ {T+} $-measurable?

Most grateful for any help provided!

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If by $\mathcal F_n \downarrow \mathcal F$ you mean$\mathcal F_{n+1} \subset \mathcal F_n$ for all $n$ and $\cap_n \mathcal F_n =\mathcal F$ then the answer is YES. Let $f =\lim f_n$. For any $N$ we can write $f=\lim \{f_N,f_{N+1},...\}$. Since $f_k$ is measurable w.r.t. $\mathcal F_N$ for al $k \geq N$ we see that $f$ is measurable w.r.t. $\mathcal F_N$. If $a$ ia real number the $\{f<a\}$ is therefore in $\mathcal F_N$. Since this is true for each $N$ it follows that $\{f<a\}$ is therefore in $\mathcal F$ for all $a$.

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  • $\begingroup$ Yes, that is what I mean! Will add it to the question! $\endgroup$ – MrFranzén Nov 15 '19 at 9:47

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