# If $f_n$ is measurable w.r.t. $\mathcal{F}_n$ and $\mathcal{F}_n \downarrow \mathcal{F}$, is $\lim f_n$ measurable w.r.t. $\mathcal{F}$?

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!

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 $$\{f is therefore in $$\mathcal F_N$$. Since this is true for each $$N$$ it follows that $$\{f is therefore in $$\mathcal F$$ for all $$a$$.