# Are $\sigma$-algebras/filtrations closed by continuity?

Let $$X,Y$$ be 2 different random variables and $$Z_t$$ a continuous stochastic process on $$[0,1]$$ such that $$Z_1=0$$ everywhere.

Consider the process

$$V_t=(Z_t + X) 1_{[0,1)}(t) + Y 1_{[1,+\infty]}(t)$$

and its filtration $$(\mathcal F_t)$$. At 1 from the left, the process is equal to $$X$$ and from the right it is equal to $$Y$$. Clearly $$\sigma(Y) \subset \mathcal F_1$$.

My question: is $$\sigma(X) \subset \mathcal F_1$$ ?

## 1 Answer

Yes, $$X = \lim_{n \rightarrow \infty} V_{1-\frac 1n}$$ and $$V_{1-\frac 1n} \in \mathcal F_1$$ for all $$n \in \mathbb{N}$$ so $$X$$ is the pointwise limit of $$\mathcal F_1$$-measurable random variables and hence $$\mathcal F_1$$-measurable. This implies $$\sigma(X) \subset \mathcal F_1$$.

• Where can I find the result that you are using ? – W. Volante Jun 10 at 20:50
• @W.Volante I think the only result I'm using is that the pointwise limit of measurable functions is measurable, which can be found in most measure theory texts. There is also a proof on this site here for the Borel sigma algebra case, but the proof generalizes. – user6247850 Jun 10 at 20:57
• I see, thank you for clarifying. So what makes it work here is that $Z_1=0$ everywhere, if we change that to a.s, then this becomes false right ? – W. Volante Jun 10 at 21:20
• Yes, it would be false in general. We frequently work with the filtrations augmented by the null sets of $\mathcal F_\infty$, though, which I think would be enough to make it true. – user6247850 Jun 10 at 21:29