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 $\sigma(X) \subset \mathcal F_1$ ?


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 $\sigma(X) \subset \mathcal F_1$.

  • $\begingroup$ Where can I find the result that you are using ? $\endgroup$ – W. Volante Jun 10 at 20:50
  • $\begingroup$ @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. $\endgroup$ – user6247850 Jun 10 at 20:57
  • $\begingroup$ 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 ? $\endgroup$ – W. Volante Jun 10 at 21:20
  • $\begingroup$ 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. $\endgroup$ – user6247850 Jun 10 at 21:29

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