MHB Filtration of a probability space

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In a probability space $(\Omega,\mathcal{F},\mathbb{P})$ with a filtration $\{\mathcal{F}_t\}_{0\leq t<\infty}$, the question arises whether $\mathcal{F}_{\infty} = \mathcal{F}$, where $\mathcal{F}_{\infty} = \sigma\left(\bigcup_{t} \mathcal{F}_t \right)$. It has been noted that $\mathcal{F}_{\infty}$ does not equal $\mathcal{F}$ if strict inclusion is not assumed in the definition of filtration. A counterexample is provided where $\mathcal{F}_t$ is a strictly contained $\sigma$-algebra $\mathcal{G}$. Further references on this topic may be needed for deeper understanding. The discussion highlights the nuances of filtration in probability theory.
gnob
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Hi again, I hope you can clarify me on this:

Consider a probability space $(\Omega,\mathcal{F},\mathbb{P})$ and let $\{\mathcal{F}_t\}_{0\leq t<\infty}$ be a filtration on it. Define
$\mathcal{F}_{\infty} = \sigma\left(\bigcup_{t} \mathcal{F}_t \right)$ where $t \in [0,\infty).$

My question: Is $\mathcal{F}_{\infty} = \mathcal{F}$?

I came across with an answer that it is not, but I forgot the source nor I remember if there is a counterexample. Can anyone please help me? Thanks. Please also give me some reference on it.

Thanks, thanks, thanks...:o
 
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If in the definition of filtration we don't assume strict inclusion we get a counter example taking $\mathcal F_t=\mathcal G$, where the latest is a strictly contained $\sigma$-algebra.
 

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