Let $\{\mathcal{F}_t\}$ be a filtration and let $\mathcal{G}_{t} = \mathcal{F}_{t+}$. Now, I want to prove that $\mathcal{G}_{t-} = \mathcal{F}_{t-}$ for $t>0$.
I am quite new to filtrations, but the definition of a filtration is not helping me any further in proving the above statement. So, I can use some of your help.
A filtrations is defined as a collection of $\sigma$-fields $\{\mathcal{F}_t:t \in \mathbb{R}_{+}\}$ that satisfy $\mathcal{F}_s \subset \mathcal{F}_t \subset \mathcal{F}$ for all $0<t<\infty$.
Given a filtration $\{\mathcal{F}_t\}$, define the $\sigma$-algebras \begin{align} \mathcal{F}_{t+} = \bigcap_{s:s>t} \mathcal{F}_s. \end{align} $\{\mathcal{F}_{t+}\}$ is a new filtration, we say $\{\mathcal{F}_t\}$ is right-continuous. Under the assumption that the filtration is right-continuous, we want to show that $\{\mathcal{F}_t\}$ is left-continuous aswel. Therefore, I can image that we will look at an the union instead of the intersection of $\sigma$-algebras. However, the exact approach is for me still unclear.
Therefore, we define left-continuity by \begin{align} \mathcal{F}_{t-} = \sigma \bigg( \bigcup_{s:s<t} \mathcal{F}_s \bigg). \end{align} Since the union of $\sigma$-fields is not necessarily a $\sigma$-field, $\mathcal{F}_{t-}$ needs to be defined as the $\sigma$-field generated by the unions of of $\mathcal{F}_s$ over $0 < s <t$.
So, now it should be trivial (why exactly?) that, \begin{align} \mathcal{F}_{t-} = \sigma \bigg( \bigcup_{s:s<t} \mathcal{F}_s \bigg) \subseteq \bigcup_{s:s<t} \bigcap_{k:k>s} \mathcal{F}_k = \mathcal{G}_{t-}. \end{align} And how to find the other inclusion?