girdav said:
Check that the unions are the same using the fact that if $s<t$ then $\mathcal F_s\subset\mathcal F_t$. Do the same for intersections.
Thanks for the suggestion. Please give your comment on the proof I produced. Thanks.
Recall that $\sigma\left(A\right)$ is the smallest $\sigma$-field that contains the set (or family of subsets) $A.$ Hence, for the first question regarding unions, it suffices to show that
$$
\bigcup_{s < t} \mathcal{F}_s =
\bigcup_{q < t} \mathcal{F}_q =
\bigcup_{\substack{n= n_k\\ \left(t-\frac{1}
{n_k}\right)>0}}^{\infty} \mathcal{F}_{t -
\frac{1}{n}}.
$$
To establish the first equality above, we consider an arbitrary $\sigma$-field
$\mathcal{F}_{s_0} \subseteq \bigcup_{ s <t} \mathcal{F}_s.$ Since $\mathbb{Q}$ is dense in $\mathbb{R},$ we can find $q_0 \in \mathbb{Q}$ such that
$s_0 < q_0 <t.$ Because $\{\mathcal{F}_t \}$ is an increasing sequence of $\sigma$-fields, we then have $\mathcal{F}_{s_0} \subseteq \mathcal{F}_{q_0} \subseteq \bigcup_{q <t} \mathcal{F}_q.$ Hence, $\bigcup_{s <t} \mathcal{F}_s \subseteq \bigcup_{q <t} \mathcal{F}_q.$ Conversely, let $\mathcal{F}_{q_0} \subseteq \bigcup_{q <t} \mathcal{F}_q.$ Note that $\mathbb{Q} \subseteq \mathbb{R}$ and so $\mathcal{F}_{q_0} \subseteq \bigcup_{s <t} \mathcal{F}_s,$ which implies that $\bigcup_{q <t} \mathcal{F}_q \subseteq \bigcup_{s <t} \mathcal{F}_s.$
Next, we establish that the extreme unions above are equal. First, observe that $t-\frac{1}{n} \in \mathbb{R}$ for every $n \in \mathbb{N}$ and so $\mathcal{F}_{t-\frac{1}{n}} \subseteq \bigcup_{s<t} \mathcal{F}_s,$ which shows that
$$
\bigcup_{\substack{n= n_k\\ \left(t-\frac{1}{n_k}\right)>0}}^{\infty}
\mathcal{F}_{t - \frac{1}{n}} \subseteq \bigcup_{s <t} \mathcal{F}_s.
$$
For the reverse inclusion, consider an arbitrary $\sigma$-field $\mathcal{F}_{s_0} \subseteq \bigcup_{s <t} \mathcal{F}_s$ and note that $s_0 < t.$ Let $\delta_{0}:= \frac{t - s_0}{2} \in\mathbb{R}_+.$ By the Archimedean property
of $\mathbb{R},$ we can find an integer $n_{k_0} \in \mathbb{N}$ for which $\frac{1}{n_{k_0}} < \delta_0$ so that $s_0 < t - \frac{1}{n_{k_0}} <t$ and
$$
\mathcal{F}_{s_0} \subseteq \mathcal{F}_{t-\frac{1}{n_{k_0}}}
\subseteq \bigcup_{\substack{n= n_k\\ \left(t-
\frac{1}{n_k}\right)>0}}^{\infty} \mathcal{F}_{t - \frac{1}{n}}.
$$
It follows by transitivity that all three unions above are equal, and consequently the last two equalities hold.
For the Second Question (on Intersections): We need to show that
$$
\mathcal{F}_{t+} := \sigma\left(\bigcap_{\epsilon > 0}
\mathcal{F}_{t + \epsilon} \right) = \bigcap_{\epsilon > 0}
\mathcal{F}_{t + \epsilon} = \bigcap_{q >0}
\mathcal{F}_{t + q} = \bigcap_{n=1}^{\infty} \mathcal{F}_{t + \frac{1}{n}}.
$$
It is a standard example in Measure Theory that the intersection of a family of $\sigma$-fields (defined on the same space) is again a $\sigma$-field. So, the first equality is trivial.
We now establish the equality $\bigcap_{\epsilon > 0} \mathcal{F}_{t + \epsilon}= \bigcap_{q >0} \mathcal{F}_{t + q}.$ If $A \in \bigcap_{\epsilon > 0} \mathcal{F}_{t + \epsilon}$ then $A \in \mathcal{F}_{t+\epsilon},\,\,\forall\, \epsilon >0.$ Thus, if we chose all $\epsilon$'s to be positive rationals ($\epsilon = q \in \mathbb{Q}_+$) we also have $A \in \mathcal{F}_{t+q},\,\,\forall\, q >0,$ which in turn implies that $A \in \bigcap_{q >0} \mathcal{F}_{t+q}.$ On the other hand, $B \in \bigcap_{q >0} \mathcal{F}_{t+q}$ implies that $B \in \mathcal{F}_{t + q},\,\,\forall\, q>0.$ Our goal is to show that $B \in\mathcal{F}_{t+\epsilon},\,\,\forall\,\epsilon >0.$ But for a fixed $\epsilon_0>0,$ since $\mathbb{Q}$ is dense in $\mathbb{R},$ we can find a sequence $\{ q^{(0)}_i\}_{i=1}^{\infty}$ such that $q^{(0)}_i \downarrow \epsilon_0.$ Hence,
$$
\mathcal{F}_{t+\epsilon_0} = \lim_{q^{(0)}_i \downarrow\epsilon}
\mathcal{F}_{t+q^{(0)}_i} = \left(\bigcap_{i=1}^{\infty}
\mathcal{F}_{t+ q^{(0)}_i}\right) \ni B.
$$
Consequently, we have $B\in \Big(\bigcap_{\epsilon >0} \mathcal{F}_{t+\epsilon}\Big).$
Lastly, we show that $\bigcap_{\epsilon > 0} \mathcal{F}_{t + \epsilon}= \bigcap_{n \in \mathbb{Z}_+} \mathcal{F}_{t + \frac{1}{n}}.$ But we can use similar argument as above to conclude that $\bigcap_{\epsilon > 0} \mathcal{F}_{t + \epsilon}\subseteq \bigcap_{n \in \mathbb{Z}_+} \mathcal{F}_{t + \frac{1}{n}}.$ For the reverse inclusion, a similar argument can be done by choosing a sequence $\{ \frac{1}{n_i(\epsilon)}\}_i^{\infty} \subseteq \mathbb{Q}$ such that $\frac{1}{n_i(\epsilon)} \downarrow \epsilon$ (Note: WLOG, we can take $\epsilon>0$ to be small since we are taking intersection and that $\{ \mathcal{F}_t\}$ is an increasing family).
