$\mathcal F_t=\mathcal F_{t^+}$ $\iff $ $(X_t)$ is right continuous Let $\mathcal F_t=\sigma (X_s\mid s\leq t)$ where $(X_t)$ is a stochastic process. We say that $\mathcal F_t$ is right continuous at $t$ if $$\mathcal F_t=\mathcal F_{t^+}:=\bigcap_{s>t}\mathcal F_s.$$
I was wondering, is $\mathcal F_t$ right continuous $\iff$ $(X_t)$ is right continuous at $t$ ?


*

*Suppose $\mathcal F_t=\mathcal F_{t^+}$. I want to prove that $\lim_{h\to 0^+}X_{t+h}=X_t$.

*Conversely, suppose $\lim_{t\to h^+}X_{t+h}=X_t$. The fact that $\mathcal F_t\subset \mathcal F_{t^+}$ is obvious. To show the converse inequality, let $A\in \mathcal F_{t^+}$. How can I show that $A\in \mathcal F_t$ ?
 A: In your title you ask whether $(\mathcal  F_t)_{t\ge0}$ is right-continuous iff $(X_t)_{t\ge0}$ is right-continuous, but then you ask whether $(\mathcal  F_t)_{t\ge0}$ is right-continous iff $(\mathcal F_{t+})_{t\ge0}$ is right-continous.
Both answers are negative.
The first (and more interesting) one: do we have $(\mathcal F_t)_{t\ge0}$ right-continuous iff $(X_t)_{t\ge0}$ is right-continuous ?
No. Let $\varepsilon$ be a Bernoulli random variable with success probability $1/2$. Let $X:\Omega\to\mathbb R_+^{\mathbb R}$ be a stochastic process defined for all $t\ge0$ by
$$
X_t=\varepsilon t+(1-\varepsilon)t^2.
$$
Since $X_0=0$, we have $\mathcal F_0=\{\emptyset,\Omega\}$. Let
$$
A=\left\{\sum_{n\in\mathbb N^*}\left\vert X_{1/n}\right\vert<+\infty\right\}.
$$
It is clear that $A\in\mathcal F_{0+}$. However, $A=\{\varepsilon=0\}$ so $\mathbb P(A)=1/2$ and $A\notin\mathcal F_0$, hence $\mathcal F_0\neq\mathcal F_{0+}$. We deduce that $(\mathcal F_t)_{t\ge0}$ is not right-continuous whereas $(X_t)_{t\ge0}$ is right-continuous.
Conversely, let $X:\Omega\to\mathbb R_+^{\mathbb R}$ be a stochastic process defined for all $t\ge0$ by
$$
X_t=\left\{\begin{array}{rcl}0&\textrm{if}&t=0\\1&\textrm{if}&t>0\end{array}\right..
$$
Then for all $t\ge0$, $\mathcal F_t=\{\emptyset,\Omega\}$ so $(\mathcal F_t)_{t\ge0}$ is right-continuous whereas $(X_t)_{t\ge0}$ is not right-continuous.
And now the second question: do we have $(\mathcal F_t)_{t\ge0}$ right-continuous iff $(\mathcal F_{t+})_{t\ge0}$ is right-continuous ?
No. Indeed, $(\mathcal F_{t+})_{t\ge0}$ is always right-continuous. Consider the first example above where $(X_t)_{t\ge0}$ is right-continuous but $(\mathcal F_t)_{t\ge0}$ is not. In particular, $(\mathcal F_t)_{t\ge0}$ is not right-continuous whereas $(\mathcal F_{t+})_{t\ge0}$ is right-continuous.
