Let $\mathcal{F}_t$ be a filtration, $\overline{\mathcal{F}}_t$ be a completion of $\mathcal{F}_t$, and $\overline{\mathcal{F}}_{t+}$ be a completion of $\mathcal{F}_{t+}:=\bigcap_{s>t}\mathcal{F}_s$.
Suppose $\mathrm{E}[\mathbb{1}_A\ |\overline{\mathcal{F}}_t]=\mathbb{1}_A$ for any $A\in\overline{\mathcal{F}}_{t+}$. Then, $\overline{\mathcal{F}}_{t+}\subset\overline{\mathcal{F}}_t$. Consequently, we get $\overline{\mathcal{F}}_{t+}=\overline{\mathcal{F}}_t$, complete filtration is right-continuous.
I don't understand why we can say $\overline{\mathcal{F}}_{t+}\subset\overline{\mathcal{F}}_t$ by assumption.
Thank you for your cooperation.
If $E[\mathbb{1}_A \mid \overline{\mathcal F}_t] = \mathbb{1}_A$, then $A$ is $\overline{\mathcal F}_t$-measurable, i.e. $A \in \overline{\mathcal F}_t$. So the assumption implies that for all $A \in \overline{\mathcal F}_{t+}$, we have $A \in \overline{\mathcal F}_t$, or equivalently $\overline{\mathcal F}_{t+} \subset \overline{\mathcal F}_{t}$.