Suppose on a probability space $(\Omega,\mathcal{A},\mathbb{P})$ we have $\mathcal{F}$ and $\mathcal{G}$, independent sub-sigma-algebras of $\mathcal{A}$. Consider $F\cap G\in \sigma(\mathcal{F},\mathcal{G})$ where $F\in \sigma(\mathcal{F},\mathcal{G})$ and $G\in\mathcal{G}$. If we can show, due to the structure of $\mathcal{G}$, that $F\cap G\in \mathcal{F}\cap G$ (where $\mathcal{F}\cap G$ denotes the trace sigma-algebra on the event $G$) can we conclude that $F\in\mathcal{F}$?
Although I do not think so, to give some context, I came across two papers that make essentially the above reasoning, with the sole difference that the final claim is that $\mathbb{1}_F$ is $\mathcal{F}$-measurable 'on $G$' (a claim that does not seem to make sense to me, as I have never met a notion of measurability restricted to an event, and therefore I interpreted as given above) and then go on to use the fact in computing $\mathbb{E}(X\mathbb{1}_F\mathbb{1}_G|\mathcal{F})=\mathbb{P}(G)X\mathbb{1}_F$, having exploited that $X$ is a $\mathcal{F}$-measurable random variable. Clearly this step is what suggest my interpretation of their claim. However, I failed to prove the theoretical fact above. Any ideas?