# Martingale preservation under independent enlargement of filtration

I think this is probably a very easy question but I haven't worked with $\sigma$-algebras in depth for a long time now so am finding myself a little rusty. Would be very grateful if someone could give me a (careful) proof of the following (I'm pretty sure it's true!). I guess I'm missing the right way of characterising the elements of the join of two $\sigma$-algebras appropriately.

Let $M_t$ be a martingale with respect to the filtration $\mathcal{F}_t$ on some probability space $(\Omega, \mathcal{F}, P)$. Assume that $\mathcal{G}_t \subseteq \mathcal{F}$ where for each $t \geq 0$, $\mathcal{G}_t$ is independent of $\mathcal{F}_t$ and let $\mathcal{H}_t := \mathcal{F}_t \vee \mathcal{G}_t$. Then $M_t$ is also a martingale with respect to $\mathcal{H}_t$.

Thanks!

Here $$\mathcal F_t\vee\mathcal G_t$$ denotes the smallest (for the inclusion) $$\sigma$$-algebra that contains both $$\mathcal F_t$$ and $$\mathcal G_t$$.
Integrability condition is already verified, so we just need to check that for all $$s\leqslant t$$, we have $$\mathbb E\left[M_t\mid \mathcal F_s\vee\mathcal G_s\right]=M_s.$$ Let $$\mathcal B:=\left\{F\cap G,F\in \mathcal F_s,G\in \mathcal G_s \right\}$$. It can be checked that $$\mathcal B$$ is $$\pi$$-system which generates $$\mathcal F_s\vee\mathcal G_s$$,. The set of elements $$B$$ of $$\mathcal F$$ such that $$\int_B M_tdP=\int_BM_sdP$$ is a $$\lambda$$-system. So we just have to show that for all $$F\in\mathcal F_s$$, $$G\in\mathcal G_s$$, we have $$\int_{F\cap G}M_tdP=\int_{F\cap G}M_sdP.$$ Since $$\mathcal F_t$$ is independent of $$\mathcal G_t$$, $$\mathcal F_t$$ is also independent of $$\mathcal G_s$$. Conclude using the fact that $$M_t\chi_F$$ is $$\mathcal F_t$$ measurable.