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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!

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Integrability condition is already verified, so we just need to check that for all $s\leqslant t$, we have $$E[M_t\mid \mathcal F_s\vee\mathcal G_s]=M_s.$$ Let $\mathcal B:=\{F\cap G,F\in \mathcal F_s,G\in \mathcal G_s \}$. It's a $\pi$-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 $$\int_{F\cap G}M_tdP=\int_{F\cap G}M_sdP.$$ As $\mathcal F_t$ is independent of $\mathcal G_t$, it's also independent of $\mathcal G_s$. Conclude using the fact that $M_t\chi_F$ is $\mathcal F_t$ measurable.

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