Show that $\mathbb{E}(X|\mathcal{F}_t)$ is a square-integrable martingale

Question

For a BSDE :

$$y_t = \xi + \int_{t}^{T}g_0(s)ds - \int_{t}^{T}z_sdB_s$$

Which has a fixed $\xi \in L^2(\mathscr{F}_T)$ and $g_0(\cdot)$ satisfying $E(\int_{0}^{T}|g_0(t)|dt)^2 < \infty$. There exists a unique pair of process $(y., z.)\in L_{\mathscr{F}}^2(0,T;R^{1+d})$ satisfies the BSDE shown before.

We define

$$M_t = E^{\mathscr{F}_t}[\xi + \int _0 ^T g_0(s)ds]$$

The paper says that $M$ is a square integrable $(\mathscr{F}_t)$-martingale.

So my problem is that how to prove that?

I think maybe we need to show that for every $t\ge0, E(M_t^2) < +\infty$.

But I have no idea about the detail, or maybe we need to find another way to prove $M$ is a square integrable $(\mathscr{F}_t)$-martingale?

Answer 1

For any r.v. $X$ with $E|X|<\infty$, $M_t=E(X|\mathcal F_t)$ is a nice (uniformly integrable) martingale. If $X \in L^{2}$ then $M_t \in L^{2}$ by Jensen's inequality since $x \to x^{2}$ is a convex function.