Why are martingales always adapted to a filtration?

Question

Let $(M_t)$ a martingale and $(\mathcal F_t)$ a filtration. Then, $\mathbb E|M_t|<\infty $ for all $t$ and $$\mathbb E[M_t\mid \mathcal F_s]=M_s,$$ when $s\leq t$. Why this last condition tell us that it's always adapted to $\mathcal F_t$ ? (i.e. that $\{M_t\leq x\}\in \mathcal F_t$ for all $t$).

Answer 1

By the very definition of conditional expectation,

$$\mathbb{E}(X \mid \mathcal{F}) \quad \text{is $\mathcal{F}$-measurable} \tag{1} $$

for any integrable random variable $X$ and any $\sigma$-algebra $\mathcal{F}$. Since

$$\mathbb{E}(M_t \mid \mathcal{F}_s) = M_s \qquad \text{for all $s \leq t$}$$

we have in particular

$$M_t = \mathbb{E}(M_t \mid \mathcal{F}_t).$$

By $(1)$, this implies that $M_t$ is $\mathcal{F}_t$-measurable.