Where is an error in my deduction? (question about martingales)

Question

Suppose we have a filtration $\{\mathcal{F_{t}},t\geq 0\}$ and a stochastic process $\{ X_{t},t\geq 0\}$ which is adapted to this filtration and also integrable. All we need for this process to be a martingale is to hold $ E[X_{t} | \mathcal{F_{s}} ] = X_{s}$ for every $t > s$. But suppose we take a trivial filtration i.e. $\mathcal{F_{t}}=\mathcal{F}$ for every $t\geq 0$ where $\mathcal{F}$ is some sigma-algebra in our probability space. Then obviously $ E[X_{t} | \mathcal{F_{s}} ] = X_{t}$ for every $t > s$ which breaks our martingale property. So adaptability and integrability for $\{ X_{t},t\geq 0\}$ and $ E[X_{t} | \mathcal{F_{s}} ] = X_{s}$ is not enough to be a martingale? Do we need some extra assumtions concerning $\{\mathcal{F_{t}},t\geq 0\}$?

Answer 1

Suppose that $\left(X_t\right)_{t\geqslant 0}$ is a martingale with respect to the filtration $\left(\mathcal F_t\right)_{t\geqslant 0}$ where $\mathcal F_t=\mathcal F$ for all $t$.

The condition of adaptedness implies that for all $t$, $X_t$ is $\mathcal F$-measurable.

The condition $\mathbb E\left[X_t\mid\mathcal F_s\right]=X_s$ for $0\leqslant s\lt t$ reads therefore $X_s=\mathbb E\left[X_t\mid\mathcal F\right]=X_t$, since $X_t$ is $\mathcal F$-measurable hence $X_t=X_0$ for all $t$.