On a property of a converging positive martingale

Question

There is a result about convergence which says that if $(X_t : t ≥ 0)$ is a martingale with right continuous sample paths and $X_t$ converges almost surely to $X_\infty$ then we have that $X_s=\mathbb{E}[X_\infty|F_s]$. If we take a positive contunuous martingale converging to $0$ a.s then isn't it true that $X_s=\mathbb{E}[0|F_s]=0$. Would be grateful if anyone could explain where my mistake is.

Answer 1

In general, the assertion is wrong. Consider for example a Brownian motion $(B_t)_{t \geq 0}$ the process

$$M_t := \exp \left( B_t - \frac{t}{2} \right), \qquad t \geq 0.$$

$(M_t)_{t \geq 0}$ is a continuous positive martingale. Moreover, the strong law of large numbers gives

$$\frac{B_t}{t} \xrightarrow[]{t \to \infty} 0$$

implying

$$M_t = \exp \left( t \left[ \frac{B_t}{t} - \frac{1}{2} \right] \right) \xrightarrow[]{t \to \infty} 0=: M_{\infty}.$$

However, we clearly have $M_t \neq 0 = \mathbb{E}(M_{\infty} \mid \mathcal{F}_t)$.