Suppose $M^i_t = X^i_t - X^i_0 - \int_0^t b_i(s,X)\, ds$ where $b_i:[0,\infty)\times \Omega \to \mathbb{R}$ is a progressively measurable functional and $X^i_t: C[0,\infty)^d \to \mathbb{R}$ ( $X(t\omega)$is a continuous function in $t$
The question is: Show that if $M^i_t$ is a continuous local martingale then
$$\mathbb{P} \bigg[\int_0^t |b_i(s,X)|\, ds < \infty; 0 \leq t <\infty \bigg] = 1$$
I tried the following: Let $E$ be the expectional set $E = \bigg[\omega:\int_0^{t_0} |b_i(s,\omega)|\, ds = \infty \text{ for some } t_0 >0 \bigg]$. If $P(E) >0$ then $M^i_t$ is not a local martingale.
To be a local martingale means that there is a sequence of stopping times $\tau_n$ ($\tau_n \to \infty$ almost surely) such that $M^i_{t \wedge \tau_n}$ is a martingale.
The idea is to show that on the set $E$, $\tau_n < t_0$. This I think would be easier if $b_i>0$ but in the case where there is a an absolute value, I can't quite show that the sequence diverges and that the condition $M^i_{t\wedge \tau_n} \in \mathbb{R}$ forces $\tau_n < t_0$
thanks for your help
for reference: this is used in proposition 4.6 (chapter 5 pg 315) of Karatzas and Schreve - Brownian motion and Stochastic calculus