First hit of a martingale I came across this result somewhere and I don't grasp its proof in its entirety.
Let $M$ be a continuous martingale such that $M_0 = 0$. Define $\tau_x = \inf\{t\geq 0: M_t =x \}$. Then,
$$P\{\tau_a < \tau_b  \}\leq \frac{b}{b-a} \leq P\{\tau_a \leq \tau_b  \}$$
The proof goes like this. Both $\tau_a$ and $\tau_b$ are stopping times. Hence, so is $\tau_a \wedge\tau_b$. Then, Doob's optional stopping theorem gives
$E[M_{\tau\wedge t}] = 0$. The author lets $t$ go to infinity and claims $E[M_{\tau}] = 0$ by dominated convergence. This is all good but wasn't he supposed to show $\tau$ is a.s. finite first?
Anyway, then he writes:
$$0 = aP\{M_{\tau} = a\} + bP\{M_{\tau} = b\} + E[M_{\infty}\mathbb{1}_{\tau=\infty}]$$
Clearly, $P\{M_{\tau} = a\} = P\{\tau_a <\tau_b\}$ and $P\{M_{\tau} = b\} = P\{\tau_b <\tau_a\}$. Then the next line reads
$$0 \leq aP\{\tau_a <\tau_b\} + bP\{\tau_b \leq\tau_a\} $$
From this the first inequality in the claim follows. Where did $E[M_{\infty}\mathbb{1}_{\tau=\infty}]$ go?
I also don't see how the second inequality must be shown. Any help appreciated..
 A: *

*In general, we cannot expect $\tau<\infty$ almost surely. Just consider e.g. the trivial martingale $M_t := 0$, $t \geq 0$, then $\tau =\infty$ almost surely.

*Set $$M_{\infty} := \lim_{t \to \infty} M_{t \wedge \tau}.$$ Then, by the very definition of the stopping time, we have $M_{\infty} \in [a,b]$ and $M_{\infty}(\omega) = a$ for $\omega \in \{\tau_a< \tau_b\}$ and $M_{\infty}(\omega)=b$ for $\omega \in \{\tau_b<\tau_a\}$. Using $\mathbb{E}(M_{t \wedge \tau}) = 0$ and applying Fatou's lemma, we get $$\begin{align*} 0 &= \limsup_{t \to \infty} \mathbb{E}(M_{t \wedge \tau})  \\ &\leq \mathbb{E} \left( \limsup_{t \to \infty} M_{t \wedge \tau} \right) \\ &= a \mathbb{P}(M_{\tau}=a) + b \mathbb{P}(M_{\tau}=b) + \mathbb{E}(\underbrace{M_{\infty}}_{\in [a,b]} \underbrace{1_{\{\tau=\infty\}}}_{\geq 0}) \\ &\leq a \mathbb{P}(\tau_a<\tau_b) + b \mathbb{P}(\tau_b<\tau_a) + b \mathbb{P}(\tau=\infty) \\ &= a \mathbb{P}(\tau_a<\tau_b) + b \mathbb{P}(\tau_b<\tau_a) + b \mathbb{P}(\tau_a = \tau_b) \tag{1} \\ &= a \mathbb{P}(\tau_a<\tau_b) + b \mathbb{P}(\tau_b \leq \tau_a). \end{align*}$$ This gives the first inequality. In $(1)$ we have used that the continuity of $(M_t)_{t \geq 0}$ implies $$\{\tau_a = \tau_b\} = \{\tau=\infty\} \tag{2}$$ for any $a \neq b$.

*A simular reasoning as in step 2 shows $$\begin{align*} 0 &\geq a \mathbb{P}(\tau_a<\tau_b) + b \mathbb{P}(\tau_b<\tau_a) + a \mathbb{P}(\tau=\infty) \\ &\stackrel{(2)}{=} a \mathbb{P}(\tau_a \leq \tau_b) + b \mathbb{P}(\tau_b<\tau_a). \end{align*}$$ This implies the second inequality.

