Continuous in probability of hitting times

Question

Let $(B_t)$ be a standard Brownian motion. How can we show that the process $$ \tau_t := \inf \{ s \geq 0 : B_s >t \}$$ satisfies continuity in probability? $$\bigg( \text{i.e. } \quad \lim_{h \rightarrow 0^{+}} \mathbb{P} ( |\tau_{a+h} - \tau_{a} | > \epsilon ) =0, \quad \forall a \geq 0 , \epsilon>0 . \bigg)$$

Answer 1

Hints:

1. Prove the claim for $a=0$. To this end, note that $\tau_a=\tau_0=0$ almost surely and therefore $$\mathbb{P}(|\tau_{a+h}-\tau_a|>\epsilon) = \mathbb{P}(|\tau_h|>\epsilon) \leq \mathbb{P} \left( \left| \sup_{s \leq \epsilon} B_s \right| \leq h \right).$$
2. By the strong Markov property, $W_t := W_{t+\tau_a}-W_{\tau_a}$ is a Brownian motion for each $a \geq 0$. Moreover, $$\tau_{a+h}-\tau_a = \tilde{\tau}_h$$ for $$\tilde{\tau}_h := \inf\{s \geq 0; W_s>h\}.$$ Use the first step, to conclude that $$\lim_{h \to 0} \mathbb{P}(|\tau_{a+h}-\tau_a|>\epsilon) =0.$$