Path properties of Brownian Motion: relation between its maximum and hitting time

Question

Let $B(t)$ be a Brownian motion. $$T_a=\inf\{t>0,B(t)=a\}$$ $$M(t)=\max_{0\le s\le t} B(s)$$

There is a statement in Durrett's textbook (3rd last line in page 318, 4th edition): $$\{T_a<t\}=\{M(t)>a\}$$ I don't quite understand why this holds. Inclusion $\supseteq$ is easy to get.

But why is $\subseteq$ true? It excludes the event "$M(t)=a$", is there something wrong?

The identity holds only up to the null set ${M(t)=a}$. – Did Dec 26 '14 at 08:55 — Did, Dec 26 '14 at 08:55
@Did oh,I see, the author does take the $\mathbb P$ then. – Lookout Dec 26 '14 at 08:57 — Lookout, Dec 26 '14 at 08:57

score 2 · Accepted Answer · edited Apr 13 '17 at 12:21

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The two sets $\{T_a<t\}$ and $\{M(t)>a\}$ are equal to up to a null set, because the event $M(t)=a$ has probability zero. Indeed, both $T_a$ and $M(t)$ are continuously distributed random variables: Distribution of hitting time of line by Brownian motion.

edited Apr 13 '17 at 12:21

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answered Jan 03 '15 at 21:45

Path properties of Brownian Motion: relation between its maximum and hitting time

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