question on Brownian Motion stopping time and end state I came across this equation in my lecture notes, which states:
$P(T_a < t , W_t \ge a) = P(W_t \ge a)$  where $T_a = \min\{t \ge 0, W_t \ge a\}$.
I'm really confused by this equation: as far as I understand it, the R.H.S $P(W_t \ge a)$ is the probability that at the end of duration $t$, the Brownian Motion hits level $a$ and it doesn't really care about what happened within the $t$ time frame (meaning the Brownian Motion may or may not hit level a before the end of $t$). Whereas the L.H.S $P(T_a < t , W_t \ge a)$ clearly requires the the Brownian Motion hits a before the end of t. So shouldn't $P(T_a < t , W_t \ge a)$ be strictly less than $P(W_t \ge a)$? since R.H.S also includes such processes where $a$ is not hit a single time before the end of $t$.
Could someone please enlighten me on this? Thanks a lot in advance.

Remember that Brownian motion is (almost surely) continuous.  So if it is at or above level $a$ at time $t$, then by the intermediate value theorem, it (almost surely) must have hit level $a$ at some time at or prior to $t$.  (This assumes $W_0 = 0$ and $a > 0$.)  That gives you $P(T_a \le t, W_t \ge a) = P(W_t \ge a)$.  To replace $T_a \le t$ with $T_a < t$, note that if $T_a = t$ then $W_t = a$ which happens with probability 0.
