= Solution
For each deterministic $a>0$, the <Brownian reflection principle> gives
$$
\mathbb P(H_a>t)=\mathbb P\!\left(\sup_{s\leq t}X_s<a\right)
=2\Phi(a/\sqrt t)-1\longrightarrow0,
$$
so $H_a$ is finite <almost surely>; $H_0=0$. Here $\Phi$ is the standard normal <cumulative distribution function>. The same reflection formula at levels decreasing to zero gives $\mathbb P(\sup_{s\leq\varepsilon}X_s\leq0)=0$ for every $\varepsilon>0$. Taking a countable intersection over $\varepsilon=1/n$ shows that Brownian motion exceeds zero arbitrarily soon after its start.
By the <Strong Markov property> at the finite <stopping time> $H_a$, the process $X_{H_a+t}-a$ is fresh standard <Brownian motion>. It therefore exceeds zero arbitrarily soon. Path continuity gives $S_a\geq H_a$, and the preceding fact gives the reverse inequality. Consequently
$$
\boxed{\mathbb P(H_a=S_a)=1\quad\text{for each fixed }a\geq0.}
$$
The exceptional event may depend on $a$; this does not imply simultaneous equality for all levels.
To exhibit the difference, set $A=\max_{0\leq t\leq1}X_t$, a random level. The reflection formula gives $A>0$ almost surely. Also $A>X_1$ almost surely: if $X_1$ were the maximum, the reversed increment process $Y_s=X_1-X_{1-s}$ would be nonnegative throughout $[0,1]$. This process has the law of standard <Brownian motion>, whose probability of staying nonnegative from zero is zero by reflection and symmetry. The level $A$ is attained at some time strictly before one, so $H_A<1$. It is not exceeded by time one, and continuity with $X_1<A$ gives a further positive interval before it can be exceeded. Thus
$$
\boxed{H_A<1<S_A\quad\text{almost surely}.}
$$
All these passage times exist simultaneously on the probability-one event where the Brownian path is continuous and unbounded above, obtained by intersecting the finite-hitting events for positive integer levels.
\Image[/past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2010/iii/paper-28-passage-inverses.png]
{title=A schematic excursion below a record level separates the first hitting and first exceeding times}
The illustration shows the pathwise mechanism. The <Brownian running maximum> stays flat during an excursion below an attained record. The non-strict inverse chooses the beginning of that plateau, while the <strict generalized inverse of a nondecreasing function> chooses its end. The curve is a schematic of this mechanism, not a claim of piecewise linear Brownian paths. The fixed-level equalities give the <modification of a stochastic process> relation, while the random-level discrepancy rules out <indistinguishability of stochastic processes>. This is <fixed-level versus simultaneous Brownian passage-time equality>.
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