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For standard Brownian motion $B$ and $h>0$, let $T_h$ be its first hitting time of $h$. Define the reflected path byleaving a path unchanged if $T_h=\infty$. The Brownian reflection principle says that $\hat{B}$ has the same path probability distribution as $B$. In its endpoint form, for $t>0$ and every Borel set $A\subset (-\infty ,h)$,Here is a direct proof of preservation of the path law. For a fixed integer horizon $L$, let $T^{(m)}=2^{-m}\lceil 2^m{T}_h\rceil \wedge L$, with an infinite first term interpreted as infinity. This is a stopping time taking finitely many deterministic grid values. On $\{T^{(m)}=j{2}^{-m}\}$, the history determining that event is measurable at $j{2}^{-m}$, and subsequent Brownian motion increments are independent of that history. Those increments and their negatives have the same joint law, by symmetry of centered normal distributions and independent increments. Thus reflecting after $T^{(m)}$ around $B_{T^{(m)}}$ preserves the full path law. For any fixed horizon $t<L$, the reflected paths converge uniformly on $[0,t]$ to the displayed reflected path, by continuity as $T^{(m)}$ decreases to $T_h\wedge L$. This includes $T_h=\infty$, since reflection after $L$ has no effect before $t$. Passing to all finite-dimensional probability distributions proves the assertion; no prior recurrence claim is needed. Equivalently, the Strong Markov property gives the same proof by symmetry of the fresh increments after the hit. Reflection is an involution preserving the hitting event and mapping the indicated endpoint $y$ to $2h-y$. An endpoint above $h$ forces a hit by continuity; this proves the endpoint identity, rather than assuming it.

For $S_t={\sup }_{0\le s\le t}{B}_s$, split $\{T_h\le t\}$ according to whether $B_t$ is above or below $h$. Since $B_t$ has no atom at $h$, reflection givesBoth variables are nonnegative, so their positive tails identify the law:For $t>0$ this is the half-normal distribution of scale $\sqrt{t}$; at $t=0$ both variables vanish.

For the two separated intervals, putHere $A_0$ is measurable for ${\mathcal{F}}_b$, $Z$ has the normal distribution $N(0,c-b)$, and $H$ is a function of increments after $c$. The independent increments make these three objects mutually independent. The later absolute maximum is $B_b+Z+H$, so equality of the two maxima means $Z=A_0-H$. Conditional on $(A_0,H)$, this specifies a single value of the nondegenerate normal distribution $Z$, and hence has probability zero. ThereforeThis proves the atomless maxima on separated Brownian intervals without incorrectly treating the two absolute maxima as independent.