uniqueness of minimum of absolutely continuous stochastic process

30 Views Asked by At

I am trying to understand the arguments in the suggested proof of Remark 1.2 in this paper

https://link.springer.com/article/10.1007/BF00536298

In essence, the authors consider the process $X(t)=B(t)+\theta t$, where $B(t)$ is a Brownian motion and $\theta$ a constant, and show that, under some assumptions spelt out in the paper, it has a unique minimiser on $[0,\infty)$.

Their argument is as follows. They define the set, on the interval $[0,K]$, where the minimum is not unique, say $C_K$. Then this is majorised by

$C_K \subset \bigcup_{0 \leq r < s \leq K} \{\omega | \min_{0 \leq t \leq r}X(t,\omega)-\min_{s \leq t \leq K}X(t,\omega)\}$

for $0 \leq r < s \leq K$ \textit{rational} numbers. The authors then define

$Z_{rs}=\min_{0 \leq t \leq r}X(t)-\min_{s \leq t \leq K}X(t)\\ =\min_{0 \leq t \leq r}[B(t)-B(r)+\theta t] +[B(s)-B(r)] -\min_{s \leq t \leq K}[B(t)-B(s)+\theta t] $

The authors show that $Z_{rs}$ is absolutely continuous for all rationals $r<s$. I can follow until here.

But then they say that, if I follow correctly, such absolute continuity implies that $P(C_K)=0$, which I am not sure how to follow. There must be a consequence of absolute continuity which I am not getting.

Any help will be received with many many thanks.