local maximum of brownian motion

I have a question:

Given two disjoint intervals $[a,b]$ and $[c,d]$, how to prove almost surely we have

$$\sup_{t\in[a,b]}B_t\neq\sup_{t\in[c,d]}B_s$$

where $B$ is a standard brownian motion. I have no idea about this problem. Does someone have an idea? Thansk a lot!

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If you write $X_t = B_{a+t}-B_a$ for $0 \le t \le b-a$ and $Y_s =B_{c+s}-B_c$ for $0 \le d-c$, then $(X_t)_{t\in[0,b-a]}$ and $(Y_s)_{s\in[0,d-c]}$ are independent Brownian motions, also independent of $B_a$ and $B_c$. So their suprema $$A = \sup\limits_{t\in [0,b-a]}X_t \text{ and } C = \sup\limits_{s\in [0,d-c]}Y_s$$ are independent random variables, which are also independent of $B_a$ and $B_c$. Now $$\mathbb{P}\left[\sup\limits_{t\in[a,b]} B_t = \sup\limits_{t\in[c,d]} B_t\right] =\mathbb{P}[B_a + A = B_c + C] = \mathbb{P}[A-C = B_c - B_a].$$ This is always $0$ since $B_c-B_a$ is continuous (actually, $N(0,c-a)$-distributed) and independent of $A-C$ (which is also continuous unless both $a=b$ and $c=d$.)
Why would $A-C$ be constant? Since they are independent, that would only be the case if both $A$ and $C$ are constant, i.e., iff $a=b$ and $c=d$. – Lukas Geyer May 14 '13 at 16:24
Why $\mathbb{P}[A-C = B_c - B_a]=0$? Could you formalize this argument? – leticia Nov 8 '15 at 11:09
@leticia: I guess the easiest way to formally argue is that the distribution of the difference $(A-C) - (B_c - B_a)$ is the convolution of the distributions of $A-C$ and $B_a - B_c$ (by independence), and that the convolution of two continuous random variables is also continuous. This shows that it has no atoms and so it is $0$ with probability $0$. – Lukas Geyer Nov 9 '15 at 21:23