I'm currently reading the book Brownian Motion by Mörters and Peres and on p. 39 the following situation comes up:
We have a Brownian Motion $B(t)$ and two intervals $[a_1,b_1]$ and $[a_2,b_2]$ where $b_1<a_2$. We define $m_1 = \max_{a_1\leq t \leq b_1}B(t)$ and $m_2 = \max_{a_2\leq t \leq b_2}B(t)$. It then follows that $B(a_2)-B(b_1)$, $m_1-B(b_1)$ and $m_2-B(a_2)$ are independent.
We wan't to show that $\mathbb{P}(m_1 = m_2) = 0$ and therefore they consider the event $$\{\omega:B(a_2)-B(b_1)= m_1-B(b_1)-(m_2-B(a_2))\}$$
They say that they condition on the values of the random variables $m_1-B(b_1)$ and $m_2-B(a_2)$ and that the output from the LHS is a continuous random variable while the RHS outputs a constant.
I don't understand what they mean by conditioning in this case. What I've seen before is conditional expectation and I believe that $$\mathbb{E}[B(a_2)-B(b_1)\vert m_1-B(b_1)] = \mathbb{E}[B(a_2)-B(b_1)]$$ $$\mathbb{E}[m_1-B(b_1)-(m_2-B(a_2))\vert m_1-B(b_1)] =m_1-B(b_1)$$ holds by the independence however I don't see how this implies the above where they are dealing with a specific event.
Define $X=B(a_2)-B(b_1)$, $Y=m_1-B(b_1)$, $Z=m_2-B(a_2)$, and $m=m_2-m_1=X-Y+Z$. As noted, $X,Y,Z$ are mutually independent. Let $\mu$ be the distribution of $Y$ and $\nu$ the distribution of $Z$. Then, using the independence, because $m=X-Y+Z$, $$ P[m=0]=\int\int P[X-y+z=0]\mu(dy)\nu(dz).\qquad\qquad (1) $$ But the random variable $X$ has a continuous distribution, so $$ P[X=y-z]=0, $$ for each pair $(y,z)$. It follows that the integral on the right in (1) vanishes, hence so does the left; that is, $P[m=0]=0$.