probability of a brownian motion being equal to the running maximum Let $B$ be a standard Brownian motion on $\mathbb{R}$. I would like to show that 
$$ \mathbb{P} \bigg\{ B_1 = \max_{t \in [0,1]} B_t \bigg\} =0 .$$
I argue that since $\max_{t \in [0,1]} B_t $ has the same distribution as $|B_1|$, 
$$ \mathbb{P} \bigg\{ B_1 = \max_{t \in [0,1]} B_t \bigg\} = \mathbb{P} \bigg\{ B_1 = |B_1| \bigg\} = \mathbb{P} \bigg\{ B_1 =0 \bigg\} =0 .$$
Is this argument correct (by replacing $\max_{t \in [0,1]} B_t$ by $|B_1|$ simply due to their equality in distribution)?
I am not so convinced about that fact that for any three random variables $X, Y,Z$, if $X$ and $Y$ have the same law, then $\mathbb{P}(X=Z) = \mathbb{P} (Y=Z)$. I just can't prove it rigorously.

The joint distribution of the running maximum $M_1 = \max_{0 \leq t \leq 1}B_t$ and $B_1$ is 
$$f(m,b) = \frac{2(2m-b)}{\sqrt{2\pi}}\exp(-\frac{(2m-b)^2}{2}), \qquad  m\geq 0, b \leq m $$
Hence, $P(M_1 = B_1) = 0$
Note : In your approach, you replaced $M_1$ with $\vert B_1 \vert $ which is not true! They have same distribution but the random variables are different.
