Contradiction Optional stopping theorem : If $\tau=\inf\{t\geq 0\mid B_t=1\}$ why $\mathbb E[B_\tau]=0$? Let $(B_t)$ a standard Brownian motion. So, it's a Martingale. Let $$\tau=\inf\{t\geq 0\mid B_t=1\}.$$ In particular, $$B_{\tau}=1\ \ a.s.,$$
therefore, we could expect that $\mathbb E[B_\tau]=1.$ But according to Optionnal stopping theorem this is $$\mathbb E[B_\tau]=\mathbb E[B_0]=0,$$
how can this be possible ?  

We can apply optional stopping time theorem because taking $f\in \mathcal C_0^2(\mathbb R)$ s.t. $f(x)=x^2$ on $[0,1]$, we have by Dynkin formula $$\mathbb E[B_\tau^2]=\mathbb E\int_0^\tau \frac{1}{2}f''(u)du=\frac{1}{2}\mathbb E[\tau],$$ and thus $$\mathbb E[\tau]=2\mathbb E[B_\tau^2]=2<\infty .$$ 
 A: As mentioned in the comments, the reason why we can't conclude $E[B_\tau] = 0$ is because $\tau$ doesn't have a finite expectation.
$$\begin{align*}
\mathbb{E}\tau &= \int_0^\infty \mathbb{P}(\tau > t)\, dt 
 = \int_0^\infty (1 - \mathbb{P}(\tau \leq t)\,dt \\
 &= \int_0^\infty (1 - 2\mathbb{P}(B_t > 1))\,dt 
 = \int_0^\infty \mathbb{P}(|B_t| \leq 1) \,dt  
 = \int_0^\infty \mathbb{P}(|B_1| \leq 1/\sqrt{t})\,dt \\
 &= \sqrt{\frac{2}{\pi}} \cdot \int_0^\infty \int_0^{1/\sqrt{t}} e^{-x^2/2}\,dx\,dt \\
 &= \sqrt{\frac{2}{\pi}} \int_0^\infty \int_0^{1/x^2} e^{-x^2/2}\,dt\,dx 
 = \sqrt{\frac{2}{\pi}} \int_0^\infty  \frac{1}{x^2} e^{-x^2/2}\,dx \\ 
 &= \mathbb{E}[Z^{-2}] = \infty
\end{align*}$$
where $Z$ is a standard normal.

As a sidenote, the standard way to extract information from the optional stopping theorem for a stopping time with infinite expectation is to remember that the minimum of two stopping times is a stopping time, so we can take a sequence $\nu_t \to \infty$ of stopping times, each with finite expectation, and consider $\tau \wedge \nu_t$ instead.  A common choice is the truncated stopping time $\tau \wedge t.$  In this case, doing so lets you evaluate $$\mathbb{E}\left[B_t \middle| \max_{0 \leq s \leq t} B_s < 1\right]$$
