Question: If $\tau$ is a stopping time and $\omega(t)$ is Brownian Motion then the Strong Markov Theorem states that $Z(t)=\omega(t+\tau) -\omega(\tau)$ conditioned on $\{\tau <\infty\}$ is distributed as standard Brownian motion. 

I am confused on what "conditioned on $\{\tau <\infty\}$ " means.
Does this mean that when we restrict Wiener measure to $\{\tau <\infty\}$ that $Z(t)$ is distributed as Brownian Motion. What happens if $\{\tau <\infty\}$ has measure zero?

Answer: 
If $\lbrace \tau < \infty \rbrace$ has measure zero, the stopping time $\tau$ is almost surely equal to $\infty$. This is not an interesting case, and you can not really define $\omega(\tau)$ or $\omega(t+\tau)$ in this case anyway.
You can view the Brownian Motion as a random element from some probability space $ (\Omega, \mathcal{F}, \mathbb{P}) $ to the space $ C[0, \infty) $, then the event $ \{ T < \infty \}$ is an event of $ \Omega $. In the same way you can see $ Z $ as a random element, too. $ Z: \Omega \rightarrow C[0, \infty) $.

So, with this in mind, what this Markov Property is saying is that, conditioned on
$ \{ T < \infty \}$ the random elements $B$ and $Z$ have the same distribution.

Hope this can help.