What does this expectation notation in a Markov's inequality mean? I'm trying to understand Markov's inequality, and the textbook says:

Let $Z \geq 0$ be a non-negative random
  variable. Then for all $t \geq 0$ [...].
  Note that
  $P(Z ≥ t) = E[1_{\{Z \geq t\}}]$.

Why is there a $1$ in the expectation? How is it different than $E[Z \geq t]$?
 A: $Z \ge t$ is a event, so does not have a numerical value and so does not have an expectation
$1_{A}$ represents an indicator function, taking the value $1$ if the event $A$ occurs and the value $0$ if $A$ does not occur
Since $1_{Z \ge t}$ takes a numerical value, namely $1$ when $Z \ge t$ occurs and the value $0$ when $Z \not\ge t$, you can therefore find its expectation $\mathbb E\left[1_{Z \ge t}\right]$, which is $\mathbb P(Z \ge t)\times 1 + \mathbb P(Z \not \ge t)\times 0 = \mathbb P(Z \ge t)$ 
A: This is from measure theory. The 1 you see is called an indicator function. For a generic set $A$, the indicator, $1_A(x)$ of $A$ is given by $1_A(x)=1$ if $x \in A$ and $1_A(x)=0$ if $x \notin A$. Sometimes when writing an event in brackets $\{...\}$, we don't bother with the subscript and write $1\{...\}$ instead of $1_{\{...\}}$ (because we like to be lazy with our writing sometimes).
The input of the indicator is also often dropped when its argument is clear from the context. In the context of $\{Z \geq t\}=\{\omega:Z(\omega) \geq t\}$, the input is $\omega$ belonging to your probability space $\Omega$ on which $Z$ is defined.
It is a fact that indicators of events are random variables, and as it turns out, for an event $A$, we have $E(1_A)=P(A)$ (for more information read up on Lebesgue integration). So in the context of your question we have $E(1_{\{Z \geq t\}})=P(Z \geq t)$.
