If $P(E_n) = 0$, then $P(\cup E_n) =0$ What does this question mean here. If the probability of $E_n$ is zero does that mean $p(E_i) =0, i < n$? I am not sure if I should assume subset here? 
 A: I think what this should say is that if $P\left(E_{n}\right)=0$ for
all $n\in\mathbb{N}$, then $P\left(\bigcup_{n=0}^{\infty}E_{n}\right)=0$.
This follows from subadditivity:
$$
P\left(\bigcup_{n=0}^{\infty}E_{n}\right)\leq\sum_{n=0}^{\infty} P\left(E_{n}\right)=0.
$$
Edit: As an intuitive example, consider the events of the form $E_n$ in which $X$, a random variable that is normally distributed takes on the value $x_n\in\mathbb{R}$. Note that $P\left(E_n\right)=0$ (to convince yourself of this, just look at the integral $$P\left(E_n\right)=\int_{x_n}^{x_n}\varphi\left(x\right)dx=0$$
where $\varphi$ is the normal probability density function). Now, consider $\bigcup_{n=0}^\infty E_n$. This has probability
$$P\left(\bigcup_{n=0}^\infty E_n\right)=\sum_{n=0}^\infty \int_{x_n}^{x_n}\varphi\left(x\right)dx=0.$$
You can come up with any number of examples.
A: For countable $n$, the identity always holds since
$$0\leq P(\cup E_n)\leq\sum_n P(E_n)=0$$
But if $n$ can't be enumerable, then the last summation doesn't seem to true in general case. For example, pick a point randomly at interval $[0,1]$, the probability of hitting 0 is zero, but the summation of hitting $x\in[0,1]$ is obviously 1.
