X and Y are uniformly distributed independent random variables, let Z = max{X,Y}. find the PDF of Z I have a few questions regarding this problem that I do not quite understand. 
For a random variable U distributed uniformly on the interval [0,1] its CDF is given by 
$$\ F_U(u) = \begin{Bmatrix} 0, & u < 0 \\ u, & 0 \le u \le 1, \\ 1, & u > 1 \\ \end{Bmatrix} $$


*

*I don't understand why for values of u > 1 the CDF of U is 1. I would think that $\ F_U $ would be undefined outside of the interval that the random variable U is defined on.


====The solution to the problem in post title and  explanation of my lack of understanding====
Let Z = max{X,Y}. For z$\ \in $ (0,1)
$\ F_Z(z) = P(Z \le z ) $ 
=  P(X$\le$z and Y$\le$z)
= $\ F_X(z)F_Y(z)$
= $\ z^2 $


*I understand that the CDF of a u.r.v. is $\ \frac{x-a}{b-a}$ for X on [a,b] so that $\ F_X(z)$ and $\ F_Y(z) $ are equal to z, but I do not understand why P(X$\le$z and Y$\le$z) = $\ F_Z$

 A: 
I don't understand why for values of $u > 1$ the CDF of $U$ is $1$. I would think that $F_U$ would be undefined outside of the interval that the random variable $U$ is defined on.

The probability density function is defined for values above the support's supremum; it equals zero there.   (Likewise for values below the infimum.)
$U$ has a support of $(0;1)$, so $\mathsf P(U\leqslant 1)=1$ by the Law of total probability.   A random variable is certainly not-more than the supremum of its support.   Likewise it will also certainly not be greater than any higher value.   $\therefore~\forall u\geqslant 1 : \mathsf P(U\leqslant u)= 1~$. 

I do not understand why $P(X\leqslant z\text{ and }Y\leqslant z) =
F_Z(z)$

Because $F_Z(z)=\mathsf P(Z\leqslant z)$, by definition, and $Z=\max\{X,Y\}$.
Now when $Z$ is not greater than $z$, neither $X$ nor $Y$ is greater than $z$, and vice versa.   Thus the event $\{Z\leq z\}$ is the join of the events $\{X\leqslant z\}$ and $\{Y\leqslant z\}$.
$$\begin{align}F_Z(z) &= \mathsf P(Z\leqslant z) \\ & =\mathsf P(\min\{X,Y\}\leqslant z) \\ &= \mathsf P(X\leqslant z~\cap~ Y\leqslant z) \\ &= \mathsf P(X\leqslant z)~\mathsf P(Y\leqslant z) \\ &= F_X(z)~F_Y(z)\end{align}$$
