Why does the cumulative distribution function have to be nondecreasing? The cumulative distribution function $F(x)$ has the property that it is nondecreasing, i.e. $F(x) \leq F(y)$ if $ x \leq y $. Why? What if it were a decreasing function?
 A: Suppose that $F$ is not non-decreasing. Then by definition, there would exist $x \leq y$ for which $F(x) > F(y)$. If $F$ is the CDF of some random variable $X$, then this would mean:
$$P(X \leq x) > P(X \leq y)$$
This is impossible since the event $\{X \leq x\} \subset \{X \leq y\}$, and in general it is the case that if $A \subset B$ then $P(A) \leq P(B)$. So from this we see that $F$ must be non-decreasing.
It is possible for a CDF to be either increasing, or strictly increasing. A CDF will be not strictly increasing, whenever there is a `discontinuity' in the set of values the variable can take on: for instance a discrete distribution.
For example, consider $X \sim \text{Bin}(1/2)$. In this case the CDF, $F\colon [0,1] \rightarrow [0,1]$, is given to be:
\begin{align*}
F(x) = \begin{cases}
\frac12 & x \in [0,1) \\
1 & x = 1.
\end{cases}
\end{align*}
Conversely, if we consider $X \sim \text{Unif}(0,1)$, then the CDF, $F:[0,1] \rightarrow [0,1]$ is given to be:
$$F(x) = x$$
So that in this case the CDF is strictly increasing.
A: $F$ is a probability measure. When you're writing $F(x)$, what you're doing is you're calculating a probability measure on the set $(-\infty, x]$. Similarly, $F(y)$ is a probability measure on the set $(-\infty, y]$.
To note that $F$ is a probability measure, I'll write, for example, $F(x) = \mathbb{P}((\infty, x])$.
Note, however, that $(-\infty, x] \subset (-\infty, y]$ if $x \leq y$. One of the properties of a probability measure is given at Prove that if $A \subset B$ then $P(A) \leq P(B)$. Thus it follows that $\mathbb{P}((\infty, x]) \leq \mathbb{P}((-\infty, y])$, or $F(x) \leq F(y)$.
In other words, it doesn't make sense for $F(x) > F(y)$ under consequences of the probability axioms. See the link above for more details.
