Let $X$ be a random variable with probability density function $f$. Prove that the probability distribution function of $X$ is non-decreasing Can anyone please help me with this random variable question I've stumbled across.
Recall from calculus that a function $h$ is called non-decreasing if $x \le y$ implies $h(x) \le h(y)$, for every $x, y \in \mathop{\mathrm{dom}} h$.
Q1a) Let $X$ be a continuous random variable with probability density function $f$. Prove
that the probability distribution function of $X$ is non-decreasing.
I'm assuming this means show $F(x) = \int_{-\infty}^x f(y)\,dy$, is a non-decreasing function of $x$ in $\mathbb R$.
Q1b) Show that $\lim_{x\to-\infty} F(x) = 0$ and $\lim_{x\to \infty} F(x) = 1$, and explain the probabilistic meaning of these facts. 
Sorry about the layout i'm not used to using this site, hope it makes sense!
 A: One way to do 1(b): 
$F(x) = \int_{-\infty}^x f(t)\ dt$ is an improper integral, which by definition of improper integral means $\lim_{a \to -\infty} \int_a^x f(t)\ dt$.  Now $\int_a^x f(t)\ dt = F(x) - F(a)$, so
$$ F(x) = \lim_{a \to -\infty} (F(x) - F(a)) = F(x) - \lim_{a \to -\infty} F(a)$$
and you can solve for $\lim_{a \to -\infty} F(a)$.
As for $\lim_{x \to \infty} F(x) = \lim_{x \to \infty} \int_{-\infty}^x f(t)\ dt$, that is $\int_{-\infty}^\infty f(t)\ dt$, which according to the  definition of a probability density function must be $1$.
A: Yes because the density function is never less than 0 the integral from -∞ to x cannot be larger than from -∞ to y for any y>x.  So F is monotone non-decreasing.  The density function f has to go to 0 as x approaches + or - ∞.  So lim x→−∞ F(x)=0.  The fact that lim x→∞ F(x)=1 is a requirement for f to be a probability density with F as its CDF. There is no other way to see that f integrates to 1 from the information given.  All you can deduce is that it integrates to a positive constant.
