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!

Answer

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 $1$.