Properties of distribution function Let $(\Omega, \mathcal{F}, P)$ be a probability space, $X$ a random variable and $F(x) = P(X^{-1}(]-\infty, x])$. The statement I am trying to prove is

The distribution function $F$ of a random variable $X$ is right continuous, non-decreasing and satisfies $\lim_{x \to \infty}F(x) = 1$, $\lim_{x \to -\infty} F(x) = 0$.

As $F(x + \delta) = F(x) + P(]x, x + \delta])$, we have that $F$ is non-decreasing, but is the measure of an interval bounded by its length? In that case we would have right continuity as well.
For the limits, we have $F(x) + P(X^{-1}(]x, \infty]) = P(\Omega) = 1$, so $F(x) = 1 - P(X^{-1}(]x, \infty])$, so it suffices for $P(X^{-1}(]x, \infty])$ to get small as $x$ gets large and to get large as $x$ gets small. This is not true for general measures, take the Lebesgue measure for example, but maybe because we need $P(X^{-1}(\mathbb{R}))$ to be $1$?
 A: The probability assigned to an interval is certainly not bounded by its length. For example, discrete distributions assign positive probability to intervals of length $0.$
To prove right-continuity you need countable additivity.
\begin{align}
F(x) & = \Pr(X\le x) = 1 - \Pr(X>x) \\[8pt]
& = 1 - \Pr(x+1 < X \text{ or } x+\tfrac 1 2 < X\le x+1 \text{ or } x+\tfrac 1 3 < X\le x + \tfrac 1 2 \text{ or } \cdots) \\[8pt]
& = 1 - \big( \Pr(x+1< X) +\Pr(x+\tfrac 1 2 < X\le x+1) + \Pr(x+\tfrac 1 3< X\le x + \tfrac 1 2) + \cdots \\[8pt]
& = 1 - \lim_{N\,\to\,\infty} \sum_{n\,=\,0}^N \Pr( x + \tfrac 1 {n+1} < X \le x + \tfrac 1 n) \\[8pt]
& = \lim_{N\,\to\,\infty} \Pr(X\le x + \tfrac 1 {N+1}) = \lim_{N\,\to\,\infty} F(x + \tfrac 1{N+1}).
\end{align}
Given $\varepsilon>0,$ find $N$ large enough so that $F(x+\tfrac 1{N+1}) < F(x)+\varepsilon, $ and then choose $\delta= 1/N.$ Then for $x < w < x+\delta,$ you have $F(x)\le F(w)< F(x)+\varepsilon.$ The point of this paragraph is that it's not just $\lim_{N\to\infty} F(x+\tfrac 1 {N+1}) = F(x),$ but $\lim_{w\,\downarrow\,x} F(w) = F(x).$
A: It is a basic fact that for any finite measure $\mu$ the condition $A_n$  decreasing to $A$ implies that $\mu (A_n) \to \mu (A)$. [Lebesgue measure is an infinite measure and this property fails for Lebesgue measure]. This follow from the fact that $\mu(A_n^{c}) \to \mu(A^{c})$ since $A_n^{c}$ increases to $A$ and $\mu (E^{c})=\mu (\Omega)-\mu (E)$.  With this result in hand it should be easy for you to complete your arguments.
Note that $(x,x+\delta]$ decreases to empty set as $\delta$ decreases to $0$ and $(x, \infty)$ decreases to empty set as as $x$ increases to $\infty$.
A: Let $P_X:=P\circ X^{-1}$, then $(\mathbb{R},\mathcal{B}(\mathbb{R}),P_X)$ is a probability space (that is, $P_X$ is a probability measure in the Borel $\sigma $-algebra of the standard topology on $\mathbb{R}$). Now pick any sequence $(x_k)\to -\infty $, then from the reversed Fatou's lemma we have that
$$
\lim_{k\to \infty }F(x_k)=\limsup_{k\to\infty }P_X[(-\infty ,x_k]]\leqslant P_X\left[\limsup_{k\to\infty}(-\infty ,x_k]\right]=P_X[\emptyset ]=0
$$
Therefore $\lim_{x\to -\infty }F(x)=0$. Similarly you can show that $\lim_{x\to\infty }F(x)=1$ using the standard Fatou's lemma, continuity from the right follows also easily using the dominated convergence theorem, and the increasing nature of $F$ is a simple consequence of $P_X$ being a measure.
