[proofplan]
We write the [cumulative distribution function](/page/Cumulative%20Distribution%20Function) in terms of the sublevel events $A_x=\{\omega\in\Omega:X(\omega)\le x\}$. Monotonicity follows immediately from inclusion of sublevel events. Right-continuity and the two limits are consequences of countable additivity of $\mathbb P$, applied through continuity from above and continuity from below for monotone sequences of events. The endpoint limits use the facts that a real number is not bounded above by every negative integer and is bounded above by some positive integer.
[/proofplan]
[step:Define the sublevel events and verify their measurability]
For each $x\in\mathbb R$, define the event $A_x\in\mathcal F$ by
\begin{align*}
A_x=\{\omega\in\Omega:X(\omega)\le x\}=X^{-1}((-\infty,x]).
\end{align*}
The set $(-\infty,x]$ belongs to $\mathcal B(\mathbb R)$, and $X:(\Omega,\mathcal F)\to(\mathbb R,\mathcal B(\mathbb R))$ is measurable. Hence $A_x\in\mathcal F$ for every $x\in\mathbb R$. By definition,
\begin{align*}
F_X(x)=\mathbb P(A_x).
\end{align*}
Since $\mathbb P$ is a probability measure, $0\le \mathbb P(A_x)\le 1$, so $F_X:\mathbb R\to[0,1]$ is well-defined.
[/step]
[step:Prove that the distribution function is nondecreasing]
Let $x,y\in\mathbb R$ satisfy $x\le y$. If $\omega\in A_x$, then $X(\omega)\le x\le y$, so $\omega\in A_y$. Thus $A_x\subset A_y$. By monotonicity of the measure $\mathbb P$,
\begin{align*}
F_X(x)=\mathbb P(A_x)\le \mathbb P(A_y)=F_X(y).
\end{align*}
Therefore $F_X$ is nondecreasing.
[guided]
Fix $x,y\in\mathbb R$ with $x\le y$. The event $A_x$ records the outcomes for which $X$ falls at or below $x$, while $A_y$ records the outcomes for which $X$ falls at or below the larger threshold $y$. Thus every outcome counted by $A_x$ is also counted by $A_y$: if $\omega\in A_x$, then
\begin{align*}
X(\omega)\le x\le y,
\end{align*}
and hence $\omega\in A_y$. This proves the event inclusion
\begin{align*}
A_x\subset A_y.
\end{align*}
Measures are monotone on measurable sets, so
\begin{align*}
\mathbb P(A_x)\le \mathbb P(A_y).
\end{align*}
Using the definition $F_X(t)=\mathbb P(A_t)$ for $t\in\mathbb R$, this becomes
\begin{align*}
F_X(x)\le F_X(y).
\end{align*}
Since this holds for every pair $x\le y$, the function $F_X$ is nondecreasing.
[/guided]
[/step]
[step:Record continuity of probability along monotone event sequences]
We shall use the following two consequences of countable additivity.
[claim:Continuity from below and from above for a probability measure]
Let $(E_n)_{n\in\mathbb N}$ be a sequence of events in $\mathcal F$.
If $E_n\subset E_{n+1}$ for every $n\in\mathbb N$ and $E=\bigcup_{n=1}^{\infty}E_n$, then
\begin{align*}
\mathbb P(E)=\lim_{n\to\infty}\mathbb P(E_n).
\end{align*}
If $E_{n+1}\subset E_n$ for every $n\in\mathbb N$ and $E=\bigcap_{n=1}^{\infty}E_n$, then
\begin{align*}
\mathbb P(E)=\lim_{n\to\infty}\mathbb P(E_n).
\end{align*}
[/claim]
[proof]
Assume first that $E_n\subset E_{n+1}$ for every $n\in\mathbb N$. Define $D_1=E_1$, and for $n\ge 2$ define
\begin{align*}
D_n=E_n\setminus E_{n-1}.
\end{align*}
Then $(D_n)_{n\in\mathbb N}$ is a pairwise disjoint sequence of events in $\mathcal F$, and
\begin{align*}
E_n=\bigcup_{k=1}^{n}D_k
\end{align*}
for every $n\in\mathbb N$. Also,
\begin{align*}
E=\bigcup_{k=1}^{\infty}D_k.
\end{align*}
By countable additivity and finite additivity of $\mathbb P$, we have $\mathbb P(E)=\sum_{k=1}^{\infty}\mathbb P(D_k)$ and $\mathbb P(E_n)=\sum_{k=1}^{n}\mathbb P(D_k)$. The sequence of partial sums of the nonnegative series converges to the full series, so
\begin{align*}
\lim_{n\to\infty}\mathbb P(E_n)=\mathbb P(E).
\end{align*}
Now assume that $E_{n+1}\subset E_n$ for every $n\in\mathbb N$, and set $E=\bigcap_{n=1}^{\infty}E_n$. Define $G_n=E_1\setminus E_n$ for $n\in\mathbb N$, and define
\begin{align*}
G=E_1\setminus E.
\end{align*}
Then $G_n\subset G_{n+1}$ for every $n\in\mathbb N$, and
\begin{align*}
G=\bigcup_{n=1}^{\infty}G_n.
\end{align*}
By the continuity-from-below part already proved,
\begin{align*}
\mathbb P(G)=\lim_{n\to\infty}\mathbb P(G_n).
\end{align*}
Because $G_n\subset E_1$ and $G\subset E_1$, finite additivity gives
\begin{align*}
\mathbb P(G_n)=\mathbb P(E_1)-\mathbb P(E_n)
\end{align*}
and
\begin{align*}
\mathbb P(G)=\mathbb P(E_1)-\mathbb P(E).
\end{align*}
Since $\mathbb P(E_1)\le 1<\infty$, subtracting from $\mathbb P(E_1)$ yields
\begin{align*}
\lim_{n\to\infty}\mathbb P(E_n)=\mathbb P(E).
\end{align*}
[/proof]
[/step]
[step:Prove right-continuity by shrinking intervals from the right]
Fix $x\in\mathbb R$. For each $n\in\mathbb N$, define the event $B_n\in\mathcal F$ by
\begin{align*}
B_n=A_{x+1/n}.
\end{align*}
Since $x+1/(n+1)\le x+1/n$, the monotonicity of sublevel events gives $B_{n+1}\subset B_n$. Moreover,
\begin{align*}
\bigcap_{n=1}^{\infty}B_n=A_x.
\end{align*}
Indeed, if $\omega$ belongs to every $B_n$, then $X(\omega)\le x+1/n$ for every $n\in\mathbb N$, and therefore $X(\omega)\le x$. The reverse inclusion follows from $x\le x+1/n$.
By continuity from above for $\mathbb P$,
\begin{align*}
\lim_{n\to\infty}F_X(x+1/n)=\lim_{n\to\infty}\mathbb P(B_n)=\mathbb P(A_x)=F_X(x).
\end{align*}
Now let $\varepsilon>0$. Choose $n\in\mathbb N$ such that
\begin{align*}
F_X(x+1/n)-F_X(x)<\varepsilon.
\end{align*}
If $y\in\mathbb R$ satisfies $x\le y\le x+1/n$, then $A_x\subset A_y\subset A_{x+1/n}$, hence
\begin{align*}
0\le F_X(y)-F_X(x)\le F_X(x+1/n)-F_X(x)<\varepsilon.
\end{align*}
This proves
\begin{align*}
\lim_{y\downarrow x}F_X(y)=F_X(x),
\end{align*}
so $F_X$ is right-continuous.
[/step]
[step:Compute the limit at negative infinity]
For each $n\in\mathbb N$, define the event $C_n\in\mathcal F$ by
\begin{align*}
C_n=A_{-n}.
\end{align*}
Since $-(n+1)\le -n$, the monotonicity of sublevel events gives $C_{n+1}\subset C_n$. Also,
\begin{align*}
\bigcap_{n=1}^{\infty}C_n=\varnothing.
\end{align*}
To justify this, suppose $\omega\in\bigcap_{n=1}^{\infty}C_n$. Then $X(\omega)\le -n$ for every $n\in\mathbb N$, which is impossible because $X(\omega)\in\mathbb R$.
By continuity from above, $\lim_{n\to\infty}F_X(-n)=\lim_{n\to\infty}\mathbb P(C_n)=\mathbb P(\varnothing)=0$. Given $\varepsilon>0$, choose $n\in\mathbb N$ such that $F_X(-n)<\varepsilon$. If $x\le -n$, then by monotonicity,
\begin{align*}
0\le F_X(x)\le F_X(-n)<\varepsilon.
\end{align*}
Therefore
\begin{align*}
\lim_{x\to-\infty}F_X(x)=0.
\end{align*}
[/step]
[step:Compute the limit at positive infinity]
For each $n\in\mathbb N$, define the event $D_n\in\mathcal F$ by
\begin{align*}
D_n=A_n.
\end{align*}
Since $n\le n+1$, the monotonicity of sublevel events gives $D_n\subset D_{n+1}$. Also,
\begin{align*}
\bigcup_{n=1}^{\infty}D_n=\Omega.
\end{align*}
Indeed, if $\omega\in\Omega$, then $X(\omega)\in\mathbb R$, so there exists $n\in\mathbb N$ with $X(\omega)\le n$; hence $\omega\in D_n$.
By continuity from below, $\lim_{n\to\infty}F_X(n)=\lim_{n\to\infty}\mathbb P(D_n)=\mathbb P(\Omega)=1$. Given $\varepsilon>0$, choose $n\in\mathbb N$ such that
\begin{align*}
1-F_X(n)<\varepsilon.
\end{align*}
If $x\ge n$, then by monotonicity,
\begin{align*}
F_X(n)\le F_X(x)\le 1.
\end{align*}
Thus
\begin{align*}
0\le 1-F_X(x)\le 1-F_X(n)<\varepsilon.
\end{align*}
Therefore
\begin{align*}
\lim_{x\to\infty}F_X(x)=1.
\end{align*}
Together with nondecreasingness and right-continuity, this proves all asserted properties of $F_X$.
[/step]
