$\textbf{The Problem:}$ Let $F$ be the CDF of a random variable $X$. Prove that $\lim\limits_{t\to\infty}F(t)=1.$
$\textbf{My Thoughts:}$ We claim that $\{X\leq n\}\nearrow\Omega\quad\text{as }n\to\infty.$ First we note that the set $\{X\leq n\}$ are nested nondecreasing, since $X\leq n$ implies that $X\leq n+1.$ Now, we clearly have $\bigcup_n\{X\leq n\}\subseteq\Omega$. Now let $\omega\in\Omega$ be arbitrary but fixed. Then $X(\omega)$ is a real number, so there exists an integer $m$ such that $X(\omega)<m$, and hence $\omega\in\{X\leq m\}$, so $\omega\in\bigcup_n\{X\leq n\}$, whence $\bigcup_n\{X\leq n\}=\Omega.$ Then the continuity of the probability measure implies that $F(n)=P(\{X\leq n\})\to P(\Omega)=1$ as $n\to\infty$. Now we are ready to tackle the problem.
Let $\varepsilon>0$ be given and choose $n\in\mathbb N$ such that $1-\varepsilon<F(n).$ Then by the monotonicity of the CDF, we have that $t\geq n$ implies that $1-\varepsilon<F(n)\leq F(t)\leq1,$ so that $\vert F(t)-1\vert<\varepsilon.$ This proves that $\lim\limits_{t\to\infty}F(t)=1.$
Do you agree with the above proof? Any feedback is much appreciated. Thank you for your time and have a great week.