The Problem: Let $X$ be a random variable. Show that the function $G(t)=P(\{X<t\})$ is left-continuous.
My Thoughts: It suffices to show that if $(t_n)_n$ is a sequence with $t_n\nearrow t$ as $n\to\infty$, then $G(t_n)\to G(t)$ as $n\to\infty$. To see this, consider the sequence of events $\{X< t_n\}_{n=1}^\infty$. Since $t_{n}\leq t_{n+1}$, we have that $\{X<t_n\}\subseteq\{X<t_{n+1}\}$. Therefore, the sequence of sets is increasing. Now we prove that $\{X<t\}=\bigcup_{n=1}^\infty\{X<t_n\}$. Let $\omega\in\bigcup_{n=1}^\infty\{X<t_n\}$. Then $X(\omega)<t_n$ for some $n\in\mathbb N$, and since $t_n\leq t$ for all $n\in\mathbb N,$ we have that $X(\omega)<t$, so $\omega\in\{X<t\}$. Next, let $\omega\in\{X<t\}$. Then $X(\omega)<t$, and since $t_n\nearrow t$, it follows that there is some $n\in\mathbb N$ such that $X(\omega)<t_n\leq t$, and thus $\omega\in\bigcup_{n=1}^\infty\{X<t_n\}.$
Now the continuity of the probability measure implies that
$$\lim\limits_{n\to\infty}G(t_n)=\lim\limits_{n\to\infty}P(\{X<t_n\})=P\left(\bigcup_{n=1}^\infty\{X<t_n\}\right)=P(\{X<t\})=G(t),$$
and the result follows.
Do you agree with the above proof? Any feedback is much appreciated, including the possibility of a more elegant and shorter solution. Thank your for your time.