I'd like to know if this proof is correct. Thank you for your help.
Let $X$ be a random variable. A function $F_X: \mathbb{R} \to [0,1]$ defined by:
$$F_X(x) = \mathbb{P}(X \leq x) $$
is called the cumulative distribution function of X. By the definition of cadlag, it is a right continuous function with a limit on the left. That is,
$\bullet$ The limit on the left, $\, \lim_{s\uparrow t} F_X(s) = F_X(t^-) $ exists.
$\bullet$ The limit on the right, $\,\lim_{s\downarrow t} F_X(s) = F_X(t^+)$ exists and equals $F_X(t)$.
To begin, we show the continuous limit. For some decreasing sequence, $\{ x_n:\, \, x_n \downarrow t\}$, the sequence of events $\{X \leq x_n \}$, is a decreasing sequence of sets.
$$\implies \lim_{n \to \infty} \boldsymbol{1}\{X \leq x_n \} = \boldsymbol{1} \left \{ \bigcap_{n=1}^\infty X \leq x \right\} = \boldsymbol{1}\{X \leq t\}$$
$$\implies \lim_{s\downarrow t} F_X(s) = \lim_{n \to \infty} F_X(x_n) = \lim_{n \to \infty } \mathbb{P}(X \leq x_n) = \mathbb{P}(X\leq t) = F_X(t) $$
By definition, $F_X$ is right-continuous. Now for the other direction. Naturally we notice that for a sequence of sets, $\{X \leq x_n \}$, which is decreasing the complement is correspondingly increasing.
$$\boldsymbol{1}\left \{\bigcup_{n =1}^\infty X \leq x_n \right \} = \lim_{n \to \infty} \boldsymbol{1}\{X \leq x_n \} = \boldsymbol{1}\{X < x \} \implies \mathbb{P}(X < x) = \lim_{n \to \infty} \mathbb{P}(X \leq x_n)$$
If the sequence of numbers now approaches upwards, $\{ x_n:\, \, x_n \uparrow t\}$
$$ \lim_{s\uparrow t} F_X(s) = \lim_{n \to \infty } F_X(x_n) =\lim_{n \to \infty} \mathbb{P}(X \leq x_n) = \mathbb{P}(X < t)$$
I'm most concerned about having to define some behavior of the function at this point for the second part. Please let me know how this may be improved.
Best Answer
You said. $$\lim_{s\uparrow t} F_X(s) = 1 - \lim_{s \downarrow t }F_X(s) $$ This is not correct. This would mean that if $F$ is a continuous CDF then $F(t)=1-F(t)$ or $F(t)=\frac 1 2$ for all $t$!.
If $x_n$ strictly increases to $x$ then $(X \leq x_n)$ increases to $X<x$ and $P(X \leq x_n) \to F(x-)$
Other parts of your answer seem OK.