# Proving a CDF is cadlag

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.

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 $$X and $$P(X \leq x_n) \to F(x-)$$