# Why left continuity does not hold in general for cumulative distribution functions?

Definition: The c.d.f. $F$ of a random variable $X$ is a function defined for each real number $x$ as follows:$$F(x)=\Pr(X\leq x) \text{ for } -\infty<x<\infty$$

Let $$F(x^-)=\lim_{y\rightarrow x,\,y<x}F(y)$$ and $$F(x^+)=\lim_{y\rightarrow x,\,y>x}F(y)$$

Property of cumulative distribution function: A c.d.f. is always continuous from the right; that is , $F(x)=F(x^+)$ at every point $x$.

Proof: Let $y_1>y_2>\dots$ be a sequence of numbers that are decreasing such that $$\lim_{n\rightarrow \infty}y_n=x.$$Then the event $\{X\leq x\}$ is the intersection of all the events $\{X\leq y_n\}$ for $n=1,2,\dots$ .Hence, $$F(x)=\Pr(X\leq x)=\lim_{n\rightarrow \infty} \Pr(X\leq y_n)=F(x^+).$$

Now I think the left inequality can also be proved in the similar way as:

Let $y_1<y_2<\dots$ be a sequence of numbers that are increasing such that $$\lim_{n\rightarrow \infty}y_n=x.$$Then the event $\{X\leq x\}$ is the union of all the events $\{X\leq y_n\}$ for $n=1,2,\dots$ .Hence, $$F(x)=\Pr(X\leq x)=\lim_{n\rightarrow \infty}\Pr(X\leq y_n)=F(x^-).$$

Where am I wrong?

• Look at $F=1_{[\alpha,\infty)}$ (point mass at $\alpha$). $F$ is a cdf, but is not continuous from the left. You can see that if $y_n \uparrow \alpha$, but $y_n < \alpha$, then $\lim_n F(y_n) = 0$, but $F(\alpha) = 1$. – copper.hat Oct 11 '13 at 6:51

Then the event $\{X\leq x\}$ is the union of all the events $\{X\leq y_n\}$ for $n=1,2,\dots$.
If $y_n\lt x$ for every $n$ and $y_n\to x$, then $\bigcup\limits_n\{X\leqslant y_n\}$ is equal to $\{X\lt x\}$, not to $\{X\leqslant x\}$.
You might want to check that $x$ is in $(-\infty,y_n]$ for no $n$ whatsoever hence $x$ is not in $\bigcup\limits_n(-\infty,y_n]$, and in fact $\bigcup\limits_n(-\infty,y_n]$ is equal to $(-\infty,x)$, not to $(-\infty,x]$.