I want to prove that a distribution function $F$ is continuous at all points in its domain. By definition, $F$ is right-continuous, so it is enough to prove that it is left-continuous.
Basically we want the limit from the left to agree with $F(x)$: $$\lim_{n \to \infty} F(x - \frac{1}{n}) = P(X < x) = F(x) - P(X=x) = P(X \leq x) = F(x).$$
My problem is this: we can conclude $F$ is continuous at a point $x$ if $P(X=x) = 0$. But suppose we have a distribution with some jump discontinuity at $x$. Clearly $F$ is not continuous at $x$, but aren't all individual points in an interval assigned a probability of zero? Wouldn't that imply that $F$ is continuous at the point where it has a jump discontinuity (obviously wrong)?
Basically my issue is that if you have a probability measure that assigns $0$ to all individual points, then wouldn't this argument imply our $F$ is continuous (even though we only know it to be right-continuous)?