Suppose that $f: S \rightarrow T$ is increasing, therefore $f(x-)$ and $f(x+)$ exist when $x \in S$. The fact to show is that $f$ has at most countably many jump points. A proof is given as follows on Page 4 in Chung's A Course in Probability Theory: $x_1$ is a jump point such that $f(x_1-) < f(x_1+)$. Let $I_{x_1} = (f(x_1-), f(x_1+))$. Let $x_2$ denote another jump point and $x_2 > x_1$, then by monotonicity, we have $f(x_1+) \leq f(x_2-)$. Let $I_{x_2} = (f(x_2-), f(x_2+))$ Therefore, $I_{x_1}$ and $I_{x_2}$ are disjoint. For each jump point, such an interval exists and all of them are disjoint. Then, for any two jump point $x_1$ and $x_2$, Chung says that one can find two rationals, $r(x_1) \in I_{x_1}$ and $r(x_2) \in I_{x_2}$ such that $r(x_1) < r(x_2)$. Therefore the sets of these jump points are in a one-to-one correspondence with the rationals.
My question is: Since all of these intervals are nonempty and disjoint, we can also find two irrationals $ir(x_1)\in I_{x_1}$ and $ir(x_2)\in I_{x_2}$ such that $ir(x_1) < ir(x_2)$. Therefore $ir$ is a one-to-one correspondence from the set of jump points to the set of irrationals. Could anyone help me: What mistake did I make?