Exercise 4.17

Following the hint, let $E_1$ be the set of points in $(a,b)$ on which $f(x-)<f(x+)$. With each $x\in E_1$ associate a triple $(p,q,r)$ of rational numbers such that $f(x-)<p<f(x+)$, $f(y)<p$ for $a<q<y<x$, and $f(y)>p$ for $x<y<r<b$. Each triple is associated with at most one point of $E_1$. Otherwise there would be $x_1$ and $x_2$ in $E_1$ such that $f(y)<p$ for $q<y<x_1$ and $q<y<x_2$, and $f(y)>p$ for $x_1<y<r$ and $x_2<y<r$, which is a set of contradictory conditions on $f(y)$ for $y$ between $x_1$ and $x_2$. Hence $E_1$ can be mapped in a one-to-one fashion to a subset of the set of all rational triples, which is countable, and so $E_1$ is at most countable. Similarly, the set $E_2$ of points in $(a,b)$ on which $f(x-)>f(x+)$ is also at most countable.

Let $F_1$ be the set of points in $(a,b)$ on which $f(x-)=f(x+)<f(x)$. With each $x\in F_1$ associate a triple $(p,q,r)$ of rational numbers such that $f(y)<p<f(x)$ for $a<q<y<x$ and $x<y<r<b$. Each triple is associated with at most one point of $F_1$. Otherwise there would be $x_1$ and $x_2$ in $F_1$ such that $f(y)<p<f(x_1)$ for $q<y<x_1$ and $x_1<y<r$, and $f(y)<p<f(x_2)$ for $q<y<x_2$ and $x_2<y<r$, which is a set of contradictory conditions on $f(x_1)$ and $f(x_2)$. Hence, as with $E_1$, $F_1$ is at most countable. Similarly, the set $F_2$ of points in $(a,b)$ on which $f(x-)=f(x+)>f(x)$ is also at most countable.

Since the set of points in $(a,b)$ at which $f$ has a simple discontinuity is $E_1\cup E_2\cup F_1\cup F_2$, the set of such points is at most countable.