I feel as if my proof went in the wrong direction. Would someone be willing simply to verify that I've gone in the right or wrong direction, and if it is indeed the wrong direction, offer a recourse? Thank you :) (NEW TO ANALYSIS, please don't judge ><).
Consider a monotone nondecreasing function $f$ on a closed interval $[a,b].$ Show that it... is continuous except at countably many points. Show that the same is true over the unbounded interval $(-\infty, \infty).$
Consider the sets \begin{align} S^+ = \{f(x+h): h \in (0, b-x]\} \\ S^- = \{f(x+h): h \in [a-x, 0)\} \end{align} I claim that $$\displaystyle{\lim_{h \to 0^+} f(x+h) = \inf S^+}.$$ Suppose not. Then, there exists an $\epsilon > 0$ such that for all $\delta > 0$, there is some $h \in (0, \delta)$ from which it follows that $$|f(x+h) - \inf S^+| \geq \epsilon.$$ Consider all $\delta > 0$ such that $\delta \leq b - x.$ Notice that if $h \in (0, \delta)$ for all such $\delta,$ then $h \in (0, b-x]$ and $f(x+h) \in S^+.$ Indeed, $$S^+_{\delta} = \bigcup \limits_{\delta \leq b-x} \{f(x+h): h \in (0, \delta)\} \subset S^+.$$ In fact, $S^+ = S^+_{\delta} \cup \{f(b)\}$ and, by the monotonicity of the function $f,$ $f(b) \geq s$ for all $s \in S^+.$ It follows, then, that $\inf S^+ = \inf S^+_{\delta}$ and $\inf S^+ \leq s_{\delta}$ for all $s_{\delta} \in S^+_{\delta}.$ However, by the earlier assumption that $$\displaystyle{\lim_{h \to 0^+} f(x+h) \neq \inf S^+}$$ it also follows that $\inf S^+ + \epsilon \leq s_{\delta}$ for all $s_{\delta} \in S^+_{\delta},$ signifying that there exists a lower bound greater than $\inf S^+,$ contradicting the foremost assumption that $\inf S^+$ is the greatest lower bound. Therefore, $$\displaystyle{\lim_{h \to 0^+} f(x+h) = \inf S^+}.$$ Again, I claim that $$f(x^-) = \displaystyle{\lim_{h \to 0^-} f(x+h) = \sup S^-}.$$ Consider the function $g(x): [-b,-a] \rightarrow \mathbb{R},$ defined $$g(\alpha) := -f(-\alpha),$$ and the sets \begin{align} S^+_{g(x)} = \{g(-x+h): h \in (0, x-a]\} \\ S^-_{g(x)} = \{g(-x+h): h \in ([x-b,0)\} \end{align}
Lemma:
Consider any two sets $A, B$ such that $A = -B,$ that is, $B = \{-a: \forall a \in A\}.$ Then, $\sup A = - \inf B.$
By definition, $\sup A \geq a$ for all $a \in A.$ Therefore, $-\sup A \leq -a,$ for all $a \in A.$ Now, consider any upper bound $b_A,$ that is, any $b_A \in \mathbb{R}$ such that $b_A \geq a$ for all $a \in A.$ It is immediate that $-b_A \leq -a$ for all $a \in A.$ That is, if $b_A$ is an upper bound of $A,$ then it is a lower bound of $B.$ Moreover, if $\sup A \leq b_A$ for all upper bounds $b_A$ of $A,$ then $-b_A \leq -\sup A$ for all lower bounds $-b_A$ of $B.$ Hence, $-\sup A = \inf B,$ or $-\inf B = \sup A.$
Now, following from the definition of $g(x),$ $S^- = -S^+_{g(x)}$ and $-\inf S^+_{g(x)} = \sup S^-.$ Moreover, by the earlier argument demonstrating that $$\displaystyle{\lim_{h \to 0^+} f(x+h) = \inf S^+}$$ and by the symmetry of the sets $S^+$ and $S^+_{g(x)},$ it follows that $$\displaystyle{\lim_{h \to 0^+} g(-x+h) = \inf S^+_{g(x)}}.$$ Again, following from the definition of $g(x)$ and the symmetry shared between $g(x)$ and $f(x)$ $$-\displaystyle{\lim_{h \to 0^+} g(-x+h)} = \displaystyle{\lim_{h \to 0^-} f(x+h)},$$ and $$\displaystyle{\lim_{h \to 0^-} f(x+h)} = -\inf S^+_{g(x)}, = \sup S^-.$$
Now, I claim that $f$ is continuous if and only if $f(x^+) = f(x^-)$ and that there are countably many points at which $f(x^+) \neq f(x^-).$
Suppose $f(x^+) \neq f(x^-).$ Then $$|f(x^+) - f(x^-)| = \epsilon' \quad \textrm{for some $\epsilon' > 0$}.$$ Then it follows for $\epsilon'$ that for all $\delta > 0$ and for $h \in (-\delta, \delta),$ $$|f(x) - f(x+h)| \nless \epsilon',$$ and the function is discontinuous at $x.$ Now, assume that $f(x^+) - f(x^-) = 0.$ It follows that $$\displaystyle{\lim_{h \to 0^+} f(x+h)} = \displaystyle{\lim_{h \to 0^-} f(x+h)} = \displaystyle{\lim_{h \to 0} f(x+h)} = f(x)$$ and, therefore, that for all $\epsilon > 0,$ there exists a $\delta > 0$ such that for $h \in (-\delta, \delta),$ $$|f(x) - f(x+h)| < \epsilon,$$ and $f$ is continuous at $x.$ \newline Now, consider any jump, that is, any $x \in [a,b]$ such that $$j(x) = f(x^-) - f(x^+).$$ Consider all jumps such that $j(x) \geq 1.$ Notice that there at most $b - a$ such jumps $j(x).$ In fact, for all jumps $j(x) \geq \frac{1}{n},$ there at most $n(b-a)$ such jumps. Consider the sets $$A_n := \{x \in [a,b]: j(x) \geq \frac{1}{n}\} \quad \textrm{for all $n \in \mathbb{N}$}.$$ Then, for all $n \in \mathbb{N},$ $|A_n| \leq n(b-a),$ and $A_n$ is finite. It follows from the finitude of $A_n$ for all $n \in \mathbb{N}$ that $\bigcup_{n \in \mathbb{N}} A_n$ is countable. Therefore, the set of all points $x$ at which $j(x) \neq 0$ is countable. The quality is maintained for monotone nondecreasing functions defined on the entirety of the real numbers. Consider that $$\mathbb{R} = \bigcup_{i \in \mathbb{Z}} I_i, \quad \textrm{where } I_i = [i, i+1].$$ Notice that there are countably many intervals $I_i,$ as $|\{i: i \in \mathbb{Z}\}|$ = $|\mathbb{Z}|.$ A countable union of countable subsets is also countable, and $$\bigcup_{i \in \mathbb{Z}} \Big(\bigcup_{n \in \mathbb{N}} A_n^{I_i}\Big), \quad \textrm{where $A_n^{I_i} = \{x \in I_i: j(x) \geq \frac{1}{n}\}$}$$ is countable.
Not an answer, but too long for a comment.
It might be easier to define $o_f(x) = \lim_{y \downarrow x} f(y) - \lim_{y \uparrow x} f(y)$. Note that $o_f(x) = 0$ iff $f$ is continuous at $x$.
Let $D_n = \{ x \in [a,b] | o_f(x) \ge {1 \over n} \}$. Note that since $f(a) \le f(x) \le f(b)$, the set $D_n$ can have at most ${f(b)-f(a) \over {1 \over n}}$ points, in particular, it is finite.
Note that $D = \{ x | o_f(x) > 0 \} = \cup_n D_n$, hence $D$ is at most countable.