# Eliminating discontinuities for increasing function

Let $f:\mathbb{R}\rightarrow\mathbb{R}$ be a monotone increasing function. We know from a general fact that the set of discontinuities of $f$ is countable. Denote the set of discontinuities by $D$.

Define $g:\mathbb{R}\rightarrow\mathbb{R}$ as follows: $$g(x)=f(x^-)-\sum_{y\in D \text{ and }y<x}(f(y^+)-f(y^-))$$ It should be relatively clear that $g$ is monotone increasing. But is $g$ necessarily continuous?

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What is $g(x)$ if $f(x)=\lfloor x\rfloor$? – bof Nov 29 '13 at 4:19

Let's start from the beginning. For $a<b$, let $j(a,b)$ be the sum of all jumps strictly between $a$ and $b$, that is, the sum of $f(x^+)-f(x^-)$ over $x\in (a,b)$. The sum looks crazy (uncountable), but since the sum of any finite subcollection is bounded by $f(b^-)-f(a^+)$, the total sum makes sense and is finite (consequently, $f(x^+)-f(x^-)=0$ for all but countably many $x$). It follows that $g$ is increasing, because $$g(b)-g(a) = f(b^-)-f(a^+) - j(a,b) \ge 0$$
The finiteness of $j(a,b)$ also implies that for any fixed $x$, $j(x,y)\to 0$ as $y\to x^+$ and $j(y,x)\to 0$ as $y\to x^-$. (For the same reason that a series of positive terms with bounded partial sums converges.) Therefore, $$\begin{split}\lim_{y\to x^-} (g(x )-g(y)) & = f(x^-)-\lim_{y\to x^-}f(y^-) - \lim_{y\to x^-}j(y,x) \\&= f(x^-)- f(x^-) - \lim_{y\to x^-}j(y,x) \\&= 0 \end{split}$$ and $$\begin{split}\lim_{y\to x^+} (g(y )-g(x)) &= \lim_{y\to x^+}f(y^-) - f(x^-) - (f(x^+)-f(x^-)) - \lim_{y\to x^+}j(x,y) \\&= f(x^+)- f(x^-) - (f(x^+)-f(x^-)) - \lim_{y\to x^+}j(x,y) \\&= 0\end{split}$$