Increasing functions are Baire one

Solution 1:

I explained in a comment why your approach cannot work; here’s a start on one that does.

Let $D_0$ be the set of points at which the monotonically increasing function $f$ is discontinuous, let $D_1$ be a countable dense subset of $\Bbb R\setminus D_0$, and let $D=D_0\cup D_1$. Note that we may assume that $n,-n\in D$ for each $n\in\Bbb Z^+$. (Why?) $D$ is countably infinite, so we can enumerate it as $D=\{x_n:n\in\Bbb N\}$. For $n\in\Bbb N$ let $D_n=\{x_k:k\le n\}$.

For each $n\in\Bbb Z^+$ there is an $m_n\in\Bbb N$ such that $n,-n\in D_{m_n}$ and for each $x\in[-n,n]$ there is an $x_k\in D_{n_m}$ such that $|x-x_k|<\frac1{2^n}$. (Why?) We may further assume that $m_{n+1}>m_n$ for all $n\in\Bbb Z^+$. Let $E_n=D_{m_n}\cap[-n,n]$; we can enumerate $E_n=\{y_k^{(n)}:0\le k\le e_n\}$, where $e_n=|E_n|-1$, in such a way that $$-n=y_0^{(n)}<y_1^{(n)}<\ldots<y_{e_n-1}^{(n)}<y_{e_n}^{(n)}=n\;.$$

Let $f_n:\Bbb R\to\Bbb R$ be defined as follows:

• $f_n(x)=f(-n)$ if $x\le -n$;
• $f_n(x)=f(n)$ if $x\ge n$;
• $f_n\left(y_k^{(n)}\right)=f\left(y_k^{(n)}\right)$ for $k=0,\ldots,e_n$; and
• $f_n$ is linear on each interval $\left[y_k^{(n)},y_{k+1}^{(n)}\right]$ for $k=0,\ldots,e_n-1$.

Clearly $f_n$ is continuous. Show that $\langle f_n:n\in\Bbb Z^+\rangle$ converges pointwise to $f$.