I believe I have a proof of the following result, which I need as a lemma. Indeed I have seen oblique references to this result using Google. But I would prefer a specific citation. Can anyone provide one?
Lemma: Suppose that $f$ is continuous and non-decreasing in $[0,1]$. Then $f$ can be uniformly approximated by functions which are non-decreasing and absolutely continuous.
Let $f_n$ be defined by $f_n(x)= n\left(f\left(\frac{k+1}{n}\right)-f\left(\frac{k}{n}\right)\right)\left(x-\frac{k}{n}\right)+f\left(\frac{k}{n}\right)$ for $k=\lfloor nx\rfloor$ (i.e. each $\text{ }f_n$ is a continuous piecewise-linear approximation of $\text{ }f$). Then $\text{ }f_n$ is differentiable a.e., the derivative is Lebesgue integrable (because it is a step function), and $$f_n(0)+\int_0^xf^{\prime}_n(t)dt=f_n(0)+\sum_{k=0}^{\lfloor nx\rfloor-1}\int_{\frac{k}{n}}^{\frac{k+1}{n}}f^{\prime}_n(t)dt+\int_{\frac{\lfloor nx\rfloor}{n}}^xf^{\prime}_n(t)dt\\=f_n(0)+\sum_{k=0}^{\lfloor nx\rfloor-1}\left(f\left(\frac{k+1}{n}\right)-f\left(\frac{k}{n}\right)\right)+n\left(f\left(\frac{\lfloor nx\rfloor+1}{n}\right)-f\left(\frac{\lfloor nx\rfloor}{n}\right)\right)\left(x-\frac{\lfloor nx\rfloor}{n}\right)\\ = n\left(f\left(\frac{\lfloor nx\rfloor+1}{n}\right)-f\left(\frac{\lfloor nx\rfloor}{n}\right)\right)\left(x-\frac{\lfloor nx\rfloor}{n}\right)+f\left(\frac{\lfloor nx\rfloor}{n}\right)=f_n(x)$$
Finally, we will use the fact that $f$ is uniformly continuous on $[0,1]$. Let $\epsilon>0$ and $\delta>0$ such that $\vert x-y\vert<\delta$ implies $\vert f(x)-f(y)\vert<\epsilon$ for every $x,y$. $$\vert f_n(x)-f(x)\vert=\left\vert\left[ f_n(x)-f\left(\frac{\lfloor nx\rfloor}{n}\right)\right]+\left[f\left(\frac{\lfloor nx\rfloor}{n}\right)-f(x)\right]\right\vert\\\leq\epsilon+\epsilon=2\epsilon\text{ if $n>\frac{1}{\delta}$, regardless of $x$}$$ Thus the sequence $f_n\to f$ uniformly on $[0,1]$, and each $f_n$ is absolutely continuous and non-decreasing.