Let $\xi$ be an increasing function , and $f$ be a continuous function on the interval $[0,1]$. Take $\phi$ a smooth function such that $\int_0^1 \phi(s)\, ds= 1 $ and consider an approximation of identity $\phi_n(s) = n \phi(ns)$. define $\xi^n(s) = \xi*\phi_n(s)$ (the convolution with $\phi_n$) Is is true that $$\int f(x) \, d\xi(x) = \lim_n \int f(x) \, d\xi^n(x).$$
ps: could $\xi$ be arbitrary?
It's true if $\xi$ is continuous at $0$ and $1$. Alternatively, it's true for arbitrary $\xi$ if we restrict to $f$ with $f(0)=0=f(1)$.
I'm assuming here that $\xi$ is defined on $\mathbb R$, so that the convolution is defined. If $\xi$ is originally defined only on $[0,1]$ you can extend it.
There's a simple proof if you know about Schwarz functions and tempered distributions. There exists a positive measure $\mu$ such that $$\int f\,d\xi=\int f\,d\mu.$$This $\mu$ satisfies $$\mu=D\xi$$in the sense of distributions. Hence $$D\xi^n=D(\xi*\phi_n)=(D\xi)*\phi_n=\mu*\phi_n,$$so that $\mu*\phi_n$ is the measure corresponding to $\xi^n$: $$\int f\,d\xi^n=\int f\,d(\mu*\phi_n).$$
Now, $\mu*\phi_n\to\mu$ in the weak* topology of $C_c(\mathbb R)^*$. This shows that $$\int f\,d\xi^n\to\int f\,d\xi\quad(f\in C_c(\mathbb R)).$$Which is not quite what you asked about.
Suppose that $f\in C([0,1])$ and $f(0)=0=f(1)$. Then $f$ can be extended to a continuous function on $\mathbb R$ which has compact support contained in $[0,1]$. Hence $$\int_0^1f\,d\xi^n\to\int_0^1f\,d\xi\quad(f\in C([0,1]),f(0)=0=f(1)).$$That's the second of the two claims I made at the top.
Finally, since the formula holds for all $f\in C([0,1])$ with $f(0)=0=f(1)$, it follows that it holds for all $f\in C([0,1])$ if and only if it holds for $f(t)=1$ and $f(t)=t$. Which you easily verify directly, if $\xi$ is continuous at $0$ and $1$.