Say I have a continuously differentiable function $f: [0,1] \to \mathbb{R}^n$ (whose first derivative we denote by $f'$). Let $\psi: \mathbb{R} / \mathbb{Z} \times [0,1] \to \mathbb{R}^n$ be a continuous map which satisfies the following integral representation -
$$ f' = \int_0^1 \psi(s, \cdot)ds. $$
Where $n = \lfloor Nt \rfloor$, let $k: [0,1] \to \mathbb{R}^n$ be the step function given by $$k(t) = f(0) + \sum_{i=1}^n \int_{\frac{i-1}{N}}^{\frac{i}{N}}\psi(Ns, s)ds,$$ and let $l: [0,1] \to \mathbb{R}^n$ be the step function given by $$l(t) = f(0) + \frac{1}{N}\sum_{i=1}^nf'(i/N)$$
I want to understand why $k$ converges pointwise to $l$. By a change of variables, we can rewrite $k$ as $$k(t) = f(0) + \sum_{i=1}^n \int_{0}^{1}\psi\left(u, \frac{u+i}{N}\right)du.$$ This very much seems arbitrarily close to $f(0) + \sum_{i=1}^n \int_{0}^{1}\psi\left(u, \frac{i}{N}\right)du = l(t)$, but I can't quite prove it rigorously, though it seems elementary (because I'm running into difficulties with convergence of sums).
(For some context, this is what is asserted more generally in the proof of Proposition 3.1 of David Spring's book on Convex Integration Theory - but I think there is something off about one part of it - in particular, I can't believe assertion (3.16) in the proof of Proposition 3.1, which is what this is related to.)
I would very much appreciate help!