Limit of Riemann sum as a definite integral

Question

I have the following Riemann Sum \begin{align*} \lim_{n\to\infty}\frac{1}{nh}\sum_{i=1}^nf\left(\frac{j-i}{nh}\right), \end{align*} where $j,n\in\mathbb{N}, h>0$. How can I rewrite it as a definite integral? By the change of variables $\ell=j-i$ I get \begin{align*} \lim_{n\to\infty}\frac{1}{nh}\sum_{\ell=j-n}^{j-1}f\left(\frac{\ell}{nh}\right)=\int_0^1f(u)\text{d}u, \end{align*} but using the change of variables $r = (j-i)/h$ I get \begin{align*} \lim_{n\to\infty}\frac{1}{nh}\sum_{r=(j-n)/h}^{(j-1)/h}f\left(\frac{r}{n}\right)=\frac{1}{h}\int_0^1f(u)\text{d}u. \end{align*} Of course, this is impossible, so I must have made many mistakes in what I wrote above. What I've been implicitly doing is to use the fact that \begin{align*} \lim_{n\to\infty}\sum_{i=1}^nf(a+i\Delta)\Delta = \int_{a}^bf(u)\text{d}u,\quad b:\frac{b-a}{n}=\Delta. \end{align*} I'm afraid that my mistake comes from the fact that in the second example the step in the summation is $1/h$ and not $1$, but haven't been able to formalize it. Could anyone spot the mistake(s)? Thank you so much for you help!