Given is a function $f(E)$. If $E=\frac{1}{2}mv^2$ and thus $dv=\frac{dE}{mv}$, then the following Riemann sums are equal to each other. For simplicity's sake the intervals are considered equal so that: $$\lim_{\Delta E\rightarrow0}\sum_{n=\frac{a}{\Delta E}}^{\frac{b}{\Delta E}-1}f(n\cdot \Delta E)\Delta E=\lim_{\Delta v\rightarrow0}\sum_{n={\sqrt{\frac{2a}{m}\cdot}\frac{1}{\Delta v}}}^{\sqrt{\frac{2b}{m}}\cdot \frac{1}{\Delta v}-1}f(n\cdot \Delta v)\cdot mv\cdot \Delta v$$ I noticed that even if $\Delta E$ and $\Delta v$ are numerically infinitesimally equal, this equation still holds. So the relation $dv=\frac{dE}{mv}$ doesn't need to be taken into account for Rieman sums, as long as the intervals $\Delta E$ and $\Delta v$ approach $0$.
If the above premise is correct, then a confusion arises when writing these sums as integrals and taking one step of the integration: $$f(E)dE=f(v)mvdv$$ This equation shows that $dv$ must be equal to $\frac{dE}{mv}$ otherwise the equation is invalid. This also makes sense to me because a step in $dE$ equals a step $\frac{dE}{mv}$ in $v$ dimensions.
So how can the relation $dv=\frac{dE}{mv}$ be ignored when summed/integrated but must hold when looking at one step of the sum/integration? Is my premise wrong in the first place?