Approximation of a sum with an integral................... Let $G$ a continuous function in $C([0,1], \mathbb R)$. I think that $$\frac{1}{N}\sum_{x \;\text{odd}\in \{1,\ldots, N\}}G\Big (\frac{x}{N}\Big )\xrightarrow{N\to +\infty}\frac{1}{2}\int_0^1G(r)dr,$$
and I would like to prove that. I think to do a change of variables as follows:
$$\frac{1}{N}\sum_{x \;\text{odd}\in \{1,\ldots, N\}}G\Big (\frac{x}{N}\Big )=\frac{1}{N}\sum_{k=0}^{\frac{N-1}{2}}G\Big(\frac{2k+1}{N}\Big)=\frac{1}{2}\frac{1}{N}\sum_{z=1}^NG\Big(z\Big),$$
where in the last step I defined $z=2k+1$ and I applied a kind of change of variables for the series that usually holds for integrals ($dk=\frac{1}{2}dz$). Is that correct?
It would perhaps be worth recalling that for a sequence of partitions $ \Big  \{ 0=y_0^{(m)}&lt; y_1^{(m)}&lt;....&lt; y_{n^{(m)}}^{(m)} =1  \Big\}_{m=1}^\infty  $  of $[0,1]$ such that $\underset{k\in n^{(m)}}{\min}  \vert y_k^{(m)}-y_{k-1}^{(m)} \vert \overset {m\rightarrow \infty}{\rightarrow}0$, that:

$\underset{k=1}{ \overset{n^{(m)} }{\sum} }  G(\zeta_k^{(m)}) \cdot (y_k^{(m)}-y_{k+1}^{(m)}) \overset{m\rightarrow \infty}{\rightarrow} \int_0^1 G(t) dt $ 

where $\zeta_k^{(m)}\in [y_k^{(m)}, y_{k+1}^{(m)}]$ for all $k\in \{1,...,n^{(m)} \}$.
The points $x/N$ (add $0,1$ if needed) for odd $x$ form a partition of $[0,1]$ with norm $2/N$ and hence the Riemann sum $$\frac{2}{N}\sum_{x\text{ odd}} G(x/N) $$ tends to $\int_{0}^{1}G(r)\,dr$ as $N\to\infty $ and the proof of the result in question in complete. Note that the result holds for all Riemann integrable functions $G$ and not just for continuous functions.