discretization of probability measures Suppose I have given a probability measure $\nu$ over the positive reals. For a fixed $n\in\mathbb{N}$, we set $\lambda := \frac{1}{n}$ and $A_n:=\{\lambda k, k=0,\dots\}$. Now we look at a certain discretization of $\nu$ on $A_n$:
$$\nu_n(\{0\}):= \int_0^\lambda (1-nx)d\nu (x) \\ \nu_n(\{\lambda k\}):=\int_{(k-1)\lambda}^{(k+1)\lambda}(1-|nx-k|)d\nu(x)$$
If we have a function $f:A_n\to\mathbb{R}$, I want to show that the following equation holds:
$$\int g(x)d\nu_n(x)=\int F^n(g)d\nu(x) \tag{1}$$
where $F^n(g):=(1-\kappa)g(\lfloor nx\rfloor \lambda)+\kappa g((\lfloor nx\rfloor +1)\lambda)$. Note for continuous $g$, we have $F^n(g)\to g$ pointwise.
About $(1)$, we need for sure the result, that if you have a measure of the form $\mu_f=\int fd\mu$, then $\int gd\mu_f=\int gf d\mu$. Writing the LHS out, I do not see why this should be the RHS. Thanks in advance for your help.
 A: We have, for non-negative, measurable $g$: $\def\R{\mathbb R}\def\abs#1{\left|#1\right|}$ 
\begin{align*}
  \int_\R g(x)\,d\nu_n &= \sum_{k \ge 0} g(\lambda k)\nu_n(\lambda k)\\
   &= g(0)\int_0^\lambda (1-nx)\, d\nu + \sum_{k\ge 1} g(\lambda k) \int_{(k-1)\lambda}^{(k+1)\lambda} (1 - \abs{nx-k})\, d\nu\\
   &= g(0)\int_0^\lambda (1-nx)\, d\nu + \sum_{k\ge 1} g(\lambda k)\cdot \left(\int_{(k-1)\lambda}^{k\lambda} (1-k+nx)\, d\nu + \int_{k\lambda}^{(k+1)\lambda} (1+k-nx)\, d\nu \right)\\ 
   &= g(0)\int_0^\lambda (1-nx)\, d\nu+ \sum_{k\ge 0} g(\lambda (k+1))\cdot \int_{k\lambda}^{(k+1)\lambda} (-k+nx)\, d\nu \\&{}\quad + \sum_{k \ge 1} g(\lambda k)\int_{k\lambda}^{(k+1)\lambda} (1+k-nx)\, d\nu\\ 
  &= \sum_{k\ge 0}g(\lambda (k+1))\cdot \int_{k\lambda}^{(k+1)\lambda} (-k+nx)\, d\nu +  \sum_{k \ge 0} g(\lambda k)\int_{k\lambda}^{(k+1)\lambda} (1+k-nx)\, d\nu\\
  &= \sum_{k\ge 0} \int_{k\lambda}^{(k+1)\lambda} g\bigl((k+1)\lambda\bigr)(-k+nx) + g(\lambda k)(1+k-nx)\; d\nu
\end{align*}
Note now that for $x \in [k\lambda, (k+1)\lambda)$ we have $nx \in [k, k+1)$, hence $\def\fl#1{\lfloor#1\rfloor}\fl{nx}=k$, so we continue
\begin{align*}
  &= \sum_{k\ge 0} \int_{k\lambda}^{(k+1)\lambda} g\bigl((\fl{nx}+1)\lambda\bigr)(-\fl{nx}+nx) + g(\fl{nx}\lambda)(1+\fl{nx}-nx)\; d\nu\\
  &= \int_{\R} \{nx\}g\bigl((\fl{nx}+1)\lambda\bigr) + (1-\{nx\})g(\fl{nx}\lambda)\, d\nu.
\end{align*}
So if you choose your $\kappa$ as $\kappa = \{nx\} := nx - \fl{nx}$, then 
$$ \int_{\R} g\, d\nu_n = \int_{\R} F^n(g)\, d\nu $$
holds for non-negative $g$ and (by linearity) hence for all integrable $g$.
A: Both sides of (1) are linear functionals of the sequence $(g(x))_{x\in A_n}$, the LHS because the support of $\nu_n$ is included in $A_n$ and the RHS because $F^n(g)$ depends on $(g(x))_{x\in A_n}$ only. 
Fix some $k\geqslant0$. The coefficient of $g(k/n)$ on the LHS is $\nu_n(\{k/n\})$.
The coefficient of $g(k/n)$ on the RHS is $(1-\kappa)\nu(B_k)+\kappa\nu(B_{k-1})$, where $B_i=\{x\mid\lfloor nx\rfloor=i\}=[i/n,(i+1)/n)$. By definition,
$$
\nu_n(\{k/n\})=\int_{-1/n}^{1/n}(1-n|x|)\mathrm d\nu(x).
$$
I fail to see how this can coincide with
$$
(1-\kappa)\nu(B_k)+\kappa\nu(B_{k-1})=\int_{-1/n}^{1/n}u_\kappa(x)\mathrm d\nu(x),
\qquad u(x)=(1-\kappa)\mathbf 1_{x\gt0}+\kappa\mathbf 1_{x\lt0}.
$$
