If $\frac1n\sum_{i=1}^n1_{[0,\:a)}(x_i)\to a$ for all $a$, can we conclude $\frac1n\sum_{i=1}^nf(x_i)\to\int_{[0,\:1)}f$ Let $(x_n)_{n\in\mathbb N}\subseteq[0,1)$ with $$\frac1n\sum_{i=1}^n1_{[0,\:a)}(x_i)\xrightarrow{n\to\infty}a\;\;\;\text{for all }a\in[0,1]\tag1$$ and $f:[0,1)\to\mathbb R$ be bounded and Borel measurable.

Let $\lambda$ denote the Lebesgue measure on $\mathcal B(\mathbb R)$. Under which additional assumption on $f$ can we show that $$\frac1n\sum_{i=1}^nf(x_i)\xrightarrow{n\to\infty}\int_{[0,\:1)}f\:{\rm d}\lambda?\tag2$$

Since it shoulbe be possible that $x_n\in\mathbb Q$ for all $n\in\mathbb N$ and $g:=1_{\mathbb Q\cap[0,\:1)}$ is a possible choice for $f$ for which $$\frac1n\sum_{i=1}^ng(x_i)=1\ne0=\int_{[0,\:1)}g\:{\rm d}\lambda\;\;\;\text{for all }n\in\mathbb N\tag3,$$ we obvioulsy need to impose an additional assumption.
Unfortunately, I was able to "prove" the claim for genral $f$, but I absolutely don't see where I've made a mistake.
First of all, if $(2)$ is true for all $f$ satisfying $\left|f([0,1))\right|\in\mathbb N$, then the result should hold for all $f$. In fact, given $\varepsilon>0$, we can find Borel measurable $g_i:[0,1)\to\mathbb R$ with $\left|g_i([0,1))\right|\in\mathbb N$, $$|g_1|\le|f|\le|g_2|$$ and $$\left|f-g_i\right|<\varepsilon.$$ Now, \begin{equation}\begin{split}\int_{[0,\:1)}f\:{\rm d}\lambda-\varepsilon&<\int_{[0,\:1)}g_1\:{\rm d}\lambda\\&=\lim_{n\to\infty}\frac1n\sum_{i=1}^ng_1(x_i)\\&\le\liminf_{n\to\infty}\frac1n\sum_{i=1}^nf(x_i)\\&\le\limsup_{n\to\infty}\frac1n\sum_{i=1}^nf(x_i)\\&\le\lim_{n\to\infty}\frac1n\sum_{i=1}^ng_2(x_i)\\&=\int_{[0,\:1)}g_2\:{\rm d}\lambda\\&<\int_{[0,\:1)}f\:{\rm d}\lambda+\varepsilon.\end{split}\tag4\end{equation}
So, I guess we cannot show that $(2)$ holds for all $f$ satisfying $\left|f([0,1))\right|\in\mathbb N$. But what goes wrong? Noting that $$\mathcal B([0,1))=\sigma\left(\left\{[a,b):0\le a<b\le 1\right\}\right)\tag5$$ it is obviously enough to show $(2)$ for $f$ satisfying $$f=\sum_{j=1}^kf_j1_{[a_{i-1},\:a_i)}$$ for some $k\in\mathbb N$, $f_1,\ldots,f_k\in\mathbb R$ and $0=a_0<\cdots<a_k=1$. But $(1)$ obviously implies that $(2)$ holds for such $f$ ... So, what am I missing?
EDIT: Another possibility would be that it is not possible that $x_n\in\mathbb Q$ for all $n\in\mathbb N$, but while I have no example of such $(x_n)_{n\in\mathbb N}$ at hand, I'm pretty sure that it can be constructed, but please let me know if this is wrong.
 A: Let $\Sigma$ denote the set of all simple, Lebesgue measurable functions on $[0,1)$. Let $\Sigma_0$ denote the set of all functions on $[0,1)$ of the form $$s=\sum_{j=1}^k f_j1_{[a_{j-1},a_j)},$$ where $k\in \mathbb{N}$, $f_1, \ldots, f_k\in \mathbb{R}$, and $0=a_0<\ldots <a_k=1$. Note that $\Sigma\neq \Sigma_0$.
You correctly state that your hypothesis implies that $$\lim_n \frac{1}{n}\sum_{i=1}^n s(x_i)=\int s$$ for all $s\in \Sigma_0$.
You also state that for any $f$ (in $L_\infty$, say), you can find $g_1,g_2\in \Sigma$ (not $\Sigma_0$) such that $|g_1|\leqslant |f|\leqslant |g_2|$ and $|f(x)-g_i(x)|\leqslant \varepsilon$ for all $x\in [0,1)$. This is also correct.
But for an arbitrary $f\in L_\infty$, in general you cannot find such $g_1,g_2\in \Sigma_0$ to satisfy this property. Indeed, let $S=\{x_i:i\in \mathbb{N}\}$ and let $f=1_S$. Then for any $g_1,g_2\in \Sigma_0$ such that $g_1\leqslant f\leqslant g_2$, $g_1\leqslant 0$ and $g_2\geqslant 1$. Of course, this $f$ is a simple function, but it cannot be well-approximated by linear combinations of functions of the form $1_I$ , $I$ an interval.
So the problem is that while the conditions imply that $\lim_n \frac{1}{n}\sum_{i=1}^n s(x_i)=\int s$ for special kinds of simple functions, it doesn't imply that this holds for all simple functions.
