Is equal natural density on intervals with matching areas but opposite signs sufficient to use fixed-width part sizes for a simple Riemann sum?

Question

Suppose we have a sequence $\theta_n$ which is dense on $\left(0,2\pi\right)$. Furthermore, if $A=(x,y)\subset(0,\pi)$ and $B=(x+\pi,y+\pi)$ for some $x,y$, and if we define the natural density of a set $X$ to be: \begin{equation} \delta(X) = \lim_{N\to\infty} \frac{\left|\{\theta_n \in X\ \big|\ n \leq N\}\right|}{N} \end{equation} we have that $\delta(A)=\delta(B)$ for all such intervals.

Is the following true? \begin{equation} \lim_{N\to\infty} \sum_{n=1}^N \sin(\theta_n) \frac{2\pi}{N} = 0 \end{equation}

Certainly if the $\theta_n$ were equidistributed, this would be true. However, it seems that since for any intervals $A,B$ as above, since the natural densities match, that any error caused by multiplying by parts of average size $\frac{2\pi}{N}$ for points in $A$ would be offset by a matching error for the points in $B$. In other words, if some interval $A$ in $(0,\pi)$ were much more (or less) dense than the average, the corresponding interval $B$ in $(\pi,2\pi)$ would be just as more (or less) dense than the average, and everything would cancel out. However, I am not clear how to prove this explicitly.

Any help/suggestions are greatly appreciated, and I thank you.

Answer 1

Let us take the $\theta_n$ in the interval $[0,2\pi)$. For every $N \ge 1$, set $$\mu_N = \frac1N\sum_{n=1}^N\delta_{\theta_n} \text{ and } \nu_N := \mathbb{1}_{(0,\pi)\cup(\pi,2\pi)}\mu_N.$$ The existence of a natural density for every open interval contained in $(0,\pi)$ or in $(\pi,2\pi)$, forces the narrow convergence of the sequence $(\nu_N)_{N \ge 1}$ because their cumulative distribution functions converge: for every $x \in [0,\pi)$, $\nu_N[0,x]$ converges to $1-\delta((x,\pi))-\delta((\pi,2\pi))$ and for every $x \in [\pi,2\pi)$, $\nu_N[0,x]$ converges to $1-\delta((x,2\pi))$.

Call $\nu$ the narrow limit of $(\nu_n)$. Then for all $x<y$ in $(0,\pi)$ or in $(\pi,2\pi)$ which are not atoms of $\nu$, therefore, for almost every $x<y$ in $(0,\pi)$, $$\nu((x,y)) = \lim_{N \to +\infty} \nu_N((x,y)) = \delta((x,y))$$ By right-continuity w.r. to $x$ and left-continuity w.r. to $y$, the equality $\nu((x,y))=\nu((\pi+x,\pi+y))$ holds for all $0<x<y<\pi$. Since $\sin$ is continuous and bounded on $[0,2\pi]$ and vanishes at $0$ and $\pi$, we derive $$\int \sin \theta d\mu_N(\theta) = \int \sin \theta d\nu_N(\theta) \to \int \sin \theta d\nu(\theta) = 0,$$ which is the desired conclusion.