According to Rajendra Bhatia in his book Fourier Series, Weyl's Equidistribution Theorem states that if $x$ is an irrational number, then for every subinterval $[a, b]$ of $(0, 1)$ we have

$\lim_{n \rightarrow \infty} \frac{1}{N}\operatorname{card}\{k : 1 \leq k \leq N, \tilde{(kx)} \in [a, b]\} = b - a$ where $\tilde{(kx)}$ is the fractional part of the number $kx$.

My question is what happens if we generalise to measurable subsets of $(0, 1)$?

Does $\lim_{n \rightarrow \infty}\frac1N\operatorname{card}\{k : 1 \leq k \leq N, \tilde{(kx)} \in A\} = \mu(A)$ where $A$ is a measurable subset and $\mu$ the Lebesgue measure function?

Further, for non-measurable subsets $V$ is the sequence $\frac{1}{N}\operatorname{card}\{k : 1 \leq k \leq N$, $\tilde{(kx)} \in V\}$ bounded above and below and if so, does it have the same set of sublimits for all irrational $x$?

After my last question that revealed I had momentarily forgotten all my undergraduate real analysis, I hope this one is worthy of MathOverflow... thanks...

Lebesgue-a.e.starting point $x$. Thus if $A$ is measurable and $E$ has positive Lebesgue measure and the limiting frequency of visits exists and is equal for every $x\in E$, then that limit must equal $\mu(A)$. $\endgroup$