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}$$\frac{1}{N}$$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}$$\frac{1}{N}$$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}$$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...

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...