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Gerry's reference turned out to be quite useful. Theorem 2.7 of Uniform Distribution of Sequences by Kuipers and Niederreiter states that if $a$ and $b$ are integers with $a < b$, and if $f$ is twice differentiable on $[a,b]$ with $|f''(x)| \geq \rho > 0$ on $[a,b]$, then $$\left|\sum_{n = a}^{b}{e^{2\pi i f(n)}}\right| \leq \left(\left|f'(b) - f'(a)\right| + 2\right)\left(\frac{4}{\sqrt{\rho}} + 3\right).$$ So if we assume that $g : [0,1] \to \mathbb{R}$ is a continuous twice-differentiable function with $\lambda = \inf_{x \in [0,1]} g''(x) > 0$, then by taking $a = 1$, $b = N$, $f(x) = m N g(x/N)$, we find that $$\left|\sum_{n = 1}^{N}{e^{2\pi i m N g(n/N)}}\right| \leq \left(m \left|g'(1) - g'(1/N)\right| + 2\right)\left(\frac{4}{\sqrt{m N \lambda}} + 3\right).$$

So now let $\mu_N$ be the probability measure on $[0,1]$ given by $$\mu_N(B) = \frac{1}{N} \# \left\{1 \leq n \leq N : N g(n/N) \in B \pmod{1}\right\}$$ for each Borel set $B \subset [0,1]$, and let $\mu$ denote the Lebesgue measure on $[0,1]$. Then the Erdős–Turán inequality states that for any positive integer $M$, the discrepency $$D(N) = \sup_{B \in [0,1]} \left|\mu_N(B) - \mu(B)\right|$$ satisfies $$D(N) \leq C \left(\frac{1}{M} + \frac{1}{N} \sum_{m = 1}^{M}{\left| \sum_{n = 1}^{N}{e^{2\pi i m N g(n/N)}} \right|}\right)$$ for some absolute constant $C > 0$ (independent of $N$ and $M$). Taking $M = \lfloor N^{1/3}\rfloor$ and using the earlier bound on $\sum_{n = 1}^{N}{e^{2\pi i m N g(n/N)}}$ shows that $$D(N) = O\left(N^{-1/3}\right)$$ and hence that $\mu_N$ converges weakly to $\mu$ as $N$ tends to infinity.

It may be possible to relax some of these conditions on $g$ by modifying the proof of this theorem in the book of Kuipers and Niederreiter, but I haven't checked too closely yet.

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Gerry's reference turned out to be quite useful. Theorem 2.7 of Uniform Distribution of Sequences by Kuipers and Niederreiter states that if $a$ and $b$ are integers with $a < b$, and if $f$ is twice differentiable on $[a,b]$ with $|f''(x)| \geq \rho > 0$ on $[a,b]$, then $$\left|\sum_{n = a}^{b}{e^{2\pi i f(n)}}\right| \leq \left(\left|f'(b) - f'(a)\right| + 2\right)\left(\frac{4}{\sqrt{\rho}} + 3\right).$$ So if we assume that $g : [0,1] \to \mathbb{R}$ is a continuous twice-differentiable function with $\lambda = \inf_{x \in [0,1]} g''(x) > 0$, then by taking $a = 1$, $b = N$, $f(x) = m N g(x/N)$, we find that $$\left|\sum_{n = 1}^{N}{e^{2\pi i m N g(n/N)}}\right| \leq \left(m \left|g'(1) - g'(1/N)\right| + 2\right)\left(\frac{4}{\sqrt{m N \lambda}} + 3\right).$$

So now let $\mu_N$ be the probability measure on $[0,1]$ given by $$\mu_N(B) = \frac{1}{N} \# \left\{1 \leq n \leq N : g(n/N) \in B \pmod{1}\right\}$$ for each Borel set $B \subset [0,1]$, and let $\mu$ denote the Lebesgue measure on $[0,1]$. Then the Erdős–Turán inequality states that for any positive integer $M$, the discrepency $$D(N) = \sup_{B \in [0,1]} \left|\mu_N(B) - \mu(B)\right|$$ satisfies $$D(N) \leq C \left(\frac{1}{M} + \frac{1}{N} \sum_{m = 1}^{M}{\left| \sum_{n = 1}^{N}{e^{2\pi i m N g(n/N)}} \right|}\right)$$ for some absolute constant $C > 0$ (independent of $N$ and $M$). Taking $M = \lfloor N^{1/3}\rfloor$ for any $\varepsilon > 0$ and using the earlier bound on $\sum_{n = 1}^{N}{e^{2\pi i m N g(n/N)}}$ shows that $$D(N) = O\left(N^{-1/3}\right)$$ and hence that $\mu_N$ converges weakly to $\mu$ as $N$ tends to infinity.

It may be possible to relax some of these conditions on $g$ by modifying the proof of this theorem in the book of Kuipers and Niederreiter, but I haven't checked too closely yet.

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Gerry's reference turned out to be quite useful. Theorem 2.7 of Uniform Distribution of Sequences by Kuipers and Niederreiter states that if $a$ and $b$ are integers with $a < b$, and if $f$ is twice differentiable on $[a,b]$ with $|f''(x)| \geq \rho > 0$ on $[a,b]$, then $$\left|\sum_{n = a}^{b}{e^{2\pi i f(n)}}\right| \leq \left(\left|f'(b) - f'(a)\right| + 2\right)\left(\frac{4}{\sqrt{\rho}} + 3\right).$$ So if we assume that $g : [0,1] \to \mathbb{R}$ is a continuous twice-differentiable function with $\lambda = \inf_{x \in [0,1]} g''(x) > 0$, then by taking $a = 1$, $b = N$, $f(x) = m N g(x/N)$, we find that $$\left|\sum_{n = 1}^{N}{e^{2\pi i m N g(n/N)}}\right| \leq \left(m \left|g'(1) - g'(1/N)\right| + 2\right)\left(\frac{4}{\sqrt{m N \lambda}} + 3\right).$$

So now let $\mu_N$ be the probability measure on $[0,1]$ given by $$\mu_N(B) = \frac{1}{N} \# \left\{1 \leq n \leq N : g(n/N) \in B \pmod{1}\right\}$$ for each Borel set $B \subset [0,1]$, and let $\mu$ denote the Lebesgue measure on $[0,1]$. Then the Erdős–Turán inequality states that for any positive integer $M$, the discrepency $$D(N) = \sup_{B \in [0,1]} \left|\mu_N(B) - \mu(B)\right|$$ satisfies $$D(N) \leq C \left(\frac{1}{M} + \frac{1}{N} \sum_{m = 1}^{M}{\left| \sum_{n = 1}^{N}{e^{2\pi i m N g(n/N)}} \right|}\right)$$ for some absolute constant $C > 0$ (independent of $N$ and $M$). Taking $M = \lfloor N^{1/3}\rfloor$ for any $\varepsilon > 0$ and using the earlier bound on $\sum_{n = 1}^{N}{e^{2\pi i m N g(n/N)}}$ shows that $$D(N) = O\left(N^{-1/3}\right)$$ and hence that $\mu_N$ converges weakly to $\mu$ as $N$ tends to infinity.

It may be possible to relax some of these conditions on $g$ by modifying the proof of this theorem in the book of Kuipers and Niederreiter, but I haven't checked too closely yet.