Limit of integrals over $\bigcup_{z\in\Bbb Z}[z/n,(z+a)/n]$ Let $F$ be a continuous distribution function. Fix $a\in(0,1)$, and for $n\in\Bbb N$ let 
$$A_n=\{x\in\Bbb R:x\in [0,a/n]\mod 1/n\}=\bigcup_{z\in\Bbb Z}[z/n,(z+a)/n]$$
Let $I_n=\int_{A_n}F(dx)$. I am interested to know the behavior of $I_n$ in the limit $n\rightarrow\infty$.
Intuitively, $A_n$ samples a fraction $a$ of the real line increasingly regularly. I guess that
$$\frac{\int_{[0,a/n]}F(dx)}{\int_{[0,1/n]}F(dx)}\rightarrow a$$
as $n\rightarrow\infty$ assuming $F$ is strictly increasing in a neighbourhood around $0$. Based on this, I also expect that $I_n\rightarrow a$ as $n\rightarrow\infty$.
Is my intuition correct, and how should I go about computing the limit?
 A: Let us denote by $\mu$ the measure determined by the distribution function $F$. I claim that your property holds if $\mu$ is absolutely continuous, but that your property may fail otherwise.
Today I will post the proof for the absolutely continuous case, and tomorrow I will post a counterexample for the general case.
Let $\mu = f \, dx$. I first claim that
$$
\Gamma_x := \sup_{M \text{ measurable}} |\mu(M) - \mu (M + x)| \to 0 \text{ as } x \to 0\,.
$$
This is the crucial property that does not hold in general if $\mu$ is not absolutely continuous.
To see the above property, note
\begin{align*}
|\mu(M) - \mu(M + x)|
&= |\int 1_M (y) f(y) dy - \int 1^{M+x} (z) f(z) dz| \\
&= |\int 1_M (y) (f (y) - f(y+x)) dy| \\
&\leq \int |f(y) - f(y+x)| dy ,
\end{align*}
where it is well-known that the expression at the end vanishes as $x\to 0$.
It suffices to prove the claim for rational $a$, since for general $a$, we can always find $b,c$ rational with $b < a < c$ arbitrarily close to $a$, and if we denote the corresponding sets $A_n$ by $A_n(a)$, etc, then $A_n (b) \subset A_n (a) \subset A_n (c)$, so that
$$
b = \lim_n \mu (A_n (b)) \leq \liminf_n \mu (A_n (a)) \leq\limsup_n \mu (A_n(a)) \leq \lim_n \mu(A_n (c)) = c,
$$
from which we easily get the general claim.
Let us suppose for the moment that $a = 1/ N$. We then have $\Bbb{R} = \bigcup_{\ell=0}^{N-1} A_n + \ell/(nN)$, where the union is disjoint up to sets of $\mu$ measure zero. Therefore, $1 = \sum_{\ell = 0}^{N-1} \mu (A_n + \ell/(nN)$. Hence,
$$
|1 - N \mu (A_n)|
= |\sum_{\ell=0}^{N-1} \mu(A_n + \ell / (nN)) - \mu(A_n)|
\leq \sum_{\ell=0}^{N-1} \Gamma_{\ell / (nN)}
\to 0,
$$ 
as $n\to\infty$.
This proves the claim for $\alpha = 1/N$. For the general case $\alpha = k /N$, note $A_n (k/N) = \bigcup_{\ell=0}^{k-1} A_n (1/N) + \ell/(nN)$, where the union is again disjoint up to sets of $\mu$ measure $0$. Therefore,
$$
|k/N - \mu(A_n (k/N))|
\leq |k/N - \sum_{\ell=0}^{k-1} \mu(A_n (1/N))| + \sum_{\ell=0}^{k-1} |\mu(A_n (1/N)) - \mu (A_n (1/N) + \ell/(nN))|.
$$
By the properties from above, we see that the right hand side vanishes as $n\to\infty$. This proves the claim for rational $\alpha$, and thus for all $\alpha$, as seen above.

Here is the counterexample if one does not assume the measure to be absolutely continuous. Let $\omega = (\omega_m)_{m \in \mathbb N}$ be a sequence of iid random variables that are uniformly distributed in $\{0,1\}$. It is not hard to see that the map $\Phi : \{0,1\}^\Bbb{N} \to [0,1], (\epsilon_m)_m \mapsto \sum_{m=1}^\infty 2 \epsilon_m / 3^m$ is measurable (w.r.t. the product sigma algebra on the domain), well-defined, and injective(!).

Let $\mu$ be the measure associated to the random variable $\Phi (\omega)$. I am pretty sure that this is the measure associated to the Cantor function, but we will not actually need this. Note that $\mu$ is a probability measure, and that $\mu (\{x\}) = 0$ for all $x \in \Bbb{R}$; this uses the injectivity of $\Phi$, and the fact that every single realization of $\omega$ has probability zero. Hence, $\mu$ satisfies your assumption of having a continuous distribution function.
Let us choose $a = 3$. We will show that $\mu(A_{3^n}) \geq 1/2$ for all $n \in \Bbb{N}$, so that we cannot have $\mu(A_n) \to a = 1/3$.
To see this, note
$$
A_{3^n}
= \bigcup_{\ell \in \Bbb{Z}} 3^{-n} \cdot [\ell, \ell + 1/3]
= 3^{-(n+1)} \cdot \bigcup_{\ell \in \Bbb{Z}} [3\ell, 3\ell + 1] .
$$
Hence, up to a countable set of $x$ (the boundaries of the interval) (which is hence a $\mu$ null-set), the property $x \in A_{3^n}$ is equivalent to $\lfloor 3^{n+1} x \rfloor \in 3 \Bbb{Z}$.
But for any sequence $\epsilon = (\epsilon_m)_m$ which is not eventually constant (this only excludes a null-set), we have
$$
3^{n+1} \Phi(\epsilon)
= 2 \cdot \sum_{\ell=1}^{n+1} \epsilon_\ell 3^{n+1-\ell} + \sum_{\ell=n+2}^\infty 2 \epsilon_\ell (1/3)^{\ell - (n+1)},
$$
where the first sum is an integer, while the second series is an element of $[0,1)$. Thus,
$$
\lfloor 3^{n+1} \Phi(\epsilon) \rfloor
= 2 \cdot \sum_{\ell=1}^{n+1} \epsilon_\ell 3^{n+1-\ell} \, .
$$
It is not hard to see that this is divisible by $3$ if and only if $\epsilon_{n+1} = 0$.
Overall, we see (after excluding a null set) that $\Phi(\omega) \in A_{3^n}$ holds if and only if $\omega_{n+1} = 0$. This has probability $1/2$, so that $\mu(A_{3^n}) = 1/2$, as claimed.
