$n$-sample size.
For $i \in \{1,\dots,n\}$, $x_i(n)$ is a single draw from a distribution $f(x)$ on some bounded set. Associated with each $x_i(n)$ is a value $a_i(n)$, where $a_i(n)$ are such that $\sum_{i=1}^n a_i(n) = 1$, and furthermore: \begin{align*} \max_{i \in \{1,\dots,n\}} a_i \to 0 \quad as \quad n\to \infty \end{align*} The value $x_i(n)$ is drawn independently from the associated $a_i(n)$ value. I want to show that: \begin{align*} \lim_{n \to \infty} \sum_{i: \, z\leq x_i \leq y} a_i(n) = F(y)-F(z) \end{align*} Where $F(x)$ is the c.d.f corresponding to $f(x)$.
My intuition is that, since the draws of $x_i$ are independent from the associated $a_i$ values, and since each $a_i$ value is shrinking towards $0$, the above result should hold.
So it's something like this.
Consider a random variable
$S = \sum_{i=1}^m a_i(n)\mathbb{1}_{z \leq x_i(n) < y}$.
You should see for yourself that the expectation of this is unbiasedly $F(y) -F(z)$
$(E[S] = F(y) - F(z))$.
Now think about how to show the convergence in probability: Chebyshev's inequality comes to mind.
$\mathbb{P}(|S-E[S]| > t) \leq \frac{Var(S)}{t^2} = \frac{Var(I_{x_i \in [z,y]})\sum_{i=1}^n a_i^2}{t^2} \leq \frac{n*\text{max}(a_i^2)Var(I_{x_i\in[z,y]})}{t^2}$. If the max of $a_i \rightarrow 0$ and they all add up to $1$, then for $N$ large enough, $a_i <\frac{1}n$ so $a_i^2 \leq \frac{1}{n^2}$, concluding it.