Let $(X_i)_{i \geq 1}$ be i.i.d. random variables.. The strong law of large numbers ensures that for each $t$, the empirical cdf $\hat{F_n}\left(t\right)=\frac{1}{N}\sum_{i=1}^{N}\mathbb{1}_{X_i\leq t}$ converges almost surely to $P\left(X \leq t\right)$.
Now, let $w_i$ be deterministic weights. Can we say anything about the convergence of $\sum_{i=1}^{N}\frac{w_i}{\sum_{i=1}^{N}w_i}\mathbb{1}_{X_i\leq t}$ ?
If we can, what happens when the weights are not deterministic anymore, and $w_i$ is a function of $X_i$ ?
Yes, you can simply apply the strong law of large numbers to both the numerator and denominator (assuming all the necessary assumptions are satisfied). More specifically, in the setting where $w_i \equiv w(X_i)$, and denoting $q$ the distribution according to which each $X_1,\ldots,X_N$ is distributed : $$\begin{align*} \hat F_N^w(t) := \frac{\sum_{i=1}^N w(X_i) \mathbf 1\{X_i\le t\}}{\sum_{i=1}^N w(X_i)} &= \frac{\frac{1}{N}\sum_{i=1}^N w(X_i) \mathbf 1\{X_i\le t\}}{\frac{1}{N}\sum_{i=1}^N w(X_i)}\\ &\stackrel{N\to\infty}{\longrightarrow} \frac{E_q[w(X_1)\mathbf 1\{X_1\le t\}]}{E_q[w(X_1)]} \tag1 \end{align*} $$
Now, unless $w(X_1)$ is independent of $\mathbf 1\{X_1\le t\}$ (which more or less means that $w$ is constant), it seems like there is not much we can say.
There is a very important setting in which we can however prove that $(1)$ is equal to $P(X_1\le t)$ : that is in the framework of Importance Sampling, in which this type of quantities is extremely common : the problem is to estimate $E[f(X)] := \int f(x) p(x) dx $ where we do not have access to $p$, the pdf of $X$ (or maybe we do but it's expensive/difficult to sample from $p$, or maybe we only know $p$ up to a hard-to-compute constant). In our case, the function $f$ we want to integrate is $f := \mathbf 1\{\cdot \le t\}$.
The idea is thus to let $q$ be a distribution which we can easily sample from, and define the weight function $w : x\mapsto p(x)/q(x)$. We then create an i.i.d. sample $X_1,\ldots,X_n\sim q$ and define the estimator $\hat F_N^w(t)$ as above, it can be shown that $\hat F^w_N(t)$ converges almost surely to $P(X_1\le t)$. To prove it, it is enough to show that $(1)$ is equal to $P(X_1\le t)$, so let's develop :
$$\begin{align*} \frac{E_q[w(X_1)\mathbf 1\{X_1\le t\}]}{E_q[w(X_1)]} &= \frac{\int w(x) \mathbf 1\{x\le t\}q(x)dx}{\int w(x) q(x) dx}\\ &= \frac{\int \frac{p(x) q(x)}{q(x)} \mathbf 1\{x\le t\}dx}{\int \frac{p(x) q(x)}{q(x)} dx}\\ &= \frac{\int p(x)\mathbf 1\{x\le t\}dx}{\int p(x) dx}\\ &=\frac{\int p(x)\mathbf 1\{x\le t\}dx}{1}\\ &= P(X\le t) \end{align*} $$
As desired.
Note that in the above computation, $X\sim p$ while $X_1\sim q$ ! If $q=p$ then we simply recover the unweighted estimator $\hat F_n$. If you want to read more about this, you should Google "normalized importance sampling", you will find many great references on the subject.