Suppose, $G$ is a finitely generated group. Suppose $A_1$ is a finite symmetric generating set. (That means $A_1 \subset G$, $|A_1|$ is finite, $\langle A_1 \rangle = G$, $e \in A_1$, $\forall a \in A_1 (a^{-1} \in A_1)$.) Suppose the sequence $\{A_n\}_{n=1}^{\infty}$ of subsets of $G$ is defined by recurrent relation: $\forall n \in \mathbb{N} (A_{n+1} = A_nA_1)$, where $A_nA_1$ denotes subset product. Suppose $H$ is a subgroup of $G$ of finite index. Is it always true, that $\lim_{n \to \infty} \frac{|A_n \cap H|}{|A_n|} = \frac{1}{[G:H]}$?
For finite groups the answer is obvious. However, I do not know how to deal with this problem in the case, when $G$ is infinite.
Any help will be appreciated.
The answer to your question is no, in general.
Suppose that $G$ is the free group on $2$ generators, and $A_1 = \{e,x,x^{-1},y,y^{-1}\}$. Then $A_n$ --- which is taken to be a set without multiplicity --- is just the set of words of reduced words of length $n$.
Now let $H = \langle x^2,xy,y^2\rangle$. $H$ is the kernel of the map from $G$ to $(\mathbf{Z}/2\mathbf{Z})^2$. A reduced word $w \in G$ lies in $H$ if an d only if the number of $x$s and $y$s in $w$ is congruent modulo $2$ to the number of $x ^{-1}$s and $y^{-1}$s respectively. In particular, any $w \in H$ has even length. It follows that
$$H \cap A_{2n+1} = H \cap A_{2n}.$$
This will prevent $a_n = \displaystyle{\frac{|H \cap A_n|}{|A_n|}}$ from having any limit at all, let alone the limit $1/[G:H] = 1/4$. Namely, consider the ratio
$$\frac{a_{2n+1}}{a_{2n}} = \frac{|H \cap A_{2n+1}|}{|H \cap A_{2n}|} \cdot \frac{|A_{2n}|}{|A_{2n+1}|} = \frac{|A_{2n}|}{|A_{2n+1}|}.$$
If the $a_n$ tended to a non-zero limit, this ratio must tend to $1$. But $|A_n|$ is increasing exponentially. Explicitly, for $n \ge 1$, $|A_n| = 1 + 2 (3^n - 1)$, so the ratio on the RHS tends to $1/3$.
As an alternative, one could let $B_1$ denote $A_1$, and then let $B_n$ denote the set of elements of $G$ counted with multiplicity obtained after $n$ iterations. This is an easier counting problem, because we simply want to count the number of elements in each coset of $G/H$, and we don't have to account for multiplicities. In this version of the problem, we can view the process as a markov process on the coset space $G/H$. Equivalently, there is a natural graph on $G/H$ where we adjoin $xH$ and $yH$ if there is a $g \in B_1 = A_1$ such that $gxH = yH$. We are then starting at the trivial coset and counting the number of paths of length $n$ which return to $H$.
The set $B_1$ determes the transition matrix $M$. The fact that $G$ acts transitively on $G/H$ ensures that the corresponding graph is connected, and hence $1$ is the largest eigenvalue of $M$ with multiplicity one. The fact that $e \in B_1$ also ensures that it is not bipartide, and hence $-1$ is not an eigenvalue. It then follows that there is exponential convergence to the uniform distribution, that is, if $|B_1| = k$ so $|B_n| = k^n$, then
$$\frac{|B_n \cap gH|}{|B_n|} = \frac{1}{[G:H]} + O\left(\frac{1}{\lambda^n}\right)$$
for some $\lambda > 1$ depending only on $G$ and $H$ and $B_1$.