Let $F$ be a finitely generated free group, $H\_1, \dots, H\_n$ finitely generated subgroups of infinite index in $F$, and $\epsilon > 0$. Must there be an epimorphism to a finite group $\phi \colon F \to G$ such that $$\frac{|\phi(H\_1 \cdots H\_n)|}{|G|} < \epsilon ?$$ Where $H\_1 \cdots H\_n = \{h\_1 \cdots h\_n : h\_1 \in H\_1, \dots, h\_n \in H\_n\}$.


Equivalently: Is the Haar measure of the closure of $H\_1 \cdots H\_n$ in the profinite completion of $F$ equals $0$ ?


This question asks for a generalization of [Is a free group a product of f.g subgroups of infinite index?](https://mathoverflow.net/questions/190985/is-a-free-group-a-product-of-f-g-subgroups-of-infinite-index)



Measuring products of finitely generated subgroups of free groups
The answer is yes. 


Let's prove this by induction on $n$. If $n=1$, then by M. Hall's theorem there is a finite index subgroup $K \leqslant F$ such that $H\_1 \subseteq K$ and $|F:K|>1/\epsilon$. Choose any finite index normal subgroup $N \lhd F$, which is contained in $K$, let $G=F/N$ and let $\phi:F \to G$ be the natural epimorphism. Note that $|G:\phi(K)|=|F:K|$ as $N\subseteq K$. Then
$$\frac{|\phi(H\_1)|}{|G|} \le \frac{|\phi(K)|}{|G|}=\frac{1}{|G:\phi(K)|}= 
\frac{1}{|F:K|}<\epsilon .$$


Now, suppose that $n\ge 2$ and the statement has been proved for smaller $n$. By Theorem 1.1 in <http://arxiv.org/abs/1308.3192>, there is a finitely generated subgroup $H$ of infinite index in $F$, such that $H\_{n-1} \subseteq H$ and $|H\_n: (H\_n \cap H)|=k<\infty$.


Thus $H\_n \subseteq \bigcup\_{j=1}^k H f\_j$ for some $f\_1,\dots, f\_k \in F$. By the induction hypothesis, we can find an epimorphism $\phi$ from $F$ onto a finite group $G$ such that $$\frac{|\phi(H\_1\dots H\_{n-2}H)|}{|G|}< \frac\epsilon k.$$ 
Since $H\_1 \dots H\_n \subseteq \bigcup\_{j=1}^k H\_1 \dots H\_{n-2}Hf\_j$, we have
\begin{multline\*}\frac{|\phi(H\_1\dots H\_{n-2}H\_{n-1}H\_n)|}{|G|}\le \sum\_{j=1}^k \frac{|\phi(H\_1\dots H\_{n-2}H)\phi(f\_j)|}{|G|} \\ =\sum\_{j=1}^k \frac{|\phi(H\_1\dots H\_{n-2}H)|}{|G|}<k \, \frac\epsilon k =\epsilon,\end{multline\*} as required.


