Let $F$ be a finitely generated free group and let $S \subseteq F$ be a subset for which there is some $\epsilon > 0$ such that for any epimorphism to a finite group $\phi \colon F \to G$ we have that $\frac{|\phi(S)|}{|G|} \geq \epsilon$ (that is, the closure of $S$ in the profinite completion of $F$ has positive Haar measure). 


*Is it possible that the closure of $S$ with respect to the profinite topology on $F$ does not contain a coset of any finite index subgroup of $F$ ?*


This is the same as asking whether it is possible that $S$ is nowhere dense (the interior of the closure is empty) with respect to the profinite topology.



Can a positive measure subset of a free group be nowhere dense?
Yes, $S$ can be nowhere dense. The idea is to build a 'fat Cantor set' inside $F$.


Let $H\_0=F,H\_1,H\_2,\dots$ be finite-index subgroups of $F$ such that $H\_{i+1}\leq H\_i$ for each $i$ and such that $\bigcap\_{i=0}^\infty H\_i = \{1\}$. Start with $i\_0=0$ and $C\_0=\{H\_0\}$, and then iteratively do the following. Having defined $C\_n$ as a certain set of more than half the cosets of $H\_{i\_n}$, let $S\_n\subset F$ consist of one member from each coset $gH\_{i\_n}\in C\_n$, and then choose $i\_{n+1}$ so large that
$$\frac{|C\_n|}{|F/H\_{i\_n}|} - \frac{|C\_n|}{|F/H\_{i\_{n+1}}|} > 1/2,$$
and so that when we decompose each $gH\_{i\_n} \in C\_n$ into a union of cosets of $H\_{i\_{n+1}}$ we can remove at least one of these cosets without uncovering any member of $S\_0\cup \dots \cup S\_n$. Let $C\_{n+1}$ be the resulting set of cosets of $H\_{i\_{n+1}}$.


In the end let $S=\bigcap\_{n=0}^\infty \bigcup C\_n$, and let $\bar{S}$ be the closure of $S$ in $\hat{F} = \lim F/H\_i$. Then $S$ is closed in the profinite topology of $F$ and contains no coset of any $H\_i$, so $\bar{S}$ is nowhere dense. On the other hand $S$ contains $\bigcup\_{i=0}^\infty S\_i$, so the image of $S$ in $F/H\_i$ always has size at least $|F/H\_i|/2$, so $\bar{S}$ has measure at least $1/2$.


