Pettis lemma

Remarks

Since $A$ has the Baire property and is not meagre, there is a nonempty open set $U$ such that $A$ differs from $U$ by a meagre set. Since $G$ is not meagre, the Banach category theorem and homogeneity show that no nonempty open subset of $G$ is meagre. For $g$ near $e$, the open set $U \cap gU$ is nonempty, so it is not meagre. The set $A \cap gA$ differs from it by a meagre set, so $A \cap gA$ is not empty and $g \in A A^{-1}$. Applying this to $A^{-1}$, which also has the Baire property and is not meagre, gives the statement for $A^{-1}A$. It is the category analogue of the Steinhaus theorem on sets of positive measure. A standard application: a homomorphism between Polish groups that is Baire measurable is continuous. The hypothesis that $A$ is not meagre is needed. In $\R$ the set $\Z$ is closed, so it has the Baire property, but it is meagre, and $\Z - \Z = \Z$ is not a neighbourhood of $0$.

Going deeper

For a student

The lemma is the main tool of automatic continuity results in descriptive set theory. To show a homomorphism $f$ is continuous at $e$, take a neighbourhood $V$ of the identity in the target and a smaller symmetric $W$ with $W^{-1} W \subseteq V$. If $f^{-1}(W)$ has the Baire property and is not meagre, the lemma shows that $f^{-1}(V)$ contains a neighbourhood of $e$.