Proposition 18

— Let G be a locally compact group, $\Gamma$ a set equipped with a multiplication $(u, v) \mapsto uv$ and a Hausdorff topology such that: 1) the topology of $\Gamma$ is invariant under the translations; 2) the restriction of the multiplication to every compact subset of $\Gamma \times \Gamma$ is continuous.

Let $f : G \to \Gamma$ be a mapping of G into $\Gamma$ such that $f(xy) = f(x)f(y)$ for $x, y$ in G, and measurable for a Haar measure $\beta$ on G. Then $f$ is continuous.

Set $g(x) = f(x^{-1})$ for $x \in G$. Since $f$ and $g$ are $\beta$-measurable, there exists a non $\beta$-negligible compact subset K of G such that the restrictions of $f$ and $g$ to K are continuous. The mapping $(x, y) \mapsto f(xy^{-1}) = f(x)g(y)$ of $K \times K$ into $\Gamma$ is continuous because the multiplication of $\Gamma$ is continuous on $f(K) \times g(K)$; now, this mapping may be written as $\varphi \circ \psi$, where $\psi$ is the mapping $(x, y) \mapsto xy^{-1}$ of $K \times K$ onto $K \cdot K^{-1}$, and $\varphi$ is the restriction of $f$ to $K \cdot K^{-1}$. Let $R$ be the equivalence relation defined on $K \times K$ by $\psi$. The mapping $\psi'$ of $(K \times K)/R$ onto $K \cdot K^{-1}$ deduced from $\psi$ by passage to the quotient is continuous, therefore $(K \times K)/R$ is Hausdorff and $\psi'$ is a homeomorphism. Since $\varphi \circ \psi$ is continuous, one sees that the restriction of $f$ to $K \cdot K^{-1}$ is continuous. Now, $K \cdot K^{-1}$ is a neighborhood of $e$ (Prop. 17), therefore $f$ is continuous at $e$. For every $x_0 \in G$, $f(x_0x) = f(x_0)f(x)$, thus $f$ is continuous at $x_0$ because the topology of $\Gamma$ is invariant under translations.