Corollary 2

a) The group $\mathbf{R}$ is in duality with itself relative to the map $(x, y)\mapsto$ exp(2$i\pi xy$), and the dual measure of Lebesgue measure is Lebesgue measure ;

b) The groups $\mathbf{Z}$ and $\mathbf{T}$ are in duality with respect to the mapping obtained by passing to the quotient from the mapping of $\mathbf{Z}\times \mathbf{R}$ into $\mathbf{U}$ such that $(n, x)\mapsto$ exp(2$i\pi nx$). The dual Haar measure of the counting measure on $\mathbf{Z}$ is the normalised Haar measure on $\mathbf{R}/\mathbf{Z}$.

The group $\mathbf{R}$ is in duality with itself with respect to the mapping $(x, y)\mapsto$ exp(2$i\pi xy$), by Proposition 19. Let us identify $\widehat{\mathbf{R}}$ with $\mathbf{R}$. The orthogonal of $\mathbf{Z}$ in $\widehat{\mathbf{R}}=\mathbf{R}$ is then $\mathbf{Z}$, and b) results from Theorem 4 of II, p. 226.

Let $\alpha$ be the counting measure on $\mathbf{Z}$ and $\gamma$ the normalised Haar measure on $\mathbf{T}$. If $\beta$ denotes Lebesgue measure on $\mathbf{R}$, one has $\gamma =\beta /\alpha$, since these two Haar measures on $\mathbf{R}/\mathbf{Z}$ have mass 1. The Haar measure $\widehat{\alpha}$ on $\widehat{\mathbf{Z}}=\mathbf{T}$ is the normalised Haar measure (Proposition 18), and the Haar measure $\widehat{\gamma}$ is the counting measure on $\mathbf{Z}($loc. cit.). By Proposition 16 of II, p. 231, the dual measure of $\beta$ is therefore the measure $\beta$.