Uniqueness of Haar Measures

A friend of mine proposed the following definition. For a given locally compact group $G$ let's consider some Haar measure $\mu$ and the space $L^1(\mu)$ of absolutely integrable functions. This space actually does not depend on the specific choice of the Haar measure as $L^1(\lambda\cdot \mu) = L^1(\mu)$ for any $\lambda > 0$. We can then consider the subspace $$L^1_0(\mu) = \left\{ f \in L^1(\mu): \int_G f \cdot d\mu = 0 \right\}$$ which is again independent from the specific choice of $\mu$. Let's now define the "space of volumes for $G$" $$\mathrm{Vol}(G) = L^1 (\mu) / L^1_0(\mu)$$ It is a one-dimensional real vector space. For any measurable set $K$ with finite measure we can assign it a "volume" in $\mathrm{Vol}(G)$ which is trivially the indicator function $\mathbf{1}_K$ modulo any null-integral function. This construction works even for non abelian locally compact groups. For two of these we have $$\mathrm{Vol}(G \times H) = \mathrm{Vol}(G) \otimes \mathrm{Vol}(H)$$ (where equality stands for canonical isomorphism). For some LCA group we also have $$\mathrm{Vol}\left(\hat G\right) = \mathrm{Vol}(G)^*$$ Coupled with the fact these spaces are 1-dimensional, this garantees that $$\mathrm{Vol}\left(G\right) \otimes \mathrm{Vol}\left(\hat G\right) = \mathbb{R}$$ Any Fourier transform can therefore be appropriately typed $\mathcal{F} f \in L^2\left(\hat G, \mathbb{C} \otimes \mathrm{Vol}(G)\right)$ without any non-canonical choice.