# Is this a measure?

Let $$\mu: \mathscr B \left(\mathbb R^d\right) \to \mathbb R$$ be defined by

$$\mu(A) = \left\{\begin{array}{ll} \|\mathscr F(\chi_A)\|_{L^2(\mathbb R^d)}^2 & \chi_A\in L^1(\mathbb R^d) \\ \infty & \, \chi_A\not\in L^1(\mathbb R^d)\end{array}\right.$$

where $$\mathscr F(\chi_A)$$ denotes the Fourier transform of $$\chi_A$$. Is $$\mu$$ a measure?

I know $$\mu(A)\geq 0$$ and $$\mu(\emptyset)=0$$. So I have to show that

$$\mu \left( \bigcup_{n \in \mathbb N} A_n \right) = \sum_{n\in\mathbb N} \mu(A_n)$$

Plancherel tells us that $$\mu$$ is Lebesgue measure. (Or a constant multiple, depending on where we put the $$2\pi$$s in the definitions).