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

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


1 Answer 1


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


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