Measure-theoretic proof of the inverse Fourier transform I would like to prove this statement:
Proposition. (Inverse Fourier transform)  Let $u \in L^1_\mathbb{R}(\lambda^n)$ and let us define its Fourier transform as
$$\hat{u}(\xi):=(2\pi)^{-n}\int_{\mathbb{R}^n} u(x)e^{-i\langle x,\xi\rangle}dx,\,\xi \in \mathbb{R}^n$$
then if $\hat{u} \in L^1_\mathbb{C}(\lambda^n)$ we have the inversion formula
$$u(x)=\int_{\mathbb{R}^n} \hat{u}(\xi)e^{i\langle x,\xi\rangle}d\xi$$
To prove this I would like to use the following lemma:
Lemma. Let $\mu$ be a finite measure on $(\mathbb{R}^n,\mathcal{B}(\mathbb{R}^n))$ and assume
$$\hat{\mu}(\xi):=(2\pi)^{-n}\int_{\mathbb{R}^n}e^{-i\langle x,\xi\rangle}\mu(dx)\in L^1_\mathbb{C}(\lambda^n)$$
in this case, $\mu(dx)=u(x)dx$ where
$$u(x)=\int_{\mathbb{R}^n}\hat{\mu}(\xi)e^{i\langle x,\xi\rangle}d\xi$$

I thought about decomposing $u\in L^1_\mathbb{R}(\lambda^n)$ in positive and negative parts, yielding $u=u^+-u^-$. As $u^\pm\geq 0,\,u^\pm \in L^1_\mathbb{R}(\lambda^n)$, we can construct the finite measures $\mu^\pm(dx)=u^\pm(x)dx$. Assume $\widehat{\mu^\pm}\in L^1_\mathbb{C}(\lambda^n)$. Using the above lemma:
$$u(x)=\int_{\mathbb{R}^n} (\widehat{\mu^+}(\xi)-\widehat{\mu^-}(\xi))e^{i\langle x,\xi\rangle}d\xi$$
My idea is to show $\widehat{\mu^+}(\xi)-\widehat{\mu^-}(\xi)=\hat{u}(\xi)$:
$$\begin{aligned}\widehat{\mu^+}(\xi)-\widehat{\mu^-}(\xi)&=(2\pi)^{-n}\bigg(\int_{\mathbb{R}^n} e^{-i\langle x,\xi\rangle}\mu^+(dx)-\int_{\mathbb{R}^n} e^{-i\langle x,\xi\rangle}\mu^-(dx)\bigg)=\\
&=(2\pi)^{-n}\bigg(\int_{\mathbb{R}^n} e^{-i\langle x,\xi\rangle}u^+(x)dx-\int_{\mathbb{R}^n} e^{-i\langle x,\xi\rangle}u^-(x)dx\bigg)=\\
&=(2\pi)^{-n}\bigg(\int_{\mathbb{R}^n} e^{-i\langle x,\xi\rangle}(u^+(x)-u^-(x))dx\bigg)=\hat{u}(\xi)\end{aligned}$$
so assuming $\widehat{\mu^\pm}\in L^1_\mathbb{C}(\lambda^n)$ is equivalent to assuming $\hat{u}\in L^1_\mathbb{C}(\lambda^n)$. Is this approach correct, or did I miss something? Thank you very much for your help.
 A: I thought about this problem recently, i.e. Fourier inversion theorem for complex Borel measures. Your lemma can't even hold if $\mu$ is not absolutely continuous with respect to Lebesgue measure $m$. However, if $\mu << m$, then it does hold as a direct consequence of the Fourier inversion theorem:
Suppose $\mu$ is a complex measure on $B(\mathbb{R}^n)$. If $\mu << m$, then by Radon-Nikodym theorem, we can write $d\mu = f\,dm$ with $f \in L^1(\mathbb{R}^n)$, and clearly $\hat{\mu}(\xi) = \hat{f}(\xi)$. Therefore if $\hat{\mu} \in L^1(\mathbb{R}^n)$, then $d\mu = \mathcal{F}^{-1}(\hat{\mu})\,dm$, which is the conclusion of your lemma.
The problem is that a general measure $\mu$ can only be written as $d\mu = f \,dm + d\lambda$ where $\lambda \perp m$. So $\lambda$ gets in the way.
What does hold for measures however is that the Fourier transform $\mathcal{F} \colon M(\mathbb{R}^n) \to CB(\mathbb{R}^n)$ is an injective linear map, where $M(\mathbb{R}^n)$ denotes the space of complex Borel measures on $\mathbb{R}^n$, and $CB(\mathbb{R}^n)$ denotes the space of bounded continuous $f \colon \mathbb{R}^n \to \mathbb{C}$. This result is also a consequence of Fourier inversion theorem.
