Levy theorem for signed measures Levy theorem states that if $(\mu_n)_n$ is a sequence of probability measures on $\mathbb{R}^d$ such that the sequence of Fourier transforms $(\mathcal{F}(\mu_n))_n$ converges pointwise to a function $f\colon\mathbb{R}^d\to\mathbb{R}$ that is continuous at $0$, then there exists a unique probability $\mu$ with $\mathcal{F}(\mu)=f$, and $(\mu_n)$ tends weakly to $\mu$ in $\mathbb{R}^d$.
The question: Is there a version for signed bounded measures  on $\mathbb{R}^d$.
Thanks in advance.
 A: Thanks to Yemon, I see that my initial counterexample that I posted here was nonsense.
Here is a correct counterexample for the proposed statement with no additional assumptions.
For any $n\ge 2$, let $\mu_n=\delta_{n+1/n}-\delta_n$   (two atoms, one with positive mass and one with negative mass placed at points $n+1/n$ and $n$ respectively). 
Then the Fourier transform is $F(t)=e^{it(n+1/n)}-e^{itn}=e^{itn}(e^{it/n}-1)\to 0$. 
Zero is the Fourier transform of the zero measure, but $\mu_n$ does not converge to zero measure weakly. To see that, take a continuous bounded function $f$ such that $f(n)=-1$ and $f(n+1/n)=1$ for all $n\ge 2$. Then, for all $n\ge 2$, we have $\int fd\mu_n=2$ and $2$ does not converge to $0$, the integral of $f$ w.r.t. zero measure.
A: In fact, I showed that if we suppose that the Fourier transform of both $\mu_n$ and its total variation measure $|\mu_n|$ converge simply , then the answer is affirmative, for we can in this case use the Jordan decomposition decomposition of $\mu_n$ and use the classical Levy theorem.
A: There is a version for complex measures. See Theorem 2.1,2.2 in ''Baez-Duarte, L. (1993). Central limit theorem for complex measures. Journal of Theoretical Probability, 6(1), 33-56.''
