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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.

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Is that really Levy with a y or Levi with an i? – Mariano Suárez-Alvarez May 8 '11 at 22:25
@Mariano: It's Paul Pierre Lévy, with an accent on the e and with a final y. – Michael Hardy May 9 '11 at 1:42

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.

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If this is all you wanted, then this is a much simpler question than what you originally asked – Yemon Choi May 11 '11 at 21:45
This, at least, solves the problem in this case . – mostafa May 12 '11 at 5:48

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.

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$e^{it(n+1)}-e^{itn}$ does not converge to zero as $n\to\infty$ – Yemon Choi May 11 '11 at 7:23
There is a subtle issue here. One should be careful what is meant by weak convergence when leaving the context of probability measures. Here, the right notion is probably that $\mu_n \to \mu$ if $\int f d\mu_n \to \int f d\mu$ for all continuous compactly supported $f$. This is because the space of signed measures on $\mathbb{R}$ is the dual of $C_c(\mathbb{R})$, and so my definition is weak-* convergence in this space. Under this definition, $\mu_n$ does converge weakly to the zero measure. – Nate Eldredge May 11 '11 at 19:52
When dealing with probability measures $\mu_n, \mu$, one can show that if $\int f d\mu_n \to \int f d\mu$ for all $f \in C_c(\mathbb{R})$, then the same holds for all $f \in C_b(\mathbb{R})$ (i.e. bounded continuous functions), so our definitions are equivalent. For other measures (even positive measures) they are not. – Nate Eldredge May 11 '11 at 19:56
It is also worth mentioning that with your definition, $\delta_n$ does not converge weakly to the zero measure as $n \to \infty$, which is counterintuitive. Also, your definition doesn't make the closed unit ball of signed measures weakly compact. Essentially, your topology is the weak-* topology on $C_b(\mathbb{R})^*$, which is a nasty space to work in (most of its elements are not measures). – Nate Eldredge May 11 '11 at 20:01
Nate, the topology you are talking about is called "vague topology", and the convergence is called "vague convergence". My example surely exploits the difference between the "weak" and "vague" modes of convergence. – Yuri Bakhtin May 11 '11 at 20:44

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.''

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Thank you a lot. – mostafa Mar 24 '14 at 17:11

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