Let $(\Phi_n)_n$ be the characteristic functions of probability measures $(\mu_n)_n$ and let $\Phi$ be the characteristic function of a probability measure $\mu$.

Do you know an example where $\Phi_n(t)\rightarrow\Phi(t)$ for all $t\in A$ where $A$ is dense in $\mathbf{R}^d$ but $(\mu_n)_n$ does not converge weakly to $\mu$ ?

  • 1
    $\begingroup$ By measure you mean probability measure or something more general? $\endgroup$ Mar 23, 2022 at 10:55
  • $\begingroup$ Yes, $\mu_n$ and $\mu$ are probability measures. $\endgroup$
    – Tiblodocus
    Mar 23, 2022 at 11:57

2 Answers 2


There is no such example. Let $(f_n)$ be a sequence of characteristic functions of probability measures $\mu_n$ which converges a. e. to a characteristic function $f$ of a probability measure $\mu$. You can always choose a subsequence such that $\mu_n\to\mu$ weakly to some measure on the real line (first theorem of Helly). Then for this subsequence $f_n(x)\to f(x)$ for all $x$, where $f$ is the Fourier transform of $\mu$. Your assumption that $f$ is a Fourier transform of a probability measure implies that $\mu$ must be that probability measure. Since this works for every subsequence we have convergence of $f_n$ to $f$ everywhere.

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    $\begingroup$ We need to assume that $(\mu_n)_n$ is tight to get this weak convergence. $\endgroup$
    – Tiblodocus
    Mar 23, 2022 at 13:58
  • $\begingroup$ No, one does not. If the limit measure is not a probability measure, then $f_n$ converges to its Fourier transform, which is not a characteristic function of a probability measure, but this contradicts your assumption. $\endgroup$ Mar 24, 2022 at 13:32
  • $\begingroup$ Ok, thanks. I did not know this general version. $\endgroup$
    – Tiblodocus
    Mar 24, 2022 at 14:38
  • $\begingroup$ This is a great answer. However I think you are using $\mu$ and $f$ to denote two different objects, that are only proved to be the same at the very end, which is a bit confusing. (What I understand is that you start with a given limit for the c.f., then construct the c.f. of a limit, and then you prove they must be the same.) $\endgroup$
    – Pierre PC
    Mar 24, 2022 at 16:59
  • $\begingroup$ @AlexandreEremenko In the OP, the convergence holds on a dense set (not necessarily a.e.). What happens then? $\endgroup$
    – Gagar
    Oct 15, 2022 at 14:17

If $\Phi_n\to\Phi$ pointwise on a dense subset, then by the Stone-Weierstrass theorem $$\lim_{n\to\infty}\int_B g\ d\mu_n = \int_B g\ d\mu $$ on every ball centered at $0$ and every $g\in C_0(\mathbb{R}^d)$. This implies that $\mu_n\to\mu$ in the weak$^{*}$ topology of the Banach space of bounded Borel measures $M(\mathbb{R}^d)$. However, we may not have that $\mu_n\to\mu$ in the weak topology of $M(\mathbb{R}^d)$. Let's be reminded that $\mu_n\to\mu$ weakly iff $\mu_n(E)\to\mu(E)$ for every Borel set $E\subseteq\mathbb{R}^d$.

For an example, let $\delta_a$ be the unit mass at $a\in\mathbb{R}^d$, i.e., $\delta_a(E) = 1$ if $a\in E$ and $\delta_a(E) = 0$ if $a\notin E$. Let $(a_n)$ be a sequence with nonzero terms such that $a_n\to 0$. Let $\Phi_n$ be the charcteristic function of $\delta_{a_n}$ and $\Phi_0\equiv 1$ denote the characteristic function of $\delta_0$. Clearly, $$\Phi_n(\gamma) = e^{ia_n\gamma} \hspace{6mm}\forall\gamma\in\mathbb{R}^d$$ so $\Phi_n\to 1$ pointwise on $\mathbb{R}^n$. On the other hand, let $E$ be a set such that $0\notin E$ and $\{a_n:n\in\mathbb{N}\}\subseteq E$. Then, $1=\delta_{a_n}(E)\not\to \delta_0(E)=0$, so $(\delta_{a_n})$ does not converge to $\delta_0$ weakly.

  • $\begingroup$ In your example $\delta_{a_n}\longrightarrow \delta_0$ weakly (by the continuity theorem, since $\Phi_n\longrightarrow \Phi_0$ pointwise). Your reminder is incorrect: $\mu_n\longrightarrow \mu$ weakly iff $\mu_n(E)\longrightarrow \mu(E)$ for all $\mu$-continuity sets $E$. And here $E$ is clearly not a continuity set of $\delta_0$ since $0\in \partial E$. $\endgroup$
    – esg
    Mar 24, 2022 at 20:40
  • $\begingroup$ @esg After reading your comment and quick googling (e.g. Portmentau's theorem ocw.mit.edu/courses/sloan-school-of-management/…) I realize that your description is indeed the same as the weak-star convergence or measures. "Weak convergence" of measures has just a different meaning in the Banach space language. Given the context, one must stick with the definition you've given. $\endgroup$
    – Onur Oktay
    Mar 25, 2022 at 6:11

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