# Convergence almost everywhere of characteristic functions

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 $$\mu$$ ?

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

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.

• We need to assume that $(\mu_n)_n$ is tight to get this weak convergence. Mar 23, 2022 at 13:58
• 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. Mar 24, 2022 at 13:32
• Ok, thanks. I did not know this general version. Mar 24, 2022 at 14:38
• 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.) Mar 24, 2022 at 16:59
• @AlexandreEremenko In the OP, the convergence holds on a dense set (not necessarily a.e.). What happens then? 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.

• 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 $0\in \partial E$.
– esg
Mar 24, 2022 at 20:40
• @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. Mar 25, 2022 at 6:11