Convergence almost everywhere of characteristic functions

Question

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 $\mu$ ?

Answer 1

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.

Answer 2

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.