What does the limit of the characteristic function $\varphi(x)\xrightarrow{\lvert x\rvert\rightarrow \infty} 0$ tell us about $\mu(\{a\})$?

Question

If we know that $\varphi(x)\xrightarrow{\lvert x\rvert\rightarrow \infty} 0$ for the characteristic function $\varphi$ of the probability measure $\mu$ on $\mathbb{R}$, what does that tell us about $\mu(\{a\})$ for any $a$?

Because of the identity $$\mu(\{a\}) = \lim_{T\rightarrow\infty}\frac 1{2T}\int_{-T}^Te^{-ita}\varphi(t) dt$$ I think it should be $\mu(\{a\}) = 0$ for all $a$. But I don't know how to formalize this idea, if it's true.

Edit: A TA confirmed it is true that under the assumption $\mu(\{a\})=0$ for all $a$, but I still don't even know how to start a possible proof/what the key insight is.

Edit2: I have successfully shown that $$\lim_{T\rightarrow\infty}\frac 1{2T}\int_{-T}^T\varphi^2(t)dt = \sum_{x\in\mathbb{R}}(\mu(x))^2$$ But I need to show that the LHS is $0$. It is not trivial to show that the integral is even finite in the limit, how can I show that the LHS is finite?

Answer 1

This is almost the same answer as @Paresseux Nguyen's one. The main purpose of this answer is to convince the reader that Paresseux's answer is correct, by supplying some extra details. I will gladly delete my answer if Paresseux's answer is resurrected and needs to be credited.

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Since OP already knows that

$$ \mu(\{a\}) = \lim_{T\to\infty} \frac{1}{2T} \int_{-T}^{T} e^{-ita}\varphi(t) \, \mathrm{d}t \tag{1} $$

holds for any $a \in \mathbb{R}$, it suffices to show that the integral in the right-hand side converges to $0$ under the assumption

$$ \lim_{\left|x\right|\to\infty} \varphi(x) = 0. \tag{A}$$

Using this assumption, for each fixed $\epsilon > 0$, we can pick $R > 0$ such that $\left| \varphi(x) \right| < \epsilon$ whenever $\left|x\right| \geq R$. Then

\begin{align*} \left| \int_{-T}^{T} e^{-ita}\varphi(t) \, \mathrm{d}t \right| &\leq \int_{-T}^{T} \left| e^{-ita}\varphi(t) \right| \, \mathrm{d}t \\ &\leq \int\limits_{\left|x\right|\leq R} \left| \varphi(t) \right| \, \mathrm{d}t + \int\limits_{R<\left|x\right|\leq T} \epsilon \, \mathrm{d}t \\ &\leq 2R + 2\epsilon(T-R). \end{align*}

Dividing both sides by $2T$ and taking $\limsup$ as $T\to\infty$,

\begin{align*} \limsup_{T\to\infty} \left| \frac{1}{2T} \int_{-T}^{T} e^{-ita}\varphi(t) \, \mathrm{d}t \right| \leq \limsup_{T\to\infty} \frac{2R + 2\epsilon(T-R)}{2T} = \epsilon. \end{align*}

Since the limsup in the left-hand side is a fixed number which is independent of the choice of $\epsilon > 0$, letting $\epsilon \downarrow 0$ shows that this limsup is in fact $0$, which is equivalent to

$$ \lim_{T\to\infty} \frac{1}{2T} \int_{-T}^{T} e^{-ita}\varphi(t) \, \mathrm{d}t = 0. $$

So by $\text{(1)}$, we have $\mu(\{a\}) = 0$ for any $a \in \mathbb{R}$ as desired.

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Remark. A slight modification of this proof immediately yields the following well-known result:

Theorem. (Cesàro mean) Let $f: [0, \infty) \to \mathbb{C}$ be locally integrable and $f(t) \to \ell \in \mathbb{C}$ as $t\to\infty$. Then $$ \lim_{T\to\infty} \frac{1}{T} \int_{0}^{T} f(t) \, \mathrm{d}t = \ell. $$

Conversely, this theorem can be used to produce a much shorter answer.

Answer 2

If I understand this correctly, it tells us that mu([a}) is zero for every a. Because each a that has positive probability contributes a sinusiod to the characteristic function and the sum of these sinusiodal components cannot vanish in the limit. They will at worst be a so-called almost periodic function. Neither the absolutely continuous not the singular part of the distribution will counteract the effect of the discrete component. A theorem of Wiener is relevant, and a general reference is the book by Lukacs.