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

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.