Here is an exercise from Kai Lai Chung's book A Course in Probability Theory.
Let $F$ be a distribution function and $f(t)=\int_{-\infty}^{\infty}e^{itx}\,dF(x)$ be the characteristic function of $F$. Show that if $F$ is purely discontinuous, then $\limsup\limits_{t\to\infty}f(t)=1$.
Since $F$ is purely discontinuous, there are $\{a_j\}_1^\infty$ and $\{b_j\}_1^\infty$ with $b_j\geq 0$ and $\sum_{j=1}^\infty b_j=1$ such that $F(x)=\sum_{j=1}^\infty b_j\delta_{a_j}(x)$. So $$f(t)=\sum_{j=1}^\infty b_je^{ita_j}, t\in\mathbb{R}.$$ Given $\epsilon>0$, we can find $n\in\mathbb N$ such that $\sum_{j>n}b_j<\epsilon$, and so $\left|\sum_{j>n}b_je^{ita_j}\right|<\epsilon$. It suffices to show that $$\limsup_{t\to\infty}\sum_{j=1}^nb_je^{ita_j}\geq1-\epsilon,$$ but I stuck here.
Any help would be appreciated.
Let $\mu = \sum_{k=1}^{\infty} p_k \delta_{x_k}$ be a discrete probability measure with characteristic function
$$\phi(\xi) = \sum_{k=1}^{\infty} p_k e^{i x_k \xi}.$$
Without loss of generality, $x_k \neq x_j$ for $k \neq j$. Since $|\phi(\xi)| \leq 1$, it is immediate that
$$\limsup_{|\xi| \to \infty} |\phi(\xi)| \leq 1.$$
In order to prove "$\geq$", we show that
$$\limsup_{|\xi| \to \infty} \text{Re} \, \phi(\xi)\geq1. \tag{1}$$
To this end, fix $\epsilon>0$ and choose $N \in \mathbb{N}$ such that $\sum_{k \geq N+1} p_k \leq \epsilon$. Then, $$\sum_{k=1}^N p_k \geq 1-\epsilon. \tag{2} $$ Take some $k \in \{1,\ldots,N\}$ such that $x_k \neq 0$. Clearly, $\xi \mapsto f_k(\xi) := \cos(\xi x_k)$ is periodic with period $\ell_k := 2\pi/|x_k|$. Since $f_k$ is continuous, we can choose $M>0$ such that
$$|\xi| \leq \frac{\ell_k}{M} \implies |1-\cos(\xi x_k)| \leq \epsilon. \tag{3}$$
Because of the periodicity of $f_k$, this implies
$$|1-\cos((\xi-\eta)x_k)| \leq \epsilon \tag{4}$$
for all $$\xi, \eta \in A_k^m := \bigcup_{n \in \mathbb{Z}} \bigg[ n \ell_k + \frac{m-1}{M} \ell_k, n \ell_k + \frac{m}{M} \ell_k \bigg) , \qquad m=1,\ldots,M.$$ Note that the estimate $(4)$ is trivial if $x_k=0$. Since we consider only finitely many $k \in \{1,\ldots,N\}$, we can choose $M$ independently of $k \in \{1,\ldots,N\}$. As $$\bigcup_{m=1}^M A_k^m = \mathbb{R},$$ it follows that the family $$\mathcal{A} := \left\{ \bigcap_{k=1}^N A_k^{m(k)}; m(k) \in \{1,\ldots,N\}\right\}$$ covers $\mathbb{R}$ and, moreover, by construction,
$$|1-\cos((\xi-\eta)x_k)| \leq \epsilon, \qquad \xi, \eta \in A,$$
for any $A \in \mathcal{A}$. Since there are only finitely many sets in $\mathcal{A}$, there exists at least one set $A \in \mathcal{A}$ which is unbounded. Then
\begin{align*} \sum_{k=1}^{\infty} p_k \cos((\xi-\eta)x_k) &\geq \sum_{k=1}^N p_k\underbrace{\cos((\xi-\eta)x_k)}_{\geq 1-\epsilon} - \sum_{k=N+1}^{\infty} p_k \\ &\geq (1-\epsilon)^2 - \epsilon. \end{align*}
for all $\xi,\eta \in A$. Since $A$ is unbounded, we can choose $(\xi_k)_{k \geq 1} \subseteq A$ with $|\xi_k| \to \infty$. Using the above estimate for $\xi \hat{=} \xi_k$ and $\eta =\xi_1$, we get
$$\limsup_{k \to \infty} \text{Re} \, \phi(\xi_k-\xi_1) \geq (1-\epsilon)^2-\epsilon.$$
As $\epsilon>0$ is arbitrary, this proves $(1)$.