= Continuity of a characteristic function
Every characteristic function is uniformly continuous and satisfies $\varphi(0)=1$ and $|\varphi(t)|\leq1$. Continuity follows from the <dominated convergence theorem> applied to $e^{itX}$.
More explicitly, if $t_n\to t$, then $e^{it_nX}\to e^{itX}$ pointwise and $|e^{it_nX}|=1$. Apply the <dominated convergence theorem> to the <real part> and <imaginary part> separately. There is no finite <moment> hypothesis. For <uniform continuity>, the stronger estimate
$$
\sup_{t\in\mathbb R}|\varphi_X(t+h)-\varphi_X(t)|
\leq\mathbb E[|e^{ihX}-1|]\longrightarrow0
\qquad(h\to0)
$$
follows by factoring out $e^{itX}$, whose absolute value is one. The integrand tends pointwise to zero and is bounded by two, so the <dominated convergence theorem> applies again. The bound is independent of $t$, which supplies <uniform continuity> on the entire real line.<Probability: Theory and Examples (fifth edition) by Rick Durrett>[Theorem 3.3.1(d) and proof, printed p. 126]{ref}
If the first <absolute moment> is finite, $|e^{iu}-1|\leq|u|$ additionally gives $|\varphi_X(t+h)-\varphi_X(t)|\leq|h|\mathbb E[|X|]$, a <Lipschitz continuity> estimate. The earlier argument matters because <continuity> survives even when this <expected value> is infinite. For <convergence in distribution>, the separate <Lévy continuity theorem> concerns limits of whole sequences of <characteristic functions>; the argument here establishes <continuity> of one <characteristic function>.
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