 Continuity of a characteristic function

Every characteristic function is uniformly continuous and satisfies $\phi (0)=1$ and $|\phi (t)|\le 1$. Continuity follows from the dominated convergence theorem applied to $e^{itX}$.

More explicitly, if $t_n\to t$, then $e^{i{t}_{n}X}\to e^{itX}$ pointwise and $|e^{i{t}_{n}X}|=1$. Apply the dominated convergence theorem to the real part and imaginary part separately. There is no finite moment hypothesis. For uniform continuity, the stronger estimatefollows 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.^[ref]

If the first absolute moment is finite, $|e^{iu}-1|\le |u|$ additionally gives $|{\phi }_X(t+h)-{\phi }_X(t)|\le |h|\mathbb{E}[|X|]$, 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.