Continuity of the Characteristic Function of a RV Defining the Characteristic Function $ \quad  \phi(t) := \mathbb{E} \left[ e^{itx} \right] $ for a random variable with distribution function $F(x)$ in order to show it is  uniformly continuous I say
$$
|\phi(t+u) - \phi(t)| = \left |\int e^{itx}(e^{iux} - 1) dF(x) \right| \le \\ \int 1 \cdot|e^{iux} -1|dF(x) \to 0 \quad as  \quad u\to0
$$
Now my question is, does the convergence I state in the last line follow directly, or do I need to be a little carful before I conclude it is true ? (i.e. can I directly use that $|e^{itu} -1| \to 0 ? )$
 A: There is a typo: it should be 
$$|\phi(t+u)-\phi(t)|=\left|\int (e^{i(t+u)x}-e^{itx})\mathrm dF(x)\right|\leqslant\int \left|e^{iux}-1\right|\mathrm dF(x).$$
Now take a sequence $\left( u_n\right) $ which converges to $0$ and put $f_n(x):=\left|e^{iu_nx}-1\right|$. Then $f_n(x)\to 0$ for each $x$ and $\left |f_n(x)\right|\leqslant 2$ which is integrable (when $\mathbb R$ is endowed with the probability measure $\mathbb P_X$)  so we conclude by the dominated convergence theorem that $\lim_{n\to\infty}\int\left|e^{iu_nx}-1\right|dF(x)=0$ and so $\phi$ is uniformly continuous on $\mathbb R$. 
Note that if the random variable is integrable we don't need the dominated convergence theorem, since we can write 
$$\left|\phi(t+u)-\phi(t)\right|\leqslant\int \left|e^{iux}-1\right|dF(x)=2\int\left|\sin(ux/2)\right|\mathrm dF(x)\leqslant \left|u\right|\int \left|x\right|\mathrm dF(x).$$
A: You probably meant "as $u \to 0$" in that last line, right? In any case, it follows by the dominated (or bounded) convergence theorem applied to the measure $dF$.
