Prove that the characteristic function of a random variable is uniformly continuous. The definition of the characteristic function of a random variable $X$ is
$$\varphi(\lambda)=E[e^{i\lambda X}]=\int_{-\infty}^\infty e^{i\lambda x}\,dP(x),$$
where $dP$ is the distribution of the random variable.

Prove that the characteristic function of a random variable is uniformly continuous.

My attempt of the proof is:
$$|\varphi(\lambda) -\varphi(\lambda') | = |\int_{-\infty}^\infty (e^{i\lambda x} -e^{i\lambda' x} ) dP(x)|  $$
$$\leq    |\int_{-\infty}^\infty e^{i\lambda' x} (e^{i(\lambda - \lambda') x}-1)  dP(x)|  $$
$$\leq  \int_{-\infty}^\infty   |e^{i\lambda' x}|   |e^{i(\lambda - \lambda') x}-1|    dP(x)      $$
Then use the fact that the absolute value of a complex exponential is 1.
$$=  \int_{-\infty}^\infty      |e^{i(\lambda - \lambda') x}-1|    dP(x)      $$
Then use the inequality that the inequality $|e^{ihx}-1|\le |hx|$.
$$\leq  \int_{-\infty}^\infty      |(\lambda - \lambda') x|    dP(x)      $$
I don't know how to do next. The textbook said I need to use Lebesgue Dominated Convergence Theorem. I don't know how to use that. I think I am stuck that how to prove one function is uniformly continuous. I only know this definition of uniformly continuous:
Given $\epsilon > 0$, we want to find $\delta > 0$ such that
$$
|f(x) - f(y)| < \epsilon \quad  \text{whenever} \quad |x-y| < \delta
$$
and $\delta$ is independent of $x$ and $y$.
 A: You've proved that
$$|\varphi(\lambda)-\varphi(\lambda')|\leq\int_{-\infty}^\infty|e^{i(\lambda-\lambda')x}-1|dP(x).$$
Let $\phi(\lambda)=\int_{-\infty}^\infty|e^{i\lambda x}-1|dP(x)$, then $|\varphi(\lambda)-\varphi(\lambda')|\leq\phi(\lambda-\lambda')$.
We prove that $\lim_{\lambda\to0}\phi(\lambda)=0=\phi(0)$. Recalling the definition of $\phi$, since $|e^{i\lambda x}-1|\leq 2$ and $\lim_{\lambda\to0}|e^{i\lambda x}-1|=0$ for all $x$, and $\int_{-\infty}^\infty 2dP(x)=2<\infty$, the Lebesgue Dominated Convergence Theorem tells that
$$\lim_{\lambda\to0}\phi(\lambda)=\lim_{\lambda\to0}\int_{-\infty}^\infty|e^{i\lambda x}-1|dP(x)=0.$$
Hence, for any $\epsilon>0$ we can find $\delta>0$ such that $0\leq\phi(\lambda)<\epsilon$ for all $|\lambda|<\delta$; therefore, whenever $|\lambda-\lambda'|<\delta$, we have
$$|\varphi(\lambda)-\varphi(\lambda')|\leq\phi(\lambda-\lambda')<\epsilon.$$
This proves the uniform continuity of $\varphi$.
Remark. You've got a further step that $|\varphi(\lambda)-\varphi(\lambda')|\leq\int_{-\infty}^\infty|(\lambda-\lambda')x|dP(x)$, which implies that $\varphi$ is Lipschitz continuous and then uniformly continuous as long as $\int_{-\infty}^\infty |x|dP(x)<\infty$. But this is not always true. For example, the Cauchy distribution, whose characteristic function is uniformly continuous, although the first moment does not exist.
About the DCT things. Let $\{\lambda_n\}$ be an arbitrary sequence that tends to $0$ as $n\to\infty$. Since $|e^{i\lambda_n x}-1|\leq 2$ and $\lim_{n\to\infty}|e^{i\lambda_n x}-1|=0$, the standard DCT implies that
$$ \lim_{n\to\infty}\phi(\lambda_n)=\lim_{n\to\infty}\int_{-\infty}^\infty|e^{i\lambda_n x}-1|dP(x)=0.$$
Hence, $\lim_{n\to\infty}\phi(\lambda_n)=0$ for all equences $\{\lambda_n\}$ that tends to $0$ as $n\to\infty$, which implies that $\lim_{\lambda\to0}\phi(\lambda)=0$ by Heine's theorem. This is the so called Dominated Convergence Theorem with "continuous" limits, and has been discussed many times in this site, see here and here for more information.
