Characteristic function of a random variable $X$ is absolutely continuous 
Let $(\Omega, \mathcal{F}, P)$ be a measurable space and $X:\Omega \to
> \mathbb{R}$ a random variable. 
Assuming $E[|X|]<\infty$, prove that $\psi_X(t)=E[e^{itX}]$ is absolutely continuous.
[Hint: first, prove that $|\psi_X(t+h)-\psi_X(t)|\leq E[|e^{ihX}-1|]$
  for every $t, h \in \mathbb{R}$]

I had no problem proving the statement in the hint, but I'm having a difficult time to prove that: $$\forall\epsilon>0 ,\exists\delta>0\ \text{with}$$ $$\sum_{k=1}^{n}(b_k-a_k)<\delta \Rightarrow \sum_{k=1}^{n}E[|e^{i(b_k-a_k)X}-1|]<\epsilon, $$ 
where $(a_1, b_1), ...,(a_n, b_n)$ are disjoint intervals, which is what I think I'm supposed to do.
Any ideas? Thanks!
 A: There is something fishy about the problem.
The OP says that in the original source the problem was to prove that the characteristic function of any random variable is absolutely continuous. This is false; it's easy to give an example of a nowhere-differentiable characteristic function, as below.
As I pointed out this morning, if we assume that $\Bbb E[|X|]<\infty$ then it's easy to show that the characteristic function is absolutely continuous (hint: $|e^{it}-1|\le |t|$ for $t\in\Bbb R$). But I must have been asleep, suggesting that without noticing that that's a very curious problem, since it's also easy to show that $\Bbb E[|X|]<\infty$ implies that the characteristic function is continuously differentiable (with bounded derivative)!
I can't think of a "natural" "probabilistic" condition that implies the characteristic function is absolutely continuous without also implying that it is continuously differentiable. Since $$\sup_{\delta_j>0,\,\sum\delta_j=\delta}\sum\left|e^{i\delta_jt}-1\right|=\delta|t|$$it seems there is no condition weaker than $\Bbb E[|X|]<\infty$ such that it is possible to use the hint given to show that the characteristic function is absolutely continuous.

A simple although not quite self-contained counterexample to the OP as originally stated, without the assumption on the first moment. It's easy to construct a discrete $X$ so that $$\psi_X(t)=\sum_{n=1}^\infty 2^{-n}e^{2\pi i 4^nt}.$$That's a classic example of a nowhere-differentiable function.
