Take a random variable $X$ with distribution function $F(x)=\mathbb{P}[X\leq x]$ and characteristic function $\phi_X(t)$. Then one can write
$\int_{-c}^{c}\frac{1-e^{-ixt}}{it}\phi_X(t)dt=\int_{-c}^{c}\frac{1-e^{-ixt}}{it}\int_{-\infty}^{\infty}e^{itu}dF(u)dt=\int_{-c}^{c} \int_{-\infty}^{\infty}\frac{\sin tu-\sin t(u-x)}{t}dF(u)dt$
Now, according to Kendall(v1,ed5,p121) a change of integration order can be performed.
$\int_{-c}^{c} \int_{-\infty}^{\infty}\frac{\sin tu-\sin t(u-x)}{t}dF(u)dt= \int_{-\infty}^{\infty}\int_{-c}^{c}\frac{\sin tu-\sin t(u-x)}{t}dtdF(u)$
The motivation they give is that because $\int_{-c}^{c}\frac{1-e^{-ixt}}{it}\phi_X(t)dt$ is uniformaly convergent such an operation is permitted.
Questions
1. Do they imply that Fubini theorem is used?
2 If so, what type of integrals are the integrals wrt to $dt$ and wrt $dF(u)$ ? The $dF(u)$ integral needs to be Lebesgue, but the integral wrt $dt$ ?can not? be Lebesgue since $\int^{c}_{-c} \frac{\sin x}{x} dx$ is not Lebesgue integrable.
3 If the two integrals are of different type how would one use Fubini?
4 If Fubini is not used then how is the change of integration order justified?
$\frac{\sin tu-\sin t(u-x)}{t}$ has a removable singularity at $t=0$, that is $ \lim_{t\to 0} \frac{\sin tu-\sin t(u-x)}{t} $ exists and is finite. So no principle value integral is required. It is a double integral of a bounded function over two finite measure spaces, and so Fubini applies.