I have a two-sided Laplace-Stieltjes transform, $$ \int_{-\infty}^{+\infty} e^{-xt}d\mu(t) $$ that converges absolutely in $(a,b)$.
If the measure $\mu$ is finite,then $$ \int_{-\infty}^{+\infty}d\mu(t)=\mu(\mathbb{R})<\infty $$ can I conclude that $(a,b)$ MUST contain the origin?
In general, how changes the interval of convergence of a two-sided Laplace-Stieltjes transform with respect to $\mu$?
Thank you
Even if $\mu$ is not a positive measure, its total variation measure $|\mu|$ is, and "the integral converges
absolutely" means $\int_{\mathbb R} e^{-xt}\ d|\mu|(t) < \infty$.
Now $\int_{A} e^{-xt}\ d|\mu|(t)$ is a convex function of $x$ for every measurable $A \subseteq \mathbb R$, so the set where this is finite is a convex set, i.e. an interval. If $\mu$ is a finite measure, that says $\int_{\mathbb R} e^{-xt}\ d|\mu|(t) < \infty$ for $x=0$, so $0$ is in the interval.
Note, however, that the interval doesn't have to be open. For example, take the measure with density $1/x^2$ for $x > 1$, $0$ elsewhere: this has interval $[0,\infty)$.
In case $\mu$ is finite positive, let us consider its normalization $\mu'(\cdot):=\mu(\cdot)/\mu(\Bbb R)$ which is clearly a probability measure on reals. As a result, $$ f_\mu(x):=\int_\Bbb R\mathrm e^{-xt}\mu(\mathrm dt) = \mu(\Bbb R)\int_\Bbb R\mathrm e^{-xt}\mu'(\mathrm dt) = \mu(\Bbb R)f_{\mu'}(x) = \mu(\Bbb R)m_{\mu'}(-x). $$ where $m_{\mu'}$ is a moment-generating function (MGF) of the distribution $\mu'$. Clearly, the LHS is finite iff the RHS is, so that your question for positive finite $\mu$ is equivalent to studying the properties of MGF and in particular of heavy-tailed distributions. Note, the clearly $m_{\mu'}(0) = 1$ regardless of $\mu'$, but it does not have to exist anywhere else, e.g. if $\mu'$ is Cauchy distribution.
