Relation between Laplace transform and measures. The Laplace transform of a measure $\mu$ on the real line is defined by
$$f_{\mu}(s)= \int_{\mathbb{R}}e^{-st}d\mu(t), \hspace{1cm} \forall s \geqslant 0.$$
My question is ----
1)Does the Laplace transform of a measure (finite or infinite) always exists?
2)If not, can it be said that the Laplace transform of a probability measure always exists?
If the support of the measure is changed from the real line to the non-negative part of the real line, what happens to question (1) and (2)?
 A: Your definition of the Laplace transform will not converge in general, because if $\mu$ has support on the whole real line then the exponential term can blow up, depending on the exact form of $\mu$.
The common definition of the Laplace transform is for measures supported on the non-negative real line.
(Alternatively you may want to look into the definition of the two-sided Laplace transform, where $t$ in the exponential is replaced by $|t|$.)
Now, consider a $\mu$ that is supported on the non-negative real line, and that is w.l.o.g. positive.
Since $e^{-st} \le 1$ for $\Re[s]>0$, it is easy to see that
$|f_\mu(s)| \le f_\mu(0) = \int\limits_0^\infty d\mu(t) = \| \mu \|$.
Hence, if the measure is finite, then its Laplace transform exists.
In conclusion, the Laplace transform of a probability measure always exists.
(It is in fact (related to) the measure's moment generating function.)
For infinite measures, one needs some regularity conditions on $\mu$ that ensure that the integral does not blow up.
One of these would be for example, that its distribution function $F(t) := \mu([0,t))$ is exponentially bounded.
That means:
$ \exists K,C > 0: \;\; |F(t)| \le C e^{Kt}$
Then one can show that the Laplace transform exists for $\Re[s]>K$.
A: If $\mu$ has density $\frac 1 {\pi (1+x^{2})}$ the the Laplace transform exists only for $s=0$. 
