Probability measure on $(0,\infty)$ What can be a possible probability measure on $(0,\infty)$? Give an example.
For $(0,1)$ Lebesgue measure can be used and it easily satisfies all the properties of probability measure. But when the set is $(0,\infty)$, Lebesgue measure will not lie in 0 to 1 range. I am thinking that some mapping from $(0,\infty)$ to $(0,1)$ would do the trick. Am I right?
 A: Any Lebesgue integrable function $f:(0,\infty)\to[0,\infty)$ which does not vanish almost everywhere can be made into a probability measure $\mu_f$ on $(0,\infty)$ by setting $$\mu_f(S):=\frac{\int_S\,f(x)\,\text{d}x}{\int_0^\infty\,f(x)\,\text{d}x}\text{ for every measurable set }S\,.$$
Every absolutely continuous probability measure (relative to the Lebesgue measure) arises this way.  
However, there are uncountably many other probability measures.  Singular continuous measures on $(0,\infty)$ such as the Cantor distribution and discrete probability measures on $(0,\infty)$ are some of those probability measures not in the form $\mu_f$ for some Lebesgue integrable function $f$.  Of course, you can also have a convex combination of an absolutely continuous probability measure, a singular one, and a discrete one.
Additionally, if you want a probability measure $\nu$ whose essential support is precisely $(0,\infty)$, then the absolutely continuous part $\nu_\text{abs}$ of $\nu$ cannot be $0$ (since essential supports of singular and discrete probability measures are Lebesgue null sets).  In other words, $\nu_\text{abs}$ is of the form $\nu_\text{abs}=\alpha\,\mu_f$ for some $\alpha\in(0,1]$ and for some Lebesgue integrable function $f:(0,\infty)\to[0,\infty)$ which vanishes on a set of Lebesgue measure $0$.  
