Monotone likelihood ratio without densities I would like to find a generalization of the monotone likelihood ratio ordering that does not require that the probability distributions admit a density and have the same support, which would allow me to deal with degenerate probability distributions, and with the case where the supports are disjoint. 
For instance, if $T_0$ is a degenerate probability distribution localized at $0$ and $T_1$ is a probability distribution with support included in $[1,+\infty)$, I would like the definition to guarantee that $T_1$ dominates $T_0$ according to this ordering. This seems to be a natural extension of the standard definition.
I would also like to the definition to boil down to the standard criterion if the distributions admit a density and have a common support.
Are there any existing generalizations of the monotone likelihood ratio order that would be appropriate for this purpose? Thanks!
 A: A general definition is the following (see Def. 1.C.1 in Skahed and Shuntikumar, Stochastic Orders and Applications, 2007). Let $X$ and $Y$ be continuous [discrete] r.v.'s with densities [probability distributions] $f$ and $g$; then $Y \succeq X$ (in the likelihood ratio order) is
$$f(u)g(v) \ge f(v)g(u)$$ 
for all $u \le v$. An equivalent formulation is to require that the ratio
$$\frac{f(t)}{g(t)}$$
is decreasing over the union of supports of $X$ and $Y$ (interpreting $a/0$ as $\infty$ when $a > 0$).
A: If we take in full generality two probability measures $\mu , \nu$ on $\Bbb R$ we can define the likelihood ratio order as in Def. 1.C.1 in Stochastic Orders by Shaked and Shanthikumar (2007):
For two measurable sets $A,B \subset \Bbb R$ we write $A\leq B$, if $(x,y) \in A\times B$ implies $x\leq y$ for any such pair. We say $\mu$ is smaller than $\nu$ in the likelihood ratio order, if
$$A \leq B \Rightarrow \mu (A) \nu (B) \geq \mu (B) \nu (A).$$
If $\mu , \nu$ have Lebesgue densities $f, g$, this definition is equivalent to
$$f(u)g(v) \geq f(v) g(u) \qquad \text{for almost all } u\leq v.$$
