Suppose that $\{X_i: i \in I\}$ is a collection of random variables on the measurable space $(\Omega, \mathcal F)$ taking values in the extended reals $[-\infty, \infty]$.
Does there exist a measurable function $g: [-\infty, \infty] \to [a,b]$ into a bounded interval that preserves strict expectation inequalities for $\{X_i: i \in I\}$, in the sense that for all probability measures $P$ on $(\Omega, \mathcal F)$ and all $i,j \in I$, $$\int X_i dP < \int X_j P \ \ \text{implies} \ \int g(X_i)dP < \int g(X_j)dP?\tag{1}$$
I know that there are homeomorphisms between $[-\infty, \infty]$ and bounded intervals $[a,b]$, so my thought was to try one of those that is also strictly increasing, like $g(x) = \arctan(x)$. But I've been unable to see that (1) holds. Note that strictly increasing is necessary: just consider $P$ as it ranges over point masses.
Assume you have a probability measure $P$ such that for every $i\in \mathbb{N}_{\geq 2}$ exists measurable sets $A_i$ such that $P(A_i)>0$ and $$\lim_{i\rightarrow \infty} P(A_i)=0.$$ Now define $X_i = \frac{2}{P(A_i)} 1_{A_i}$, $X_1=1$ and $X_0=0$. We have that $X_0$ has expectation $0$, $X_1$ has expectation $1$ and the rest has expectation $2$. So we would need $$ g(1)=\int g(X_1) dP <\int g(X_i) dP = g(2/P(A_i)) P(A_i) + g(0) (1-P(A_i)).$$ Taking the limit $i\rightarrow \infty$ yields (as $g$ is bounded) $$ g(1)=g(0).$$ However, we also have $$ g(0)= \int g(X_0) dP < \int g(X_1) dP = g(1)$$ which gives a contradiction.
Of course we can do the exact same construction when we replace the word probability measure with finite measure.
For infinite measure we could require that those $A_i$ are contained in some $A$ with finite measure. Then we note that strict inequalities can only hold for $g(0)=0$ (otherwise most expectations will be $\pm\infty$) and everything reduces to the finite case.