Integration against a strictly increasing function Let $f : \mathbb{R} \to \mathbb{R}$ possess the following properties:

*

*$\int_\mathbb{R}f(x)\ dx = 0 $,


*$\forall x \in \mathbb{R}$ : $0 \leq \int_x^\infty f(y)\ dy \leq 1$ where the left inequality is strict on a measure nonzero subset of choices of $x$.
Let there also be a strictly increasing function $u : \mathbb{R} \to \mathbb{R}$ such that
$$\int_\mathbb{R} u(x)f(x)\ dx$$
is well-defined (but may diverge).
What additional conditions, if any, must we place on $u$ in order to achieve $\int_\mathbb{R} u(x)f(x)\ dx > 0$?

Background: In the problem I have been considering, $f = p_1 - p_2$ is the difference between two probability density functions such that at least one of $\int_\mathbb{R} u(x)p_1(x)\ dx$ and $\int_\mathbb{R} u(x)p_2(x)\ dx$ must converge, but I suspect this is superfluous. The condition on $f$ is commonly referred to as stochastic dominance, which was discussed in this paper by Hadar & Russell, but they do not consider the general case of probability distributions with non-compact support. I am concerned that their integration by parts trick doesn't work in the general case.
 A: Let $f = p_1 - p_2$, where $p_1$ and $p_2$ are probability density functions on $\mathbb{R}$, and let $u : \mathbb{R} \to \mathbb{R}$ be strictly increasing. Then the function $\tilde{u}$ defined by
$$ \tilde{u}(x) = u(x^+) = \lim_{y \to x^+} u(y) $$
is also strictly increasing, right-continuous, and coincides with $u$ except on an at most countable set of exceptional points. Then $\tilde{u}$ induces a Stieltjes measure $\mu$ such that $\mu((a, b]) = \tilde{u}(b) - \tilde{u}(a)$.
Now, we fix am arbitrary point $c \in \mathbb{R}$ and decompose $\tilde{u}$ as $\tilde{u}(x) = \tilde{u}(c) + u_c^+(x) - u_c^-(x)$, where
$$ u_c^+(x)
= \int_{\mathbb{R}} \mathbf{1}_{(c, x]}(t) \, \mu(\mathrm{d}t)
= \begin{cases}
\tilde{u}(x) - \tilde{u}(c), & x \geq c, \\
0, & x < c,
\end{cases}
$$
and
$$
u_c^-(x)
= \int_{\mathbb{R}} \mathbf{1}_{(x, c]}(t) \, \mu(\mathrm{d}t)
= \begin{cases}
0, & x \geq c, \\
\tilde{u}(c) - \tilde{u}(x), & x < c.
\end{cases} $$
Since $u^+_c(x)p_i(x)$ is always non-negative, its integral always exists. Moreover,
\begin{align*}
\int_{\mathbb{R}} u^+_c(x)p_i(x) \, \mathrm{d}x
&= \int_{\mathbb{R}} \left( \int_{\mathbb{R}} \mathbf{1}_{(c, x]}(t) \, \mu(\mathrm{d}t) \right) p_i(x) \, \mathrm{d}x \\
&= \int_{\mathbb{R}} \int_{\mathbb{R}} \mathbf{1}_{(c, x]}(t) p_i(x) \, \mathrm{d}x \, \mu(\mathrm{d}t) \tag{Tonelli} \\
&= \int_{(c, \infty]} \left( \int_{t}^{\infty} p_i(x) \, \mathrm{d}x \right) \, \mu(\mathrm{d}t).
\end{align*}
Similarly,
\begin{align*}
\int_{\mathbb{R}} u^-_c(x)p_i(x) \, \mathrm{d}x
&= \int_{\mathbb{R}} \int_{\mathbb{R}} \mathbf{1}_{(x, c]}(t) p_i(x) \, \mathrm{d}x \, \mu(\mathrm{d}t) \tag{Tonelli} \\
&= \int_{(-\infty, c]} \left( \int_{-\infty}^{t} p_i(x) \, \mathrm{d}x \right) \, \mu(\mathrm{d}t).
\end{align*}
Now, we make the following assumptions:

Assumption 1.

*

*Both $\int u(x) p_1(x) \, \mathrm{d}x$ and $\int u(x) p_2(x) \, \mathrm{d}x$ exist;

*At least one of $\int u(x) p_1(x) \, \mathrm{d}x$ and $\int u(x) p_2(x) \, \mathrm{d}x$ converges.


This assumption ensures that at most one of $\int_{\mathbb{R}} u^{e}_c(x)p_i(x) \, \mathrm{d}x$, $i \in\{1, 2\}$, $e \in \{ -, +\}$, is infinite. In particular, sums or differences between these four integrals are all well-defined. Then
\begin{align*}
\int_{\mathbb{R}} u(x)p_i(x) \, \mathrm{d}x
&= \int_{\mathbb{R}} \tilde{u}(x)p_i(x) \, \mathrm{d}x \\
&= \tilde{u}(c) + \int_{(c, \infty]} \left( \int_{t}^{\infty} p_i(x) \, \mathrm{d}x \right) \, \mu(\mathrm{d}t) + \int_{(-\infty, c]} \left( -\int_{-\infty}^{t} p_i(x) \, \mathrm{d}x \right) \, \mu(\mathrm{d}t)
\end{align*}
and
\begin{align*}
\int_{\mathbb{R}} u(x)f(x) \, \mathrm{d}x
&= \int_{\mathbb{R}} u(x)p_1(x) \, \mathrm{d}x - \int_{\mathbb{R}} u(x)p_2(x) \, \mathrm{d}x \\
&= \int_{(c, \infty]} \left( \int_{t}^{\infty} f(x) \, \mathrm{d}x \right) \, \mu(\mathrm{d}t) + \int_{(-\infty, c]} \left( -\int_{-\infty}^{t} f(x) \, \mathrm{d}x \right) \, \mu(\mathrm{d}t) \\
&= \int_{\mathbb{R}} \left( \int_{t}^{\infty} f(x) \, \mathrm{d}x \right) \, \mu(\mathrm{d}t), \tag{*}
\end{align*}
where in the last line we utilized the equality $\int_{\mathbb{R}} f(x) \, \mathrm{d}x = 0$.
Now we turn to studying the condition for this integral to be strictly positive. To this end, we impose the stochastic domination condition:

Assumption 2. $\displaystyle \int_{t}^{\infty} f(x) \, \mathrm{d}x \geq 0$ for all $ t \in \mathbb{R}$.

Let $G(t) = \int_{t}^{\infty} f(x) \, \mathrm{d}x$. Then $\text{(*)}$ shows that $\int_{\mathbb{R}} u(x)f(x) \, \mathrm{d}x > 0$ if and only if
$$U = \{t : G(t) > 0\}$$
is not a $\mu$-null set. Since $G$ is continuous, $U$ is an open subset of $\mathbb{R}$ and hence can be written as a union of at most countably many disjoint open intervals $\{(a_1, b_1), (a_2, b_2), \ldots\}$. Now, we observe:
\begin{align*}
\mu(U) = 0
&\quad\iff\quad \mu((a_i, b_i)) = 0 \quad \text{for each $i$} \\
&\quad\iff\quad \tilde{u}(b_i^-) - \tilde{u}(a_i) = 0 \quad \text{for each $i$} \\
&\quad\iff\quad u(b_i^-) - u(a_i^+) = 0 \quad \text{for each $i$} \\
&\quad\iff\quad \text{$u$ is constant on each $(a_i, b_i)$} \\
&\quad\iff\quad \text{$u$ is locally constant on $U$.}
\end{align*}
In other words, if $u$ is not locally constant on $U$, then $\mu(U) > 0$ and hence $\int_{\mathbb{R}} u(x)f(x) \, \mathrm{d}x > 0$ (and vice versa).
