Is the following function integrable?

Question

Suppose $X:\Omega\rightarrow S$ with $S\subset\Bbb R$ is a bounded random variable with the density $p(x)$ and the function $u:S\rightarrow \Bbb R$ is a concave, continuous differentiable, monotone function, which is bounded from above but not below. Does it imply that $$E[u(X)]$$ exist? What is the weakest condition on $u$ to make the expectation $E[u(X)]$ exist?

Answer 1

The simplest condition guaranteeing the existence of expectation $\mathbb E[u(X)]$ is continuity of $u(x)$ on the closure of $S$. This closure $\bar S$ is a compact, and $u(\bar S)$ is compact too by continuity, and so it is bounded. Then the set $u(S)$ is bounded too as a subset of $u(\bar S)$.

Let us show by example the role of continuity on $\bar S$.

Consider $\Omega=[0,1]$ equipped with Borel sigma-algebra and Lebesgue measure. Let $$X(\omega)=\begin{cases}\omega, & \omega\neq 0\cr 1, & \omega = 0\end{cases}$$ Then $X:\Omega \to (0,1]$ is uniformly distributed with density $p(x)=\mathbb 1_{(0,1)}(x)$.

Set $u(x)=-\frac1x$. This function is continuous on $(0,1]$, but not on $[0,1]$. It is also concave and monotone, and bounded from above. And the set $u(X)=(-\infty, -1]$ is not bounded and the expectation does not exist: $$\mathbb E[|u(X)|]=\int_0^1 \frac1x \,dx=\infty$$

Note also that the existence of expectation depends not only on $u(x)$, but also on a given density. In probability space given above, let $p(x)=2x\mathbb 1_{(0,1)}(x)$ and $u(x)=-\frac1x$. In this case the expectation $\mathbb E[u(X)]$ exists. If we want to guarantee the existence of the expectation regardless of the type of density, we need the boundedness of $u(x)$. You can achieve it either by the requirement of continuity on the closure of $S$, or by requiring directly.