Apologies for that useless modifier in the brackets in the title -I had to add that to avoid "duplicate titles". It is most natural that there will be multiple questions with the title "Is this function continuous", but apparently that is not allowed. On to the question...
Suppose $\theta$ is a random variable which has an atomless distribution over $[0,1]$, with a density $f$.
Let $U:[0,1]\rightarrow [0,1]$ be a continuous function with "no flat regions", i.e. $\nexists\; [a,b] \subseteq [0,1]$ such that $U(\theta)=U(\theta')\;\forall\;\theta,\theta' \in [a,b]$. Let $\max_{[0,1]}U(\theta)=\overline{u}$ and $\min_{[0,1]}U(\theta)=\underline{u}$. Define the following function:
$$g:[\underline{u},\overline{u}] \rightarrow [0,1], g(v)=\mathbb{E}(\theta: U(\theta)\leq v)=\frac{\int\limits_{\{\theta:U(\theta)\leq v\}}\theta f(\theta)d\theta }{\int\limits_{\{\theta:U(\theta)\leq v\}} f(\theta)d\theta } $$
My question is: Is $g(\cdot)$ continuous?
My approach: I have shown that the correspondence $\Theta_0:[\underline{u},\overline{u}] \Longrightarrow [0,1], \Theta_0(v)= \{\theta:U(\theta)\leq v\}$ is upper hemicontinuous and also lowerhemicontinuous except at the points where $v$ is the value of $U(\cdot)$ at a local minima of $U(\cdot)$. One approach is to say - okay, the integrals in the expression for $g(\cdot)$ have as their limits points $\theta$ which solve $U(\theta)=v$. These points change "continuously" as $v$ changes (new points can get "suddenly" added but that doesn't affect the integral due to atomlessness of $f$). But there is the issue of the number of solutions to $U(\theta)=v$ which may be infinite, and I don't know how to handle that.