Let $\mu$ be a probability measure and $R(p)$ an interval with the property that both $\inf(R(p)):=\underline{r}(p)$ and $\sup(R(p)):=\bar{r}(p)$ are continuous and bounded functions of $p\in [0,1]$, with $\underline{r}(p)<\bar{r}(p)$ for every $p\in (0,1)$. Consider the function $\alpha(p):=\mu(R(p))$ for $p\in (0,1)$.
Question: Can we say that $\alpha(p)$ is lower semi-continuous for every $p$ such that $\alpha(p)>0$?
Some notes: Consider any sequence $(p_n)_{t\in\mathbb{N}}$ converging to some $p$. Suppose $r\in R(p)$ (since $p\in (0,1)$ $R(p)$ is non-empty). Since the boundary points of $R(\cdot)$ are continuous at $p$, there exists $N\in\mathbb{N}$ such that for every $n\ge N$, $r\in R(p_n)$. That implies $R(p)\subseteq \liminf R(p_n)$. From Fatou's lemma, $\alpha(p)\le\liminf \alpha(p_n)$. This proves that $\alpha(p)$ is lower semi-continuous at $p$.
Is this correct? The argument does not depend on $R(p)$ being open or closed, but I think I this must matter. Under what conditions can we say that $\alpha(p)$ is upper semi-continuous?
Best Answer
Your argument is valid if $R(p)$ is open but not in general. For example the end points of $[0,1-\frac 1 n]$ converge to end points of $[0,1]$. The point $1$ belongs to $[0,1]$ but does not belong to any of the intervals $[0,1-\frac 1 n]$. When the intervals are open your argument is correct.