If $f : [-a,a] \rightarrow \mathbb{IR}$ is Scott continuous, why are $f^-$ and $f^+$ measurable?

Question

In "A domain theoretic account of Picard's theorem" (http://www.doc.ic.ac.uk/~dirk/Publications/icalp2004.pdf), the authors assert the following.

Let $\mathbb{IR}$ be the interval domain $\lbrace [a^-,a⁺] | a^- \leq a^+, a^-, a^+ \in \mathbb{R} \rbrace \cup \lbrace \mathbb{R} \rbrace$.

Suppose $f = [f^-,f^+] : [-a,a] \rightarrow \mathbb{IR}$ is Scott continuous. Then $f^-$ and $f^+$ are measurable.

Their proof consists in asserting that $f⁻$ and $f⁺$ are lower (resp. upper) semi-continuous, which I have tried without success to establish. Can anyone help with either an explanation or a reference that proves it?

Answer 1

First, we note that $\mathbb{IR}$ is ordered by $[a,b] \sqsubseteq [c,d]$ iff $[c,d] \subseteq [a,b]$, which makes $\mathbb{IR}$ into a domain. Now suppose $f: [-a,a]\to \mathbb{IR}$ is Scott-continuous (that is, $f$ preserves all directed suprema).

We want to prove that $f^-$ as defined in the comment above is lower semi-continuous. Let $(x_n)$ be an increasing sequence in $[-a, a]$. So $(x_n) \to s$ where $s = \sup\{x_n:n\in\mathbb{N}\} \in [-a, a]$. We want to show that $\lim_{n\to\infty} f^-(x_n) = f^-(s)$.

The set $$D:=\{x_n: n\in \mathbb{N}\}$$ is a directed subset of the domain $[-a,a]$ (with the ordering inherited from $\mathbb{R}$), and we have $s=\bigsqcup D$.

For $n\in \mathbb{N}$ we have $f(x_n)\supseteq f(x_{n+1})$ since $f$ is order-preserving (which is implied by $f$ being Scott-continuous). The fact that $f$ is Scott-continuous means that $$[f^-(s), f^+(s)] = f(s)=f(\bigsqcup D) = \bigsqcup f(D) = \bigcap \{ [f^-(x_n), f^+(x_n)]:n\in\mathbb{N}\}.$$

So $(f^-(x_n))_{n\in\mathbb{N}}$ is an increasing sequence in $\mathbb{R}$ and it is bounded by $f^+(x_1)$, so it converges. The equations above imply that $\lim_{n\to\infty} f^-(x_n) = f^-(s)$, so $f^-: [-a,a]\to \mathbb{R}$ is lower semi-continuous. A similar argument shows that $f^+$ is upper semi-continuous.

Answer 2

This answers @user3078439's question in his comment to my original answer.

First, a short argument to show that if $P,Q$ are domains and $f:P\to Q$ is Scott-continuous, then $f$ is order preserving.

Let $a\leq b \in P$. So $D:= \{a,b\}$ is a directed set with $b=\bigsqcup D$. Since $f$ is Scott-continuous, $f(D) = \{f(a), f(b)\}$ is also directed, so at least one of the relations $f(a) \leq f(b)$ or $f(b) \leq f(a)$ holds in $Q$. Now $f$ being Scott-continuous means $$f(b) = f(\bigsqcup D) = \bigsqcup f(D) = \sup\{f(a), f(b)\},$$ which directly implies $f(a)\leq f(b)$. So if $f:P\to Q$ is Scott-continuous, then it is order-preserving. (The converse of this statement does not hold.)

Since in your question the ordering on $Q:= \mathbb{IR}$ is given by $\supseteq$, I hope that I have answered your implicit question "if $f$ being Scott-continuous truly means that for all $x\leq y, f(x)\supseteq f(y)$" in the positive.

As to the question whether $f(-a)\supseteq f([-a,a])$ there is a negative answer: $f(-a)$ is just a real interval, namely $[f^-(-a), f^+(-a)]$. But $f([-a,a])$ is a set of intervals.

What is true, though, is that $f(-a)=[f^-(-a), f^+(-a)]$ is an interval that is a superset of every interval $f(x)=[f^-(x), f^+(x)]$ for all $x\in[-a,a]$. If that was the intent of your question in the comment, then there is a positive answer.