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 directed set. 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.

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.

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 directed set. 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 $s = \sup\{x_n:n\in\mathbb{N}\} \in [-a, a]$. 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 $(f^-(x_n))_{n\in\mathbb{N}}\to f^-(s)$, so $f^-: [-a,a]\to \mathbb{R}$ is lower semi-continuous. A similar argument shows that $f^+$ is upper semi-continuous.

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 directed set. 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.