Proof of Semicontinuity and the Semicontinuous Envelopes are Local Notions

Conventions. We use the symmetry $d(y,z)=d(z,y)$, the vanishing $d(x,x)=0$ and the triangle inequality of a metric. Elementary order facts are those of Elementary Order Arithmetic in an Ordered Field, whose claim 1 is the compatibility of the strict order with addition, the non-strict law being an axiom of the ordered field $\mathbb{R}$, whose order $\le$ is a total order, so that its reflexivity, antisymmetry and transitivity are axioms of that definition. A strict inequality $a<b$ implies $a\le b$ by the definition of the strict order.

Proof of claim 1. Suppose first that $u$ is upper semicontinuous at $x$ relative to $S$. Since $S_{r}\subseteq S$ and $x\in S_{r}$, claim 2 of Negation, Restriction, and Separated Differences of Semicontinuous Functions shows that $u|_{S_{r}}$ is upper semicontinuous at $x$ relative to $S_{r}$.

Conversely, suppose $u|_{S_{r}}$ is upper semicontinuous at $x$ relative to $S_{r}$, and let $\varepsilon\in\mathbb{R}$ be positive. By upper semicontinuity there is a positive $\delta'\in\mathbb{R}$ such that every $z\in S_{r}$ with $d(x,z)<\delta'$ satisfies $u(z)<u(x)+\varepsilon$. By claim 9 of Elementary Order Arithmetic in an Ordered Field there is $\delta\in\mathbb{R}$ with $\delta\le\delta'$, $\delta\le r$ and $\delta=\delta'$ or $\delta=r$; in either case $\delta$ is positive. Let $y\in S$ satisfy $d(x,y)<\delta$. Then $d(y,x)=d(x,y)<\delta\le r$, so $d(y,x)<r$ by the mixed transitivity of claim 2 of Elementary Order Arithmetic in an Ordered Field and hence $d(y,x)\le r$, that is $y\in S_{r}$; and likewise $d(x,y)<\delta'$. Therefore $u(y)<u(x)+\varepsilon$. As $\varepsilon$ was an arbitrary positive real, $u$ is upper semicontinuous at $x$ relative to $S$.

Proof of claim 2. Let $-u:S\to\mathbb{R}$ be the function whose value at $z$ is $-u(z)$, so that $(-u)|_{S_{r}}=-\bigl(u|_{S_{r}}\bigr)$. By claim 1 of Negation, Restriction, and Separated Differences of Semicontinuous Functions, $u$ is lower semicontinuous at $x$ relative to $S$ if and only if $-u$ is upper semicontinuous at $x$ relative to $S$, which by claim 1 above holds if and only if $-\bigl(u|_{S_{r}}\bigr)$ is upper semicontinuous at $x$ relative to $S_{r}$, which by claim 1 of Negation, Restriction, and Separated Differences of Semicontinuous Functions again holds if and only if $u|_{S_{r}}$ is lower semicontinuous at $x$ relative to $S_{r}$.

Proof of claim 3. Let $V\subseteq S$ satisfy $S_{r}\subseteq V$; then $x\in V$, and $\bigl(u|_{V}\bigr)|_{S_{r}}=u|_{S_{r}}$. If $u|_{V}$ is upper semicontinuous at $x$ relative to $V$, then claim 2 of Negation, Restriction, and Separated Differences of Semicontinuous Functions, applied with $S_{r}$ in the role of the smaller set, shows that $u|_{S_{r}}$ is upper semicontinuous at $x$ relative to $S_{r}$, whence $u$ is upper semicontinuous at $x$ relative to $S$ by claim 1. The lower semicontinuous case is identical, using claim 2 in place of claim 1.

Proof of claim 4. For $y\in S$ let $A_{u}(y)$ be the set of those $c\in\mathbb{R}$ for which there is a positive $\rho\in\mathbb{R}$ with $u(z)\le c$ for every $z\in S$ satisfying $d(z,y)\le\rho$, as in the definition of the upper semicontinuous envelope, and for $y\in S_{r}$ let $A_{u|_{S_{r}}}(y)$ be the corresponding set formed from $u|_{S_{r}}$ on $S_{r}$.

Let $y\in S_{r}$ and let $c\in A_{u}(y)$, with witness $\rho$. Since $S_{r}\subseteq S$, every $z\in S_{r}$ with $d(z,y)\le\rho$ lies in $S$ and satisfies $u|_{S_{r}}(z)=u(z)\le c$; hence $c\in A_{u|_{S_{r}}}(y)$. As $A_{u}(y)$ is nonempty by the hypothesis that $u$ is bounded above near each point of $S$, the set $A_{u|_{S_{r}}}(y)$ is nonempty as well. Thus $u|_{S_{r}}$ is bounded above near each point of $S_{r}$ and $\bigl(u|_{S_{r}}\bigr)^{*}$ is defined.

Now let $y\in S$ satisfy $d(y,x)<r$; then $d(y,x)\le r$, so $y\in S_{r}$, and by the previous paragraph $A_{u}(y)\subseteq A_{u|_{S_{r}}}(y)$. For the reverse inclusion let $c\in A_{u|_{S_{r}}}(y)$, with witness $\rho'$. Adding $-d(y,x)$ to both sides of $d(y,x)<r$ and using claim 1 of Elementary Order Arithmetic in an Ordered Field shows that $r-d(y,x)$ is positive, so by claim 9 of Elementary Order Arithmetic in an Ordered Field there is a positive $\rho\in\mathbb{R}$ with $\rho\le\rho'$ and $\rho\le r-d(y,x)$. Let $z\in S$ satisfy $d(z,y)\le\rho$. Adding $d(y,x)$ to the two inequalities $d(z,y)\le\rho$ and $\rho\le r-d(y,x)$, which preserves $\le$ by the compatibility axiom of the ordered field $\mathbb{R}$, and using the triangle inequality, we get

so $z\in S_{r}$ by transitivity of $\le$; moreover $d(z,y)\le\rho\le\rho'$, so $u(z)=u|_{S_{r}}(z)\le c$. Hence $c\in A_{u}(y)$, and $A_{u}(y)=A_{u|_{S_{r}}}(y)$. Both sets are nonempty, and each is bounded below by the value of the function at $y$, as clause Upper and Lower Semicontinuous Envelopes of a Real-Valued Function §upper records; equal sets have the same greatest lower bound, so $\bigl(u|_{S_{r}}\bigr)^{*}(y)=u^{*}(y)$.

Proof of claim 5. The argument is the one just given, with $A_{u}(y)$ replaced by the set $B_{u}(y)$ of the definition of the lower semicontinuous envelope, the inequality $u(z)\le c$ replaced by $c\le u(z)$, and the greatest lower bound by the least upper bound; the choice of the radii $\rho$ and $\rho'$ and the use of the triangle inequality are unchanged. $\blacksquare$