Proof of Functions with Closed Superlevel Sets: Sequential Characterisation, Semicontinuity, Perturbation and Limits

Throughout we use Sequential Characterization of Closed Subsets of a Metric Space: a subset of $X$ is closed in $(X,\mathcal{T}_{d})$ if and only if it contains the limit of every convergent sequence of its elements.

A remark on shifted sequences. If $(x_{m})_{m\in\mathbb{N}}$ is a sequence in $A$ converging to $x\in X$ and $M\in\mathbb{N}$, then the sequence $(y_{m})_{m\in\mathbb{N}}$ with $y_{m}=x_{M+m}$ is a sequence in $A$ converging to $x$, and $M\le M+m$ for every $m$. Indeed, $m\le M+m$ and $M\le M+m$ by claim 6 of Properties of the Order on the Natural Numbers together with claim 1 of that lemma, so for a positive $\varepsilon$ and an $N$ with $d(x_{n},x)<\varepsilon$ for $N\le n$ we get $d(y_{m},x)=d(x_{M+m},x)<\varepsilon$ whenever $N\le m$.

Claim 1. Suppose $u$ has closed superlevel sets in $X$. Let $(x_{m})$ be a sequence in $A$ converging to $x\in X$ and let $t\in\mathbb{R}$ satisfy $t\le u(x_{m})$ for every $m\in\mathbb{N}$. Then every term of $(x_{m})$ lies in $\{u\ge t\}$, which is closed in $(X,\mathcal{T}_{d})$, so $x\in\{u\ge t\}$; by the definition of the superlevel set this says $x\in A$ and $t\le u(x)$.

Conversely, assume the stated condition and let $t\in\mathbb{R}$. Let $(x_{m})$ be a sequence in $\{u\ge t\}$ converging to some $x\in X$. Then $(x_{m})$ is a sequence in $A$ with $t\le u(x_{m})$ for every $m$, so the condition gives $x\in A$ and $t\le u(x)$, that is, $x\in\{u\ge t\}$. Hence $\{u\ge t\}$ is closed in $(X,\mathcal{T}_{d})$, and as $t$ was arbitrary, $u$ has closed superlevel sets in $X$.

Claim 2. Let $x\in A$ and let $\varepsilon\in\mathbb{R}$ be positive; put $t=u(x)+\varepsilon$ and $F=\{u\ge t\}$, a closed subset of $X$. We have $x\notin F$: otherwise $u(x)+\varepsilon\le u(x)$, which by claim 3 of Elementary Arithmetic in an Ordered Field would give $\varepsilon\le0$, contradicting $0<\varepsilon$. So $x$ lies in $X\setminus F$, which is open in $(X,d)$ by the definition of a closed subset together with The Open Subsets of a Metric Space Form a Topology. Hence there is a positive $\delta\in\mathbb{R}$ such that every $y\in X$ with $d(x,y)<\delta$ lies in $X\setminus F$.

Let $y\in A$ with $d(x,y)<\delta$. Then $y\notin F$, so $t\le u(y)$ fails; since the order of $\mathbb{R}$ is total, $u(y)<t=u(x)+\varepsilon$. This is exactly upper semicontinuity of $u$ at $x$ relative to $A$, and as $x\in A$ was arbitrary, $u$ is upper semicontinuous on $A$.

Claim 3. Let $w:A\to\mathbb{R}$ be given by $w(x)=u(x)-g(x)$; we verify the condition of claim 1 for $w$. Let $(x_{m})$ be a sequence in $A$ converging to $x\in X$ and let $t\in\mathbb{R}$ satisfy $t\le w(x_{m})$ for every $m$, equivalently $t+g(x_{m})\le u(x_{m})$ by claim 3 of Elementary Arithmetic in an Ordered Field. Since $g$ is continuous on $X$, the sequence $(g(x_{m}))$ converges to $g(x)$ by Continuity Between Metric Spaces is Equivalent to Sequential Continuity.

Let $\eta\in\mathbb{R}$ be positive. Choose $M\in\mathbb{N}$ with $|g(x_{m})-g(x)|<\eta$ for every $m$ with $M\le m$; by claim 9 of Properties of the Absolute Value in an Ordered Field this gives $g(x)-\eta<g(x_{m})$, hence $t+g(x)-\eta\le t+g(x_{m})\le u(x_{m})$ for such $m$, using claim 3 of Elementary Arithmetic in an Ordered Field. By the remark on shifted sequences, $(y_{m})$ with $y_{m}=x_{M+m}$ is a sequence in $A$ converging to $x$ with $t+g(x)-\eta\le u(y_{m})$ for every $m$. Claim 1 now gives $x\in A$ and $t+g(x)-\eta\le u(x)$.

Since this holds for every positive $\eta$, Comparison of Real Numbers with Arbitrary Positive Slack gives $t+g(x)\le u(x)$, that is, $t\le w(x)$. Thus the condition of claim 1 holds for $w$, and $w$ has closed superlevel sets in $X$.

Claim 4. Let $\varepsilon\in\mathbb{R}$ be positive. Since $(u(x_{m}))$ converges to $L$, there is $M\in\mathbb{N}$ with $|u(x_{m})-L|<\varepsilon$ for every $m$ with $M\le m$, hence $L-\varepsilon<u(x_{m})$ for such $m$ by claim 9 of Properties of the Absolute Value in an Ordered Field. By the remark on shifted sequences, $(y_{m})$ with $y_{m}=x_{M+m}$ is a sequence in $A$ converging to $x$ with $L-\varepsilon\le u(y_{m})$ for every $m$. Claim 1 gives $x\in A$ and $L-\varepsilon\le u(x)$. As $\varepsilon$ was an arbitrary positive real number, Comparison of Real Numbers with Arbitrary Positive Slack yields $L\le u(x)$.