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Let $X$ be a metric space. For any extended real-valued function $f:X\rightarrow \mathbb{R} \cup \{ -\infty,\infty \}$, we define its upper semicontinuous envelope $Uf$ as follows:

For each $x \in X,$ $Uf(x)=\inf \{ \sup_{y \in U} f(y): U \text{ is a neighbourhood of } x\}.$

Prove that for any two extended real-valued functions $f,g$ defined on $X$, upper semicontinuous envelope satisfies monotonicity: If $f \leq g,$ then $Uf \leq Ug.$

According to this post, I manage to prove the monotonicity by using sequence definition:

Let $(x_n)$ be a sequence converges to $x$ such that $f(x_n)$ converges to $L = Uf(x)$. Then $g(x_n) \geq f(x_n)$. Hence, we have $Ug(x) \geq L = Uf(x).$ Since $x$ is arbitrary, we have $Uf \leq Ug.$

Question: How to prove the monotonicity using infimum-supremum definition?

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  • $\begingroup$ It might be better to use $V$ to represent a neighborhood, instead of $U$, so it does not get mixed up with $Uf$. $\endgroup$ – Michael Apr 11 '17 at 1:50
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Let $\mathcal{N}(x)$ denote the set of neighborhoods of $x$. For each neighborhood $V \in \mathcal{N}(x)$ we have $$f(y) \leq g(y) \quad \forall y \in V$$ now take the $\sup_{y \in V}[\cdot]$ of both sides, then take $\inf_{V \in \mathcal{N}(x)}[\cdot]$ of both sides.


If you want to justify taking the $\inf$ or $\sup$ of both sides of an inequality, consider the general case when we have an index set $\mathcal{R}$ and extended-real-valued functions $w(r)$, $z(r)$ such that $$ w(r) \leq z(r) \quad , \forall r \in \mathcal{R}$$ Then show (for example, using definition of $\sup$ as least upper bound and $\inf$ as largest lower bound):

(i) $\sup_{r \in \mathcal{R}} w(r) \leq \sup_{r \in \mathcal{R}} z(r)$.

(ii) $\inf_{r \in \mathcal{R}} w(r) \leq \inf_{r \in \mathcal{R}} z(r)$

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  • $\begingroup$ Got it. By the way, is my proof using sequential acceptable? $\endgroup$ – Idonknow Apr 11 '17 at 1:58
  • $\begingroup$ I guess the sequence way works if you assume the fact about sequences from that other post. You might fill in the detail on why $\limsup_{n\rightarrow\infty} g(x_n) \leq Ug(x)$. $\endgroup$ – Michael Apr 11 '17 at 7:08

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