TheoremStatement: Literature-sourcedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-21

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Lower semicontinuity is equivalent to a closed epigraph and upper semicontinuity to a closed hypograph

Statement

Assume the Axiom of Countable Choice (The Axiom of Countable Choice ( ${A C}_{ω}$ )). Let $n \geq 1$ , let $A \subseteq R^{n}$ , and let $f : A \to R$ . The function $f$ is lower semicontinuous on $A$ if and only if $epi f$ is closed in $A \times R$ . It is upper semicontinuous if and only if $hypo f$ is closed there. The product and relative topologies are those of For $n \geq 1$ the product topology on $n$ copies of the usual topology of $R$ is the metric topology of $d_{\infty}$ on $R^{n}$ , and hence also of $d_{1}$ and $d_{2}$ , so $R^{n}$ as a product and $R^{n}$ as a metric space are one space, Subspace topology: the traces of the open sets, its closed sets and its bases, the continuity of the inclusion, and the characteristic property of a map into a subspace.

Facts & Assumptions

Given: The function, domain, and countable choice in the Statement, with sequential closure in Euclidean metric spaces as in A point lies in the closure of $A$ iff some sequence in $A$ converges to it, and a set is closed iff it is sequentially closed and semicontinuity as in Upper and lower semicontinuity on subsets of $R^{n}$ .

[A1]

The Axiom of Countable Choice supplies a choice function for every family of nonempty sets indexed by $N$ (The Axiom of Countable Choice ( ${A C}_{ω}$ )).

[L1]

Lower semicontinuity is equivalent to relative openness of every strict superlevel set and to relative closedness of every weak sublevel set (Semicontinuity on $R^{n}$ is characterized by strict open level sets and weak closed level sets).

[F1]

The epigraph of $f : A \to R$ is $epi f = {(x, s) \in A \times R : f (x) \leq s}$ (The epigraph and hypograph of a real-valued function).

[L2]

Under ${A C}_{ω}$ , a point belongs to the closure of a metric subspace exactly when a sequence from that subspace converges to it (A point lies in the closure of $A$ iff some sequence in $A$ converges to it, and a set is closed iff it is sequentially closed).

Proof

technique · direct

1.1

For the forward epigraph implication, assume $f$ lower semicontinuous and take $(x, t)$ outside [F1], so $t < f (x)$ . Choose $t < α < f (x)$ . By [L1], a relative neighbourhood of $x$ lies in ${f > α}$ ; its product with a short vertical interval below $α$ is an open neighbourhood of $(x, t)$ disjoint from the epigraph. Thus the epigraph complement is open.

L1F1algebra

1.2

For the reverse epigraph implication, suppose lower semicontinuity fails at $x$ . Then for some $ε > 0$ every relative ball about $x$ meets $S = {y \in A : f (y) \leq f (x) - ε}$ , so $x \in \overline{S}$ . By [A1] and [L2], choose $x_{j} \in S$ with $x_{j} \to x$ . The horizontal points $(x_{j}, f (x) - ε)$ lie in [F1] and converge to $(x, f (x) - ε)$ outside it, contrary to closedness. Hence the lower condition [L1] holds.

A1L1L2F1givenchoose

2.1

The reflection $(x, t) \mapsto (x, - t)$ sends the hypograph of $f$ to the epigraph of $- f$ . Applying steps 1.1 and 1.2 to $- f$ and using the exchange between upper semicontinuity of $f$ and lower semicontinuity of $- f$ proves the hypograph equivalence.

step 1.1step 1.2F1algebra∎

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Sources

D. Bertsekas, MIT 6.253 Convex Analysis and Optimization, Lectures 2 and 4 (standard reference, not scraped)
S. Boyd and L. Vandenberghe, Convex Optimization, §3.1.7 (standard reference, not scraped)