Characterization of Lower Semicontinuity

Theorem
Let $f: S \to \overline \R$ be an extended real valued function.

Let $S$ be endowed with a topology $\tau$.

The following are equivalent:


 * $(1): \quad$ $f$ is lower semicontinuous (LSC) on $S$.
 * $(2): \quad$ The epigraph $\map {\operatorname{epi}} f$ of $f$ is a closed set in $S \times \R$ with the product topology.
 * $(3): \quad$ All lower level sets of $f$ are closed in $S$.

Proof
The proof is carried out in the following three steps:

LSC implies Closed Epigraph
Let $f: S \to \overline \R$ be LSC on $S$.

Let $\operatorname{epi} \left({f}\right)$ denote the epigraph of $f$.

Take a sequence $\sequence{\tuple{x_n, a_n}}_{n \mathop \in \N} \in \map {\operatorname{epi}} f$ such that:
 * $\tuple{x_n, a_n} \to \tuple{\bar x, \bar a} \in S \times \R$

as $n \to \infty$.

This implies that $x_n \to \bar x$ and $a_n \to \bar a$ by definition of the product topology on $S \times \R$ and because a Continuous Mapping is Sequentially Continuous.

Consequently:

Thus:
 * $\tuple{\bar x, \bar a} \in \map {\operatorname{epi}} f$

Closed Epigraph implies Closed Level Sets
Let $\map {\operatorname{epi}} f$ be a closed set in $S \times \R$.

Let $a \in \R$.

Then the $\alpha$-lower level set:


 * $\displaystyle \operatorname{lev} \limits_{\mathop \le a} = \map {\operatorname{epi}} f \cap S \times \set{a}$

is closed in $S$ because:
 * closedness is preserved under Intersection|intersection

and:
 * $S \times \set{a}$ is a closed set with respect with the product topology of $S \times \R$.

Closed Level Sets implies LSC
Let all lower level sets of $f$ be closed.

Let $x \in S$.

Let $\displaystyle a: = \liminf_{t \mathop \to x} \map f t$.

In order to prove that $f$ is LSC it suffices to show that $a = \map f x$.

We know already that $a \le \map f x$.

Thus it suffices to show that $\map f x \le a$.

Let $a = \infty$.

As $\infty \le \infty$ the proof is complete.

Let $a < \infty$.

Then we can find a sequence $\sequence{x_n}_{n \mathop \in \N}$ such that:
 * $x_n \to x$ and $\map f {x_n} \to a$

as $n \to \infty$.

For any $b > a$:
 * $\map f {x_n} \le b$

or equivalently:
 * $x_n \in \displaystyle \operatorname{lev} \limits_{\mathop \le b} f$

which by hypothesis is closed.

Therefore $\displaystyle \operatorname{lev} \limits_{\mathop \le b} f$ contains the limit point $x$.

We have that for all $b > a$:
 * $\map f x \le b$

Therefore:
 * $\map f x \le a$

Hence the result.

So:
 * $(1) \implies (2) \implies (3)$

and the proof is complete.