Characterization of Lower Semicontinuity

Theorem
Let $f:S\to\R\cup\left\{-\infty,\infty\right\}$, be an extended real valued function and $S$ is endowed with a topology $\tau$.

The following are equivalent:


 * 1) $f$ is lower semicontinuous on $S$.
 * 2) The epigraph of $f$ is a closed set in $S\times\R$.
 * 3) 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
Assume that $f:S\to\R\cup\left\{-\infty,\infty\right\}$ is lower semicontinuous.

Take a sequence $\left\langle \left(x_n,a_n \right) \right\rangle_{n\in\N}\in\operatorname{epi}f$ so that $\left(x_n,a_n \right)\to\left(x,a\right)$.

This implies that $x_n\to x$ and $a_n\to a$ while $f(x_n)\leq a_n$.

Since $f$ is lower semicontinuous, the sequence $\left\langle f\left(x_n\right)\right\rangle_{n\in\N}$ has at least one limit point.

Equivalently $\left\langle f\left(x_n\right)\right\rangle_{n\in\N}$ has a convergent subsequence.

Let $\left\langle f\left(x_{n_k}\right)\right\rangle_{k\in\N}$ be a subsequence of $\left\langle f\left(x_n\right)\right\rangle_{n\in\N}$ so that $f\left(x_{n_k}\right)\to \beta$. Then $\beta \leq a$ and since:


 * $\displaystyle\liminf_{t\to x} f\left(t\right) = \min\left\{a\in\R\cup\left\{-\infty,\infty\right\},\ \exists \left\langle x_n \right\rangle_{n\in\N} \text{ such that } x_n \to x \right\}$

it follows that $\liminf_{t\to x} f\left(t\right)\leq\beta$.

Therefore $f\left(x\right)=\liminf_{t\to x} f\left(t\right)\leq a$ thus $\left(x,a\right)\in\operatorname{epi}f$.

Closed Epigraph implies Closed Level Sets
Let us assume that $\operatorname{epi}f$ is a closed set in $S\times\R$ and $a\in\R$. Then the set


 * $\operatorname{lev}_{\leq a}=\operatorname{epi}f\cap S\times\left\{a\right\}$

is closed in $S$ because closeness is preserved under intersection

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