Let \((X,d)\) be a metric space. A function \(f \colon X \to \mathbb{R}\) is said to be lower semi-continuous (l.s.c) if \(f^{-1}(a,\infty) = \{x \in X \, \colon \, f(x)> a\}\) is open in \(X\) for every \(a \in \mathbb{R}\). Analogously, \(f\) is upper semi-continuous (u.s.c) if \(f^{-1}(-\infty, b) = \{x \in X \, \colon \, f(x)<b\}\) is open in \(X\) for every \(b \in \mathbb{R}\).
Prove that a function \(f \colon X \to \mathbb{R}\) is continuous if and only if \(f\) is both l.s.c. and u.s.c.
Prove that \(f\) is lower semi-continuous if and only if \(\liminf_{n \to \infty} f(x_n) \geq f(x)\) whenever \(\{x_n\}_{n=1}^\infty \subseteq X\) such that \(x_n \to x\) in \(X\).
Solution
1. Continuous \(\Rightarrow\) l.s.c. and u.s.c.
if \(f\) is continuous, then \(f^{-1}(a,\infty)\) and \(f^{-1}(-\infty,b)\) are preimages of open sets, hence open.
2. l.s.c. and u.s.c. \(\Rightarrow\) continuous.
every open set \(U \subseteq \RR\) is a union of open intervals \((a,b)\); since \(f^{-1}(a,b) = f^{-1}(a,\infty) \cap f^{-1}(-\infty,b)\) is open, \(f^{-1}(U)\) is a union of open sets, hence open.
3. (Part 2, \(\Rightarrow\)) If \(f\) is l.s.c. and \(x_n \to x\), then \(\liminf_n f(x_n) \ge f(x)\).
fix \(a < f(x)\).
4. (Part 2, \(\Leftarrow\)) If \(\liminf_n f(x_n) \ge f(x)\) whenever \(x_n \to x\), then \(f\) is l.s.c.
suppose \(f\) were not l.s.c.: some \(f^{-1}(a,\infty)\) is not open.
5. Q.E.D.