How to prove this expression of lower semi-continuouss?

Question

The definition of lower semi-continuous in wiki (https://en.wikipedia.org/wiki/Semi-continuity) is \begin{equation} \mathop{\lim\inf}_{x\to x_{0}} f(x)\ge f(x_0). \end{equation} However, in some books, inequality is changed to equation, i.e. \begin{equation} \mathop{\lim\inf}_{x\to x_{0}} f(x) = f(x_0). \end{equation} I don't know the latter one is right or not, and i can't give a counterexample to prove that it is incorrect...

Answer 1

It depends on the definition of $\liminf_{x\to x_0}$. If you use $$ \liminf_{x\to x_0} f(x) := \lim_{r\to 0} \inf \left\{ f(x): d(x,x_0)<r, x\ne x_0 \right\} $$ then strict inequality could happen. For example, the function $$ f(x)=\cases{0 &; $x=0$\\ 1 &; $x\in\Bbb R\backslash\{0\}$} $$ is lower semicontinuous at $x=0$ but $\liminf_{x\to 0} f(x)=1> 0 = f(0)$.

However, if your definition does not have the restriction $x\ne x_0$, i.e. $$ \liminf_{x\to x_0} f(x) := \lim_{r\to 0} \inf \left\{ f(x): d(x,x_0)<r \right\} $$ then we can write $\liminf_{x\to x_{0}} f(x) = f(x_0) $ because we'd have $\inf \left\{ f(x): d(x,x_0)<r \right\}\le f(x_0)$ which implies that the equality sign actually holds.