I'm using the definition of lower semi continuous as :
A function $f:X \to \Bbb R$ is said to be lower semi continuous if for each real $\alpha$ the set $\{x \in X: f(x)\le \alpha\}$ is closed , where $X$ is a metric space. Now consider the following example:
Let , $g:X\to \Bbb R$ be continuous and $x_0\in X$. Define , $f:X\to \Bbb R$ by $$ f(x)=\begin{cases}g(x) &\text{, if $x\not=x_0$}\\g(x_0)-1 &\text{ ,if $x=x_0$}\end{cases}$$
Show that $f$ is lower semi continuous.
Here , $f$ is continuous except $x_0$. So , lower semi continuous in $X\setminus \{x_0\}$. Now I want to show that $f$ is l.s.c. at $x_0$.
Let, $x\in S_{\delta}(x_0)$. Since $g$ is continuous so , $|g(x)-g(x_0)|<\epsilon\implies |f(x)-f(x_0)-1|<\epsilon$. Then, $f(x) - f(x_0) <1-\epsilon=\epsilon_1$(say). So , $f$ is l.s.c. at $x_0$.
Is it correct ?