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 ?


1 Answer 1


Probably you have worked out everything in your mind. Your last inequality says $f(x) > f(x_0) + 1- \varepsilon$ for all $\varepsilon > 0$. This means $f(x) \ge f(x_0) + 1 > f(x_0) $. That is the set $\{x : f(x) > f(x_0)\}$ is open.

Another way is to note since you are in metric spaces, lower semicontinuity at point $x_0$ is equivalent to \begin{align*} \liminf_{x \to x_0} \, f(x) \ge f(x_0). \end{align*} The conclusion is then obvious.

Actually the inequality holds true for general topological spaces. We need to consider $x \to x_0$ as a net $\langle x_{\lambda} \rangle$ converging to $x_0$.


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