Let $f: \mathbb R \rightarrow \mathbb R$ and define $U_a = \{x \, | \, f(x) > a\}$. Show that $f$ is lower semi-continuous (lsc) if and only if $U_a$ is open for all $a$.
First assume $f$ is lsc so that for all $\epsilon > 0$ we have that $f(x_0) – \epsilon < f(x)$ for all $x$ in a $\delta$ neighborhood of $x_0$. Take $u \in U_a$ so that $f(u) > a$. By the lsc property, $f(u) – \epsilon < f(x)$ for all $x$ in some $\delta$ neighborhood of $u$. Taking $\epsilon$ small enough, we have that $f(x) > a$ for all $x$ in this $\delta$ neighborhood. So $U_a$ is open for each $a$.
Next, assume that $U_a$ is open for all $a$. Given a fixed $x_0$, take a fixed $u \in U_{f(x_0)}$ so that $f(u) > f(x_0)$. Since $U_{f(x_0)}$ is open, we can find a neighborhood of $u$ such that $f(x) > f(x_0)$ for every $x$ in this neighborhood. Clearly, this is greater than $f(x_0) – \epsilon$ for all $\epsilon > 0$, so $f$ is lower semicontinuous at each $x_0$.
My question
Is the reasoning here correct? I'm not sure about the last step in the second part since the same neighborhood works for all $\epsilon > 0$.
Best Answer
The statement itself is wrong. $f$ is l.s.c if and only if $\{x: f(x) >a\}$ is open for all $a$. You should take $U_a=\{x: f(x) >a\}$.
I believe you can construct a proof with this modified definition of $U_a$. [If needed, I will provide a proof].
Answer for the modified version: Your proof of the second part is wrong. You did not find any neighborhood of $x_0$ (required to prove l.s.c at $x_0)$. For a correct proof consider $U_{f(x_0)-\epsilon}$. This is a neighborhood of $x_0$. for any $x$ in this neighborhood we have $f(x)>f(x_0)-\epsilon$ which proves that $f$ is l.s.c. at $x_0$.