Show that the supremum of any collection of lower semicontinuous functions is lower semicontinuous.
Suppose $\{f_n\}$ is a sequence of lower semicontinuous functions on a topological space $X$. Define $$g_k=\sup_{n\ge k}f_n.$$ I could see that $$\{x: g_k(x)\gt \alpha\}=\bigcup_{n=k}^{\infty}\{x:f_n(x)\gt \alpha\}$$ from where it follows that $g_k$ is lower semicontinuous. Also $\{g_n\}$ is monotonically decreasing sequence of semicontinuous functions, and because $$\lim\sup_{n\rightarrow\infty} f_n(x)=\inf \{g_1(x), g_2(x), \cdot \cdot \cdot\}$$, $\lim\sup_{n\rightarrow\infty} f_n(x)$ is lower semicontinuous if it can be shown that a monotonically decreasing sequence of lower semicontinuous functions converges to a lower semicontinuous function. How can I show this?
Also, even if the above can be implemented, I would only prove that the supremum of a sequence of lower semicontinuous functions is lower semicontinuous. What's the way generalize this to "any collection" of lower semicontinuous functions?
Thanks.
Best Answer
Note that a function $f$ is lower semicontinuous iff given $x \in X$ and $r < f(x)$ there is an open neighborhood of $U$ of $x$ such that $r < f(y)$ for each $y \in U$.
Suppose that $\{ f_i \}_{i \in I}$ is any collection of lower semicontinuous functions on $X$, and let $g$ be the pointwise supremum, i.e., $g (x) = \sup_{i \in I} f_i(x)$ for all $x \in X$.
Taking $x \in X$, and any $r < g(x) = \sup_{i \in I} f_i(x)$ it must be that there is an $i \in I$ such that $r < f_i(x)$. Since $f_i$ is lower semicontinuous there is an open neighborhood $U$ of $x$ such that $r < f_i(y)$ for each $y \in U$. As $f_i(y) \leq g(y)$ for all $y$, it follows that $r < g(y)$ for each $y \in U$.