# Uniform sum of positive upper semicontinuous functions is upper semicontinuous?

Let $X$ be a metric space. A real-valued function $f : X \rightarrow \mathbb{R}$ is upper semicontinuous if it satisfies one of the followings:

$(1)$ For all $c \in \mathbb{R}$, its preimage $f^{-1}(-\infty,c)$ is open in $X$.

$(2)$ For all $x \in X$ and all $\varepsilon>0$, there exists an open neighbourhood $U$ of $x$ such that for all $y \in U,$ we have $f(y) < f(x) + \varepsilon$.

I know that finite sum of upper semicontinuous is upper semicontinuous.

Question: Suppose for each natural number $n$ , $f_n$ is a non-negative upper semicontinuous function and $(f_n)$ is decreasing. Assume that $\sum_{n=1}^{\infty}(-1)^nf_n$ converges uniformly to $g$. Is $g$ an upper semicontinuous function?

• The sum of u.s.c. function is u.s.c., but their difference need not be u.s.c. – Rigel May 6 '17 at 13:27
• How about uniform sum of usc? – Idonknow May 6 '17 at 15:12

Define $f_1=\chi_{[0,3]}$, $f_2=\chi_{[1,2]}$. Then $f_1-f_2$ is non-negative but not upper semicontinuous: $$(f_1-f_2)^{-1}(-\infty,1/2)=(-\infty,0) \cup [1,2]\cup (3,\infty),$$ which is not open.
By setting $f_n=0$ for $n>2$ all the conditions in your question are satisfied, but the resulting function is not upper semicontinuous.