True/false : If each $f_n$ is upper semicontnious ,then $\sum_{n=1}^{\infty} f_n$ is upper semicontinious Given $\{f_n\}$ is  a sequence   of real nonnegative  function on $\mathbb{R}^1$
Is the following statement is true/false?
If   each $f_n$ is upper  semicontnious ,then $\sum_{n=1}^{\infty} f_n$ is upper semicontinious
My attempt : I think  this statement is true because $\{\sum_{=1}^{\infty}f_n <\alpha\}=\{f_1+f_2+......+f_n <\alpha\}= \bigcup_{r \in \mathbb{Q}} ( \{x |f_1  < r\} \cap\{ x | f_2 < \alpha - r\}.......\cap \{x|f_n < \alpha -r\} \cap .....$
since countable union of open set is open  ,therefore  we  conclude $\{ x |\sum f_n < \alpha\}$  is open set  for  all $\alpha  \in \mathbb{R}$
Hence  $\sum_{n=1}^{\infty} f_n$ is upper semicontnious
 A: This is false. There exist continuous functions $g_n$ converging pointwise to $\chi_{(0,1)}$. Let $f_1=g_1$ and $f_n=g_n-g_{n-1}$ for $n \geq 2$. Then Each $f_n$ is continuous, hence upper semicontinuous but $\sum f_n= \chi_{(0,1)}$. Since $\{x: \chi_{(0,1)} <\frac  1 2\} =\mathbb R \setminus (0,1)$ is not open it follows that $\sum f_n$ is not upper semicontinuous.
Construction of $(g_n)$: Let $g_n(x) =0$ for $x \geq 0$ as well as for $x \geq 1$ and $g_n(x)=1$ for $\frac  1 n \leq x \leq 1-\frac  1n$. Let $g_n$ have  a straight line graph in the intervals $[0,\frac  1n]$ and $[1-\frac  1n ,1]$. Then $g_n$'s are continuous and converge to $\chi_{(0,1)}$ point-wise.
A: I do not understand your proof. For instance, I can't understand how an infinite sum became a finite one in your first equality.
Anyway, you cannot prove it, since the statement is false. Let $\{r_n\mid n\in\Bbb N\}$ be an enumeration of $\Bbb Q$ and, for each $n\in\Bbb N$, let $f_n$ be the characteristic function of $\{r_n\}$ (that is $f_n(x)=1$ if $x=r_n$ and $f_n(x)=0$ otherwise). Then each $f_n$ is upper semicontinuous, but $\sum_{n=1}^\infty f_n$ (which is $\chi_{\Bbb Q}$) isn't.
