Let $\{f_n\}$ be sequence of real nonnegative functions on $\mathbb{R}^1$, and consider the following statement:
If each $f_n$ is upper semicontinuous (USC), then $\sum \limits_{1}^{\infty} f_n$ is USC. Is this statement true or not?
Proof: Let $f_n=1_{(-\infty, -\frac{1}{n}]\cup [\frac{1}{n}, +\infty)}$. It's easy to see that $f_n$ is USC because it's a characteristic function of closed set and for any fixed $n\in \mathbb{N}$ the sum $\sum \limits_{1}^{n}f_k$ is USC. Then $$\sum \limits_{1}^{\infty}f_k=\sup \limits_{n} \sum \limits_{1}^{n}f_k$$ I'll prove that this infinite sum is not USC function. Let's consider the following set: $$\left\{x: \sum \limits_{1}^{\infty} f_n(x)<\frac{1}{2}\right\}=\left\{x: \sup \limits_{n} \sum \limits_{1}^{n}f_k(x)<\frac{1}{2}\right\}=\bigcap _{n\in \mathbb{N}}\left\{x: \sum \limits_{1}^{n}f_k(x)<\frac{1}{2}\right\}.$$ It's easy to check that $$\left\{x: \sum \limits_{1}^{n}f_k(x)<\frac{1}{2}\right\}=\left\{x: f_1(x)=\dots=f_n(x)=0\right\}=\bigcap _{k=1}^{n}\left\{x: f_k(x)=0\right\}=$$$$\bigcap _{k=1}^{n}\left(-\frac{1}{k},\frac{1}{k}\right)=\left(-\frac{1}{n},\frac{1}{n}\right).$$ Hence $$\left\{x: \sum \limits_{1}^{\infty} f_n(x)<\frac{1}{2}\right\}=\bigcap _{n\in \mathbb{N}}\left(-\frac{1}{n},\frac{1}{n}\right)=\{0\}$$
but the last set is NOT open in $\mathbb{R}^1$. So the above statement is false.
Is my counterexample correct? Would be very grateful for checking out!