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!

  • 1
    $\begingroup$ Fine, if you're allowing the value $+\infty$ for your sum. You could easily modify this example to get a finite-valued counterxample... $\endgroup$ Commented Jul 14, 2016 at 18:10
  • $\begingroup$ @DavidC.Ullrich, It's not seems easily for me. Can you little bit expand your answer? $\endgroup$
    – RFZ
    Commented Jul 14, 2016 at 18:17
  • $\begingroup$ I'll agree and plus-1 David's comment that the counter-example can be modified, but the way I have in mind seems not so easy (not simply weighting those functions by decaying coefficients, but doing a bit more), perhaps David has an easier way. PS: I was attracted to this question since I am from USC. Haha. $\endgroup$
    – Michael
    Commented Jul 14, 2016 at 18:19
  • $\begingroup$ @DavidC.Ullrich, It would be very interesting if you'll post an answer for finite case. $\endgroup$
    – RFZ
    Commented Jul 14, 2016 at 18:30
  • $\begingroup$ Oops. It's not quite as simple as I was thinking, sorry. $\endgroup$ Commented Jul 14, 2016 at 18:33

1 Answer 1


With regard to the discussion in the comments: Define

$$f_n(x) = \begin{cases} 0 & x < 0\\ nx(1-x)^n & x \in [0,1]\\ 0 & x >0\end{cases}$$

Then each $f_n$ is continuous and nonnegative on $\mathbb R.$ We easily get $\sum f_n(x) < \infty$ for all $x.$ Call the sum $f(x).$ Then $f(1/n) \ge f_n(1/n) = (1-1/n)^n \to 1/e.$ If follows that $f(0) = 0< 1/e\le \limsup_{x\to 0} f(x).$ This violates upper semicontinuity at $0.$

  • $\begingroup$ Looks good. Another way (simpler than what I was originally thinking when I wrote the comment to David) is to paste together functions $g_n(x) = 1_{x \in [1/(n+1), 1/n]}$. That is, consider $g(x)=\sum_{n=1}^{\infty} g_n(x)$. These functions overlap so that $g(x) \in \{0, 1, 2\}$ for all $x \in \mathbb{R}$, and $g(x) = 2$ at points $x \in \{1/n\}_{n=2}^{\infty}$. $\endgroup$
    – Michael
    Commented Jul 14, 2016 at 20:02
  • 2
    $\begingroup$ Your example is easier actually. We could modify it a bit and let g_n be a triangle of height 1 over the base [1/(n+1),1/n]. Then sum g_n is continuous except at 0, where we have the same kind of behavior with the limsup. $\endgroup$
    – zhw.
    Commented Jul 14, 2016 at 20:14
  • 3
    $\begingroup$ Simpler yet: $f_n=\chi_{\{1/n\}}$. $\endgroup$ Commented Jul 14, 2016 at 20:15
  • $\begingroup$ True!$\,\,\,\,\,$ $\endgroup$
    – zhw.
    Commented Jul 14, 2016 at 20:18

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