I tried to solve the following problem:

Let $(f_n)$ be a sequence of continuous real-valued functions on $\mathbb{R}$ such that:

$$ F = \{x \in \mathbb{R}: \lim_{n\to\infty}f_n(x) = \infty\}$$

Show that $F$ is a Borel set.

My attempt at a solution:

Given that the $f_n$ are all continuous,

$ \forall x \in F \exists \epsilon > 0, B(x,\epsilon) \subset F \tag{1}$

Now, given that the rationals are countable and dense in $\mathbb{R}$, we must have:

$ \forall x \in F \exists q_n \in \mathbb{Q}, x \in B(q_n,\epsilon_n) \subset F \tag{2}$

From $(2)$ it follows that:

$$ F = \bigcup_{n=1}^\infty B(q_n,\epsilon_n) \tag{3}$$

So $F$ is the countable union of open intervals and therefore Borel.

Remark: I discussed my solution with a friend who is taking the same measure theory course and they aren't entirely sure that my solution is correct.

  • 1
    $\begingroup$ Can you clarify what is the question? $\endgroup$
    – user169852
    Dec 18, 2016 at 0:58
  • 1
    $\begingroup$ What are you trying to show? $\endgroup$
    – AJY
    Dec 18, 2016 at 0:58
  • 1
    $\begingroup$ Sorry. I just clarified that I meant to show that F must be Borel. $\endgroup$
    – user93511
    Dec 18, 2016 at 1:01
  • 5
    $\begingroup$ $F = \bigcap_{N=1}^{\infty}\bigcup_{n=1}^{\infty} \bigcap_{k=n}^{\infty}\{x \in \mathbb R : f_k(x) > N\}$ $\endgroup$
    – user169852
    Dec 18, 2016 at 1:06
  • 2
    $\begingroup$ There is no guarantee that your $F$ is open, try $f_n(x)=ne^{-nx^2}$. $\endgroup$
    – Did
    Dec 18, 2016 at 1:07

1 Answer 1


Your (1) is incorrect. Consider sequence of functions: $$f_n(x)=\begin{cases} n^2x+n, x \in [-\frac{1}{n},0]\\-n^2x+n, x \in (0,\frac{1}{n})\\ 0, otherwise \end{cases}$$ Clearly $F=\{0\}$


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