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This question is motivated by this one on MathOverflow, which contains an example of a sequence $(f_n)$ of positive continuous functions on $\mathbb{R}$ such that $$f_n(x) \rightarrow \infty$$ if and only if $x \in \mathbb{Q}$.

My question is the following :

For a given sequence of positive continuous functions $(f_n)$ on $\mathbb{R}$, denote by $S((f_n))$ the set of divergence to $\infty$ : $$S((f_n)):=\{ x \in \mathbb{R} : f_n(x) \rightarrow \infty \}.$$

Is there a necessary and sufficient condition for a given set $S$ to be $S((f_n))$ for some sequence $(f_n)$?

Note that $$S((f_n)) = \bigcap_{N} \bigcup_{k} \bigcap_{n \geq k} \{x: f_n(x)>N\},$$ so a necessary condition is that $S$ must be a countable intersection of countable unions of $G_{\delta}'s$...

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A small improvement: since $S((f_n))=\bigcap_N \bigcup_k \bigcap_{n\ge k}\{x\colon f_n(x)\ge N\}$, the set $S$ is a countable intersection of countable unions of closed sets, so a countable intersection of $F_\sigma$s. Also known as $F_{\sigma\delta}$, I believe. This might also be sufficient... – user31373 May 17 '12 at 16:09
    
That's right, thank you for the comment! – Kalim May 17 '12 at 17:01
    
If $g_n:\mathbb R\to[0,1]$ for each $n$, and $f_n= \frac{1}{1-\frac{n}{n+1}g_n}$, then $(f_n(x))\to\infty$ if and only if $(g_n(x))\to 1$. My thought was that it might be easier to work with functions bounded by $1$ and the set where they converge to $1$. – Jonas Meyer May 18 '12 at 5:10
    
@Jonas Meyer : Good remark, thank you for the comment. – Kalim May 19 '12 at 1:50
    
Just to note that this question has now been cross-posted to MathOverflow: mathoverflow.net/questions/97654/… – Willie Wong May 22 '12 at 16:02

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