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My question stems from the responses to this question:

(a) There exists a positive continuous function $f$ on $\mathbb{R}$ so that $f$ is integrable on $\mathbb{R}$, but yet $\lim\sup_{x \to > \infty} f(x) = \infty$.

I understand the idea behind it: basically, make a sequence of "tent functions" where the base shrinks to measure zero while the height goes to infinity. What I don't understand is why we need to sum these tent functions?

Suppose $f_n$ is one of those tent functions and the base is $[n-\frac{1}{2^n},n+\frac{1}{2^n}]$. Then it is only nonzero on a set of measure $2^{-n+1}$. Then by construction, doesn't $\lim_{n \to \infty} f_n$ contain nonzero values on a set of measure zero, since $\lim_{n \to \infty} 2^{-n+1} = 0$? Doesn't that finish the proof?

user1691278
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    The pointwise limit of your $f_n$ functions is the zero function, which does not satisfy the desired condition in the question. – angryavian Feb 18 '19 at 02:33
  • Why is the pointwise limit the zero function? Isn't there still a point where the function is $\infty$? – user1691278 Feb 18 '19 at 02:36
  • Pick any real number $x$. For all sufficiently large $n$, the "tent" of $f_n$ is to the right of $x$, so $\lim_{n \to \infty} f_n(x) = 0$. – angryavian Feb 18 '19 at 02:44
  • @angryavian Thank you. Can you write it out as an answer? I'll accept it. – user1691278 Feb 19 '19 at 00:04

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The pointwise limit of $f_n$ is the zero function, which does not satisfy the desired condition in the question.

Specifically, for a fixed real number $x$, the sequence $(f_n(x))_n$ will eventually be zero as the support $[n - 2^{-n}, n + 2^{-n}]$ of $f_n$'s "tent" will eventually always be to the right of $x$.

angryavian
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