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Lately I've been trying to solve some of the excersises in Rudin's "Real and Complex analysis", and I came across one which I got myself truly stuck on. Specifically, it goes like this:

Find continuous functions $f_n:[0,1]\rightarrow[0,+\infty)$ such that $f_n\rightarrow0$ for all $x\in[0,1]$ as $n\rightarrow\infty$, $\int_0^1f_n(x)\ dx\rightarrow0$, but $\sup\limits_{n\in\Bbb{N}}f_n$ is not in $L^1$.

I've been trying to think of a sequence of functions $f_n$ which converges to $0$ and whose integral also converges to $0$, such that there exists a sub-sequence $f_{n_i}$ which also converges to $0$ (necessarily, as this is a property of convergent real number sequences, that is, any sub-sequence of a convergent real number sequence is also convergent and converges to the same limit), but such that $\lim\limits_{i\to\infty}\int_0^1f_{n_i}=+\infty$. Unfortunately, I can't seem to find such a series, and I would really appreciate some help!

I thank everyone in advance for their answers.

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You essentially just need a sequence $a_n > 0$ such that $a_n \to 0$ but $\sum_{n\in \mathbb N} a_n$ diverges. For example, take $f_n$ to be continuous, non-negative and supported on $(\frac{1}{n+1}, \frac 1n)$, such that $\int_0^1 f_n(x) dx = \frac 1 {\sqrt n}$ (you could make the graph of $f_n$ a triangular spike of the correct height to accomplish this). Then $f_n \to 0$ pointwise and in $L^1$, and since the supports are disjoint, you have $$\sup_{n}f_n = \sum_n f_n$$ and since $\sum_{n} \int^1_0 f_n(x) dx$ diverges, you will not have $\sup_n f_n \in L^1[0,1]$.

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  • $\begingroup$ I found your first answer quite useful. However, regarding your first edit, I would like to remark that the functions' range is limited to $[0,\infty)$, so we can't add a negative spike like you suggested. Not that it is needed: your first interpretation of the problem was indeed the right one. $\endgroup$
    – user246480
    Dec 4 '19 at 17:26
  • $\begingroup$ I see, I have re-edited to delete the edit (I didn't notice that the functions had to be non-negative). $\endgroup$
    – User8128
    Dec 4 '19 at 17:34
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Hint: Let $f_n=n\chi_{[1/(n+1),1/n]}.$ The $f_n$ do everything you want, except for continuity.

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