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I am looking for a sequence of continuous functions ${f_n}: [0,1] \rightarrow [0,\infty]$ such that $\int f_n d\mu \rightarrow 0$, $f_n(x) \rightarrow 0$ for all $x$, but $f(x) = \sup f_n(x)$ is not in $L_1$.

The "rotating tower" functions with growing heights seem to not converge to zero. A function that rises over reducing intervals, such as $f_n(x) = \sqrt{n} {\bf 1}_{[0,\frac{1}{n})}$ wouldn't satisfy the last condition…

Would you have any suggestions on how I could find such functions?

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Here is a related problem. – David Mitra Jul 31 '13 at 18:47
38 minutes. $ $ – Did Aug 1 '13 at 15:53
up vote 4 down vote accepted


$$f_n(x) = \begin{cases}\frac{1}{x} &, \frac{1}{n+1} \leqslant x < \frac{1}{n}\\ (n+1)2^{n+1}\left(x -\frac{1}{n+1} + \frac{1}{2^{n+1}}\right) &, \frac{1}{n+1} - \frac{1}{2^{n+1}} \leqslant x < \frac{1}{n+1}\\ n2^n\left(\frac1n + \frac{1}{2^n} - x\right) &, \frac{1}{n} < x < \frac1n + \frac{1}{2^n}\\ 0 &, \text{ otherwise}. \end{cases}$$

Then $f_n \to 0$ pointwise, $\int_0^1 f_n(x)\,dx = \log (1 + 1/n) + \frac{n}{2^{n+1}} + \frac{n+1}{2^{n+2}} \to 0$, but $\sup_n f_n(x) = \frac1x \notin L^1$.

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Wow, that did take some effort… I'll try to work it out on my own as well. Thanks. – kumanna Jul 31 '13 at 18:43
Basically, you take a non-integrable function, and cut it into parts, so that the integrals over the parts tend to $0$. With the additional continuity requirement, you need to add steep enough ramps that the integrals still tend to $0$. – Daniel Fischer Jul 31 '13 at 18:49

Try $f_n(x)=(n-n^2\cdot|1-nx|)^+$ for every $n\geqslant1$. Each $f_n$ is the tent function centered at $\frac1n$ with width $\frac2{n^2}$ and height $n$ hence $f_n(x)\to0$ for every positive $x$ since $f_n(x)=0$ for every $n$ large enough, and $\int_0^1f_n=\frac1n\to0$ when $n\to\infty$.

However, $f(x)=\sup\limits_nf_n(x)$ is such that $f(x)\sim\frac1x$ when $x\to0$ hence $f$ is not integrable.

More generally, every $f_n(x)=n\cdot f_1(n^2x-n)$ works, where $f_1$ is continuous with compact support (and not zero).

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