# Example with $\|f_n\|_\infty=1/n$, integral of $f_n$ is $1$, but $F=\sup\limits_n f_n$ is not integrable

I came across the following challenging problem in my self-study:

Let $\{f_n: \mathbb{R} \rightarrow [0,1]\}$ be functions such that $\mbox{sup}_{x\in \mathbb{R}} f_n(x) = 1/n$ and $\int_{\mathbb{R}} f_n(x)dx = 1$. Set $$F(x) = \mbox{sup}_{n \in \mathbb{N}} f_n(x).$$ Prove that $\int_{\mathbb{R}} F(x) dx = \infty$.

I am having trouble thinking of where to begin in proving this result, and wanted to see if anyone visiting had some suggestions on how to proceed.

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Assume that $F$ is integrable. Then the sequence $(f_n)$ is dominated by an integrable function (namely, $F$) and converges to $f=0$ (since $0\leqslant f_n\leqslant1/n$ uniformly and $1/n\to0$). Lebesgue dominated convergence theorem guarantees that the sequence of integrals of $f_n$ converges to the integral of $f$. But the integral of $f$ is $0$ while the integral of every $f_n$ is $1$. The latter cannot converge to the former hence $F$ is not integrable.

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Another approach: For any integer $m>0$:

$\int_{\mathbb R - [-m,m]} \sup_n f_n(x) d x \geq \int_{\mathbb R -[m,m]} f_{3m}(x) dx = 1 - \int_{-m}^{m} f_{3m}(x) dx \geq 1 - \frac{2}{3} = \frac{1}{3}$

yet $\int_{\mathbb R - [-m,m]}$ should tend to 0 with increasing $m$.

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This looks like a neat problem. I haven't been able to solve it, but it seems like we want to get the harmonic series in on the act somewhere. These were my thoughts:

Let $g_n=\sup_{1\leq i\leq n} f_i$. Then $g_n\nearrow F$, so by monotone convergence, $$\int_\mathbb{R} F(x)dx=\lim_{n\to\infty} \int_\mathbb{R} g_n(x)dx$$

Because $\sup_{x\in \mathbb{R}} f_n(x) = 1/n$ and $\int_{\mathbb{R}} f_n(x)dx = 1$, we must have that $m(\operatorname{supp}\,f_{n})\geq n$. That is, the smaller the $\sup$ is forced to be, the more spread out $f_n$ has to be if its integral will always be 1.

If we could show that for any integers $k$ and $N$, eventually the $f_n$'s spread out so much that for a sufficiently large integer $M\geq N$, $$\int_{(\operatorname{supp} g_{\,N})^c} f_M\geq\frac{1}{k}$$ then by an induction argument we could find a sequence of integers $d_k$ such that $$\large\int_\mathbb{R} g_{d_{\,k+1}}=\int_{\operatorname{supp} g_{\,d_k}} g_{d_{\,k+1}}+\int_{(\operatorname{supp} g_{\,d_k})^c} g_{\,d_{\,k+1}}\geq\left(\sum_{j=1}^k\frac{1}{j}\right)+\frac{1}{k+1}$$ so that $$\int_\mathbb{R} F(x)dx=\lim_{n\to\infty} \int_\mathbb{R} g_n(x)dx=\infty.$$ Perhaps this will help spur someone else to come up with the right idea, or maybe finish off my proposed argument.

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