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Find continuous functions $f_n(x):\;[0,1]\;\to\;[0,\infty)$ such that $f_n(x)\;\to\;0$ for all $x\in[0,1]$ as $n\to\infty$, $\int_{0}^{1}f_n(x)\,dx\to0$, but $\sup_n f_n$ is not in $\mathcal{L}^1$.

To find such sequence of functions, it necessarily obey the conditions of Lebesgue dominated convergence theorem. So there is no Lebesgue measurable function that can restrict this sequence. But how to construct such a sequence explicitly. This is an exercise from $\textit{Real and Complex Analysis}$ Rudin.

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  • $\begingroup$ Try $f_n(x)=n(n+1)((n+2)x-1)\left[\frac1{n+2}\lt x\lt\frac1{n+1}\right]+n\left[\frac1{n+1}\lt x\lt\frac1n\right]+n^2(1-(n-1)x)\left[\frac1n\lt x\lt\frac1{n-1}\right]$ $\endgroup$ – robjohn Apr 24 '19 at 1:54
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Possible Hint: Try something like $$f_n(x) = \begin{cases}\sqrt{n}\left(1-|2nx-1|\right) & \text{if} \ 0 \le x \le \tfrac{1}{n} \\ 0 & \text{if} \ \tfrac{1}{n} < x \le 1\end{cases}.$$

To help visualize this, the graph of each $f_n$ looks like a triangle with base $\tfrac{1}{n}$ and height $\sqrt{n}$.

It shouldn't be too hard to prove that $f_n(x) \to 0$ for all $x \in [0,1]$ and that $\int_0^1f_n(x)\,dx \to 0$. Computing $\sup_{n}f_n(x)$ exactly is hard, but you might be able to get a lower bound that isn't in $\mathcal{L}^1$.


EDIT: Both robjohn and zhw gave examples for which it is much easier to prove that $\sup_n f_n \not\in \mathcal{L}^1$. However, neither of these examples satisfy the requirement that the functions $f_n$ are continuous.

Here is an example where each $f_n$ is continuous $$f_n(x) = \begin{cases}\dfrac{1}{x}\left(1-\left|2n(n+1)x - (2n+1)\right|\right) & \text{if} \ \tfrac{1}{n+1} \le x \le \tfrac{1}{n} \\ 0 & \text{else}\end{cases}.$$

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