Find continuous functions $f_n:[0,1]\to [0,\infty)$ such that $f_n\to 0$ for all $x\in [0,1]$ as $n\to \infty$, $\int_0^1 f_n(x)dx\to 0$, but $\sup_nf_n$ is not in $L^1$.
My attempt: Define $f_n:[0,\infty)\to [0,\infty)$ as : $f_n (x)=n^3 x$ if $0\leq x\leq \frac{1}{n^2}$, $f_n(x)=n$ if $\frac{1}{n^2}\leq x \leq \frac{2}{n^2}$, and $f_n(x)=-n^3(x-\frac{3}{n^2})$ if $\frac{2}{n^2}\leq x\leq \frac{3}{n^2}$, and $f_n(x)=0$ otherwise. (This looks like a trapezoid)
Then it is clear that $f_n$ is continuous, $f_n\to 0$ pointwise on $[0,1]$, and $\int_0^1 f_n(x) dx\to 0$ as $n\to \infty$, but I can't show that $\sup_n f_n$ is not in $L^1$. Is my example wrong?
Remark that $f_n(2/n^2) = n$, can be written $f(x_n) = \frac{\sqrt{2}}{\sqrt{x_n}}$ for $x_n = 2/n^2$, so I think that in your case $\sup_n f_n = \frac{\sqrt{2}}{\sqrt{x_n}}$ which is integrable. What I say actually only proves that $\sup_n f_n(x)\geq\frac{\sqrt{2}}{\sqrt{x}}$ so one should compute more precisely $\sup_n f_n$ to be sure.
But starting from that, if you want $\sup_n f_n$ to be non integrable, you would like it to be more like $1/x$, so that you replace the height of your trapezoid by $n^2$ instead of $n$. But then the area of the trapezoid $∫f_n$ is not going to $0$! So you can add a little term goind slowly to $0$ and take the height of the trapezoid to be for example $\frac{n^2}{\ln(n)}$. And now the area of the trapezoid is of order $\frac{1}{\ln(n)}$ and so is converging to $0$, but $$\sup_n f_n(x)≥ \frac{2}{x \,(\ln(2)-\ln(x))} = \frac{2}{x \,(\ln(2)+|\ln(x)|)}$$ which is not integrable.