I'm having trouble with a problem I came across.
It asks to find a sequence of continuous functions $f_n : [0,1] \rightarrow [0,\infty) $ such that $f_n(x) \rightarrow 0 $ $ \forall x \in [0,1] $, $\int_0^1f_n(x)dx \rightarrow 0$, and that satisfies that $sup_n(f_n) \notin L^1$
I tried constructing functions defined in intervals which decrease as $n \rightarrow \infty$, and that form a trapezium, with increasing height and defined $0$ everywhere else , but I can't get the convergence to $0$ , as the center of the interval in which the function increases won't converge to $0$. I tried solving it moving that interval with $n$, making it cycle over the interval, but that eliminates the convergence in all points, as the function will be arbitrarily big for some $n$.
Their form is this:
$g_{n,i}, i \in \{0..n^2-1\}$
$g_{n,i}(x) = n(n^2x-i+1) $, $x \in [\frac{i-1}{n^2},\frac{i}{n^2}] $
$g_{n,i}(x) = n $, $x \in [\frac{i}{n^2},\frac{i+1}{n^2}] $
$g_{n,i}(x) = n(i+2-n^2x) $, $x \in [\frac{i+1}{n^2},\frac{i+2}{n^2}] $
$g_{n,i}(x) = 0 $ elsewhere
Define $g_n:=n\chi _{[1/(n+1),1/n)}$ then note that all $g_n$ have disjoint support and that $\lim_{n\to \infty }g_n(x)=0$ for all $x\in[0,1]$, and also that $$ \int_0^1g_n(x)\,\mathrm d x=\frac1{n+1}\implies \lim_{n\to \infty }\int_0^1g_n(x)\,\mathrm d x=0 $$ Finally note that $\sup_n g_n=\sum_{n\geqslant 1}g_n$ and so $\sup_{n}g_n\notin L^1$.
Now you can make a very small modification to each $g_n$ to make it continuous, that is, you can define $f_n=g_n$ except in the intervals $[1/(n+1),1/(n+1)+\epsilon _n )$ and $[1/n-\epsilon _n ,1/n)$ where for suitable small $\epsilon _n$ all the properties stated for the sequence $(g_n)$ holds also for $(f_n)$, by example you can set $\epsilon _n:=100^{-n}$.