Suppose $(f_n)$ is a Cauchy sequence in $L^1(X,\Sigma,\mu)$ where $(X,\Sigma,\mu)$ is a measure space. Prove that for each $\epsilon >0$ there is a $\delta >0$ such that if $\mu(E)<\delta$ then $\int_E |f_n|d\mu<\epsilon$ for all $n$.
Can one check my attempt? I think that I am missing some point.
Let $\epsilon >0$. I know that $\exists \delta_n>0$ s.t. $\mu(E)<\delta_n \Rightarrow \int_E|f_n|d\mu<\epsilon$ for all n. Since the set $\{\delta_n>0 : n\in\mathbb{N}\}$ has a lower bound, it has also infimum $\delta$. Then if $\mu(E)<\delta$ then $\int_E |f_n|d\mu<\epsilon$ for all $n$. But I didn't use Cauchy condition.
If we use this with $f_n:=nf$ where $f$ is some integrable function, then the set $\{\delta_n,n\in\mathbb N\}$ has a lower bound, but it is zero. The problem is that this sequence is not Cauchy.
However, if we fix $\varepsilon$ then there is an $N_0$ such that $\int_X |f_n-f_{N_0}|\mathrm d\mu\lt\varepsilon$ for each $n\geqslant N_0$.
The $\delta$ which works for $f_{N_0}$ will also work for $f_n$, $n\geqslant N_0$. Then the finite set $\{\delta_n,n\lt N_0\}$ has a positive lower bound.