This question originated from a problem on Baire spaces: Contradiction to Baire's theorem with $\{f\in L^2([0,1]):\int_0^1|f_j|^2\leq j\}$?
Show $L^2([0,1])$ is not complete with the $L^1$-norm.
So I want to find functions $f_n \in L^2([0,1])$ which are Cauchy wrt $L^1$-norm, i.e $\int |f_n-f_m|\to 0$ as $n,m\to \infty$ but whose limit $f\notin L^2([0,1])$. I know $x^{-1/2} \notin L^2([0,1])$ so I tried $f_n(x)=x^{-1/2+1/n}$ but I don't think this is Cauchy as $|f_n(x)-f_m(x)|=|x^{-1/2+1/m}(x^{1/n-1/m}-1)|$ and $x^{-1/2+1/n}$ is unbounded in $[0,1]$. @Ian suggested me to find a family of functions $f_{\epsilon,A}$ such that $||f_{\epsilon,A}||_{L^1}<\epsilon$ and $||f_{\epsilon,A}||_{L^2}=A$ but I couldn't do that.
Let $f_n(x)=0$ for $x\in [0,1/n)$ and $f_n(x)=1/x^{-1/2}$ for $x\in [1/n,1].$
Let $f(0)=0$ and $f(x)=x^{-1/2}$ for $x\in (0,1].$
Details. Each $f_n$ is bounded and Lebesgue-measurable so each $f_n$ belongs to $L^1[0,1] $ and to $L^2[0,1].$ And $\int_0^1|f(x)|dx=2<\infty$ so $f\in L^1[0,1].$
We have $\|f-f_n\|_1=\int_0^{1/n}|x^{-1/2}|dx=2n^{-1/2}\to 0$ as $n\to \infty.$
But $f\not \in L^2[0,1]$ because $\int_0^1|f(x)|^2dx=\int_0^1 x^{-1}dx=\infty.$