Can somebody tell me what's wrong with the following argument?
If $f$ is $L^1$ Lebesgue-integrable, say $f$ positive, then it is bounded almost everywhere by some bound $M$. Then $f^2 < M\cdot f$ which is in $L^1$, then $f$ is in $L^2$ and $L^1$ lies in $L^2$. It seems to me that the map $x^{-1/2}$ is $L^1$ but not $L^2$ on $(0,1)$, hence a counterexample...
So I'm a bit confused.