TITLE: Question about $L^1$-$L^2$ integrable functions
QUESTION [5 upvotes]: 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.

REPLY [6 votes]: Your argument shows that if $f\in L^1$ and it is bounded a.e., then it is in $L^2$. But, as your own example shows, not every function in $L^1$ is bounded.