Question about $L^1$-$L^2$ integrable functions

Question

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.

Answer 1

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.

Answer 2

$f \in L^2(a,b) \Rightarrow f \in L^1(a,b)$, but the converse is false! you just gave a counterexample.