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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.

  • 7
    Integrable functions may not be almost everywhere bounded.2012-07-26
  • 0
    @superM: no, it's not.2012-07-28

2 Answers 2