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If $ f, g \in L(I), $ does $ fg \in L(I)$?

I tried to solve this that way:

$f$ and $g$ are Lebesgue integrable for each function there is a sequence of step function ${f_n}$ and ${g_n}$ that satisfy:

  • ${f_n}$ and ${g_n}$ are increasing, bounded and $ f_n \rightarrow f , g_n \rightarrow g $
  • $ \int_{I} f_n, \int_{I} g_n $ is bounded

so the sequence $\{f_n g_n\}$ satisfy those conditions and therefore $fg \in L(I)$

Am I right?

thanks

Community
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Barak Ohana
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  • The issue is the second bullet: the integrals of the products needn't be bounded. – Ian Jun 30 '17 at 15:12

1 Answers1

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Assuming that $I$ is an interval of $\mathbb R$, then the answer is negative. Take $I=(0,1]$, and $f(x)=g(x)=\frac1{\sqrt x}$.

José Carlos Santos
  • 427,504
  • Note that even if we ask for $I$ to be closed, we can just set $f(0)=g(0)=0$, since we don't require any kind of continuity. – Wojowu Jun 30 '17 at 15:03