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Possible Duplicate:
Is the pointwise maximum of two Riemann integrable functions Riemann integrable?

Let $f$ and $g$ be two integrable real functions. Is this leads that $\max\{f,g\}$ is integrable too?

Any proof?

Thanks

Rodrigo de Azevedo
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tomerg
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    If you mean Riemann integrable, it was answered in this question: http://math.stackexchange.com/questions/72844/is-the-pointwise-maximum-of-two-riemann-integrable-functions-riemann-integrable – Martin Sleziak Nov 27 '11 at 13:08
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    I voted to close as duplicate, but I might be mistaken. The OP has not clarified whether Riemann or Lebesgue integration is intended. [Future voters: Please wait for some clarification from the OP.] – Srivatsan Nov 27 '11 at 15:16

2 Answers2

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$\max (f,g) = (f+g + |f-g|)/2$, so in the Lebesgue theory max(f,g) is integrable because linear combinations and absolute values of integrable functions are integrable.

Matthew Towers
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    Better to say: Because of this identity, it suffices to prove the special case: If $f$ is integrable, then $|f|$ is integrable. – GEdgar Nov 27 '11 at 17:53
  • @GEdgar Sorry, I don't quite understand the sufficiency. Can you explain a bit why the special case would lead to the identity. (For those who may not know how to prove the special case, here is a proof of it. If $f$ is Lebesgue-integrable, then $\lvert f \rvert$ is integrable by definition. If $f$ is Riemann-integrable, then $f$ is bounded and so $\lvert f \rvert$ is integrable(since Riemann integrals are defined over bounded intervals). ) – Sam Wong Feb 27 '20 at 15:53
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$$ |\max(f,g)|\leqslant\max(|f|,|g|)\leqslant|f|+|g| $$

Did
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    For full credit, needs more explanation. – GEdgar Nov 27 '11 at 17:52
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    GEdgar, thanks for your interest but I like it as is. Kind of similar to the point @Srivatsan made in a comment to the post, if you like. (What do you call full credit, by the way?) – Did Nov 27 '11 at 18:03