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Assume $X$ is Sigma finite. Assume $f$ is an $M$- Measurable function , $1\leq p \leq \infty$ and

$g \in L^{p} \implies fg\in L^1$

Prove that $f\in L^P $

I was trying to solve this problem on my own but I don't have the complete proof as I am failing to write it down in a proper manner , ( Also there is a claim that I am not so sure of ) Can you please help me formalise the proof properly or point out any mistakes in my ideas

My basic idea ( help needed in formalization )

Assume $f$ does not belong to $L^p$. Defining a new measure $v = \int_A |f|^P du$

Here I intend to show that $X,M,v$ is $\sigma $ finite . Then I was hoping that I could obtain a function h such that

$h \in L^{p} (X,M,v)$ such that $ h$ does not belong to $ L^1(X,M,v). $ Then I will let $ g = h|f|^{p-1} $ leading to final step of my proof. However I still have no clue on how to proceed with the case of $ p = \infty$

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  • $\begingroup$ I don't understand, what's $p'$? $\endgroup$ – man_in_green_shirt Apr 13 '17 at 8:12
  • $\begingroup$ $p'$ was right, the "corrected" statement is wrong (and $1/p+1/p'=1$). $\endgroup$ – user138530 Apr 15 '17 at 0:12
  • $\begingroup$ Yes you are correct @ChristianRemling $\endgroup$ – Noob101 Apr 15 '17 at 6:32

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