Reverse of Holder's Inequality

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 $p = \infty$

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