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I've been working on a problem. I found a couple of related questions on the site, but I was having a little bit of trouble clarifying everything. The goal is to show that given $f\notin L^p(E)$, there exists a $g\in L^{p'}(E)$ such that $fg\notin L^1(E)$. Note this is the second part of Wheeden and Zygmund's Introduction to Real Analysis textbook.

To show that for $f\notin L^p(E)$, there exists a $g\in L^{p'}(E)$ such that $fg\notin L^1(E)$, consider the following. Without loss of generality, we can assume all functions are nonnegative. Now, suppose there is a sequence $\{g_k\}_{k=1}^\infty\subseteq L^{p'}(E)$ with $\Vert g_k\Vert_{p'}=1$ and $$ \int_E fg_k>3^{k}. $$ Set $$ g=\sum_{k=1}^\infty 2^{-k}g_k $$ and observe that, by Minkowski's inequality, $\Vert g\Vert_{p'}\leq 1$. Note that $$ \int_E fg=\int_E f\sum_{k=1}^\infty 2^{-k}g_k>\int_E\sum_{k=1}^\infty \left(\frac{3}{2}\right)^{k}=\infty. $$ Thus, $g\in L^{p'}(E)$ but $fg\notin L^1(E)$. Hence, we have reduced the problem to showing that such a sequence exists. First note that if $f=\infty$ on any set $A$ of positive measure, then we can simply take

$$g=\frac{1}{|B|}\text{ when }x\in B$$ where $B\subseteq A$ has positive, finite measure, and $g=0$ otherwise. Thus $g$ has the desired properties. This implies we can assume $f$ is finite a.e., and in particular, for any positive real number $c$, we can find a set $F$ with finite measure such that $\int_F f=c$. With this in mind, we can find a nested sequence of sets $\{E_k\}_{k=1}^\infty$, each with finite measure, such that $\bigcup_{k=1}^\infty E_k=E$ and $\int_{E_k}f>3^k$. Now, take $$g_k=\frac{1}{|E_k|^{\frac{1}{p'}}} :\ x\in E_k$$ By construction, $\Vert g_k\Vert_{p'}=1$, and it is easy to check that $\int fg_k\longrightarrow\infty$ as $k\to\infty$. Hence, we have the existence of the sequence, and we've demonstrated the existence of such functions $g$. Does this seem sufficient? The last part seems to be a little bit lacking, but I'm not quite sure why I feel it's unjustified. Thanks in advance!

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On which measure space are you working? Because if $f\equiv 1$ and we have counting measure, the property that for all $c$ we can find $F$ of finite measure such that $\int_Ff=c$ doesn't hold when $c$ is not integer. So we have to replace $=$ by $>$. And in any case it deserves more details. –  Davide Giraudo Dec 1 '12 at 13:21
    
We're using Lebesgue measure on a Lebesgue measurable subset of $\mathbb{R}^n$. –  Clayton Dec 1 '12 at 15:23
    
As it's the crux of the argument, I definitely think you should be more explicit about the construction of the $E_k$. In particular, how you jump from finite ae to the existence of sets with the desired property. –  anonymous Dec 1 '12 at 17:14
    
Since it is a measurable set, we can approximate it by an $F_\sigma$ set. Taking a subsequence as necessary would provide us with such a set as what I've claimed to be $E_k$. From a previous exercise, we can in fact force the set to be compact, thus giving us the measure of the set is finite also. –  Clayton Dec 1 '12 at 19:11
    
Same question from same book, although something different is being asked here. –  leo Dec 2 '12 at 22:39
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