Theorem 6.14 - Let $p$ and $q$ be conjugate exponents. Suppose that $g$ is a measurable function on $X$ such that $fg\in L^1$ for all $f$ in the space $\sum$ of simple functions that vanish outside a set of finite measure, and the quantity $$M_q(g) = \sup\{|\int fg|:f \in \sum \ \text{and} \ \|f\|_{p} =1\}$$ is finite. Also, suppose either that $S_g = \{x: g(x)\neq 0\}$ is $\sigma$-finite or that $\mu$ is semifinite. Then $g\in L^q$ and $M_q(g) = \|g\|_{q}$
Problem 6.2.17 - With notation as in Theorem 6.14, if $\mu$ is semifinite, $q < \infty$, and $M_q(g) < \infty$, then $\{x: |g(x)| > \epsilon \}$ has finite measure for all $\epsilon > 0$ has finite measure for all $\epsilon > 0$ and hence $S_g$ is $\sigma$-finite.
As you can tell by the plethora of posts I have on $L^p$ spaces I am really struggling to grasp these concepts and solve these problems. Any advice or suggestions is greatly appreciated.
I've been struggling with this problem too, but here's my attempt mostly based from the proof of Theorem 6.14. A couple of details I'm a little iffy on, so please tell me if something is off!
Let $g \in L^q$ and suppose $M_q(g)= \sup\{|∫fg|:f∈∑ \text{and} ||f||_p=1\} < \infty.$ For some notational convenience later, assume $M_q(g)= m$.
Define $G_\epsilon := \{ x : |g(x)| > \epsilon\}$.
Note that $S_g$ is $\sigma$-finite if each $G_\epsilon$ has finite measure since $S_g = \displaystyle \bigcup_{n=1}^\infty G_{\frac{1}{n}}$.
Assume there is an $\epsilon > 0$ such that $\mu(G_\epsilon) = \infty$.
Since we assume $\mu$ is a semifinite measure, there exists an $F \in \mathcal{M}$ with $F \subseteq G_\epsilon$ with $0 < \mu(F) < \infty$.
Set $c := \mu(F)^{-1/p}$, and $f := c(\overline{sgn(g)})\mathbf{1}_F$.
$\| f \|_{L^p} = (\int |f|^pd\mu)^{\frac{1}{p}} = (\int|c|^p |\mathbf{1}_F|^p d\mu)^\frac{1}{p} = c\mu(F)^{\frac{1}{p}} = 1 $
This means that $f \in \Sigma$ since $f$ restricted to $F^c$ is $0$.
Now, from an earlier exercise we had in Folland (14 from chapter 1?), since $\mu$ is semifinite, for any $N > 0$, $\exists K \subset G_\epsilon$ such that $N < \mu(K) < \infty$. We can simply rechoose $F$ so that it is $K$. (We may be able to just state this for $F$ without needing to rechoose, but I need to think about that more. Any thoughts?)
Let $N = (\frac{m}{\epsilon})^q$, so that $\frac{m}{\epsilon} < \mu(F)^{\frac{1}{q}}$.
So:
$m = \epsilon(\frac{m}{\epsilon}) < \epsilon\mu(F) = c\epsilon\mu(F) = |\int_F c\epsilon| < |\int_F c|g|\mathbf{1}_F|$
(Where the last holds because $F \subset G_\epsilon$)
Continuing,
$|\int_F c|g|\mathbf{1}_F| = \big|\int_F c(\overline{sgn(g)})\mathbf{1}_F sgn(g)|g|\big| = |\int_F fg| \leq |\int fg| \leq m < \infty$.
But this gives $m < m$, which is a contradiction.
Thus $G_\epsilon$ must have finite measure, and we're done.