Real Analysis, Folland Problem 6.2.17 Dual of $L^p$ 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. 
 A: 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.
