Folland Theorem 6.14: Why is $ \lim |\int f_n g | \leq M_q(g)$? My question comes from Theorem 6.14 of Folland's Real Analysis:

6.14 Theorem 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
$$ M_q(g) = \sup \left\{\left|\int fg \right|: f \in \sum \text{ and } \|f\|_p = 1 \right\} $$
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 $M_q(g) = \|g \|_q.$

My main question is on the first part of the proof:

Proof. First, we remark that if $f$ is a bounded measurable function that vanishes outside a set $E$ of finite measure and $\| f\|_p = 1$, then $|\int fg | \leq M_q(g)$. Indeed, by Theorem 2.10 there is a sequence $\{f_n\}$ of simple functions such that $|f_n| \leq |f|$ (in particular, $f_n$ vanishes outside $E$) and $f_n \rightarrow f$ a.e. Since $|f_n| \leq \|f\|_{\infty} \chi_E$ and $\chi_E g \in L^1$, by the dominated convergence theorem we have $|\int fg| = \lim |\int f_n g| \leq M_q(g)$...

My question: Why is $\lim |\int f_n g| \leq M_q(g)$? The supremum in $M_q(g)$ is taken over all $f \in \sum$ such that $\|f_p\| = 1$. Clearly $f_n \in \sum$, but $\|f_n\|_p$ need not be 1. I'm thinking that the following proposition may be useful in answering this question:

6.7 Proposition For $1 \leq p < \infty$, the set of simple functions $f = \sum_{i=1}^{n} a_j \chi_{E_j}$, where $\mu(E_j) < \infty$ for all $j$, is dense in $L^p$.

So by this proposition, I assume we can pick a sequence of simple functions $\{f_n\} \subset L^p$ such that $|f_n| \nearrow |f|$ and $\|f_n\|_p \leq \|f \|_p = 1$ for all $n$. (Right?) I'm also somewhat bothered by the fact that the absolute value in the supremum is outside the integral, rather than inside--does this actually matter? i.e. can we say that $\sup \left\{\left|\int fg \right|: f \in \sum \text{ and } \|f\|_p = 1 \right\} = \sup \left\{\int |fg|: f \in \sum \text{ and } \|f\|_p = 1 \right\}$? How would we then put these facts together to conclude that $\lim |\int f_n g| \leq M_q(g)$? I have a feeling I'm overlooking something obvious...
 A: Figured it out. No density arguments needed, just a simple normalization of the $f_n$'s and an application of the Monotone Convergence Theorem: 
Let $\tilde{f}_n := \frac{f_n}{\|f_n \|_p}$. (We can assume $\|f_n \|_p \neq 0$.) Then $\{\tilde{f}_n\} \subset \sum$ and $\|\tilde{f}_n \|_p = 1$ for all $n$. It follows that
\begin{align*}
  \lim_{n \to \infty} \left| \int f_n g \right| &= \lim_{n \to \infty} \left| \int (\tilde{f}_n  \|f_n \|_p) g \right| \\[5pt]
 &= \lim_{n \to \infty} \left(\|f_n \|_p \left| \int \tilde{f}_n g  \right| \right) \\[5pt]
  &\leq \lim_{n \to \infty} \|f_n \|_p M_q(g)  && (\text{Since } \left| \int \tilde{f}_n g \right| \leq M_q(g) \text{ for all } n)  \\[5pt]
  &= M_q(g) \lim_{n \to \infty} \|f_n \|_p \\[5pt]
  &= M_q(g) \|f \|_p && (\text{Monotone Convergence Theorem}) \\[5pt]
  &= M_q(g) && (\|f \|_p = 1)
 \end{align*}
as desired. $\qquad \square$

I'm still curious though: Is 
$$ \sup\left\{ \left| \int fg \right|: f \in \sum \text{ and } \|f\|_p = 1 \right\} = \sup\left\{ \int |fg|: f \in \sum \text{ and } \|f\|_p = 1 \right\} \;? $$
Obviously, we have LHS $\leq$ RHS. I'm wondering if a density argument can be used to show the reverse inequality. 
