Background
I am self-studying Donald Cohn's Measure Theory second edition. I got stuck on a step of the proof of Theorem 4.5.1. Here is the theorem:
Theorem$\quad$Let $(X,\mathscr{A},\mu)$ be a measure space, let $p$ satisfy $1\leq p<+\infty$, and let $q$ be defined by $\frac{1}{p}+\frac{1}{q}=1$. If $p=1$ and $\mu$ is $\sigma$-finite, or if $1<p<+\infty$ and $\mu$ is arbitrary, then the operator $T$ defined above is an isometric isomorphism of $L^q(X,\mathscr{A},\mu)$ onto $(L^p(X,\mathscr{A},\mu))^*$.
Proof$\quad$ Since we know that $T$ is an isometry (Proposition 3.5.5), we need only show that it is surjective.
Let $F$ be an arbitrary element of $(L^p(X,\mathscr{A},\mu))^*$. First suppose that $\mu(X)<+\infty$ and that $p$ satisfies $1\leq p<+\infty$. We define a function $\nu$ on the $\sigma$-algebra $\mathscr{A}$ by means of the formula $\nu(A)=F(\langle\chi_A\rangle)$. If $\{A_k\}$ is a sequence of disjoint sets in $\mathscr{A}$ and if $A=\bigcup_kA_k$, then the dominated convergence theorem implies that $\lim_{n\to\infty}\left\|\chi_A-\sum_{k=1}^n\chi_{A_k}\right\|_p=0$; since $F$ is continuous and linear, this implies that $F(\langle\chi_A\rangle)=\sum_k^{\infty}F(\langle\chi_{A_k}\rangle)$ and hence that $\nu(A)=\sum_{k=1}^{\infty}\nu(A_k)$. Thus $\nu$ is countably additive and so is a finite signed or complex measure. It is clear that $\nu$ is absolutely continuous with respect to $\mu$. Hence the Radon-Nikodym theorem provides a function $g$ in $\mathscr{L}^1(X,\mathscr{A},\mu)$ that satisfies $\nu(A)=\int_Agd\mu$ for each $A$ in $\mathscr{A}$. We will show that $g$ belongs to $\mathscr{L}^q(X,\mathscr{A},\mu)$ and that $F(\langle f\rangle)=\int fgd\mu$ holds for each $f$ in $\mathscr{L}^p(X,\mathscr{A},\mu)$.
For each positive integer $n$ let $E_n=\{x\in X:|g(x)|\leq n\}$. Then $g\chi_{E_n}$ is bounded and so belongs to $\mathscr{L}^q(X,\mathscr{A},\mu)$ (recall that $\mu$ is finite.) Define a function $F_{E_n}$ on $L^p(X,\mathscr{A},\mu)$ by $F_{E_n}(\langle f\rangle)=F(\langle f\chi_{E_n}\rangle)$. Consider the relation\begin{align} F_{E_n}(\langle f\rangle) = \int fg\chi_{E_n}d\mu\tag1\end{align}If $f$ is the characteristic function of an $\mathscr{A}$-measurable set $A$, then both sides of (1) are equal to $\nu\left(A\bigcap E_n\right)$; thus (1) holds if $f$ is the characteristic function of an $\mathscr{A}$-measurable set and hence if $f$ is an $\mathscr{A}$-measurable simple function. Since the $\mathscr{A}$-measurable simple functions determine a dense subspace of $L^p(X,\mathscr{A},\mu)$, equation (1) holds for all $\langle f\rangle$ in $L^p(X,\mathscr{A},\mu)$. ....
My Question
Why does the fact that the $\mathscr{A}$-measurable simple functions determine a dense subspace of $L^p(X,\mathscr{A},\mu)$ imply that equation (1) holds for all $\langle f\rangle$ in $L^p(X,\mathscr{A},\mu)$?
What I Have Tried So Far
What I have tried is the following:
Let $\langle f\rangle\in L^p(X,\mathscr{A},\mu)$. Then there is a sequence $\{\langle f_k\rangle\}$ of elements of $L^p(X,\mathscr{A},\mu)$, where for each $k$ the function $f_k\in\mathscr{L}^p(X,\mathscr{A},\mu)$ and $f_k$ is simple, such that$$\lim_{k\to\infty}\|\langle f_k\rangle - \langle f\rangle\|_p = \lim_{k\to\infty}d(\langle f_k\rangle,\langle f\rangle) = 0;$$that is, $\lim_{k\to\infty}\langle f_k\rangle=\langle f\rangle$. Then\begin{align*}F_{E_n}(\langle f\rangle) &= F_{E_n}\left(\lim_{k\to\infty}\langle f_k\rangle\right)\\&= \dots\end{align*}
I got stuck here. I don't know how to proceed next. I wanted to put the limit outside, that is, I wanted to write $$F_{E_n}(\langle f\rangle) = F_{E_n}\left(\lim_{k\to\infty}\langle f_k\rangle\right) = \lim_{k\to\infty}F_{E_n}(\langle f_k\rangle).$$ But I don't think that is legit, I cannot justify this equality.
Could someone please help me out? Thanks a lot in advance!