Restrict $g$ to a dense subclass in $\|f\|_p=\text{sup}\{|\int_Xfg|:\|g\|_{p'}\leq 1\}$ We know that if $1\leq p \leq \infty$ and $f \in L^p(X)$, then $\|f\|_p=\text{sup}\{|\int_Xfg|:\|g\|_{p'}\leq 1$} where $p'$ is the dual exponent of $p$. We want to show that the same expression holds when $g$ is restricted to be a simple function (indicator functions with infinite measure support are alowed). Here is my attempt:
Let $S$ be the space of space of simple functions which are  $p'$ integrable for all $1\leq p\leq\infty$. We know that $S$ is dense in $L^{p'}$. Let $g_n\in S$ be such that $g_n\rightarrow g$. We can choose $g_n$ such that $\|g_n\|\leq1$. We know that $\int_X fg_n\rightarrow \int_X fg$. This is because $\|\int_X f(g-g_n)\|_1\leq\|f\|_p\|g-g_n\|_{p'}$. Hence, $|\int_X fg_n|\rightarrow |\int_X fg|$ and the two supremums should be equal. Is this correct? 
The text which I am following says that we need to argue the case $p=1$ more carefully by using the $\sigma$-finite hypothesis, which is why I am thinking that I am doing something wrong.
 A: I would suspect that you actually don't allow simple functions with infinite measure support in your text. The only case this would be relevant is when $p=1$ i.e. $p'=\infty$, as if $p'<\infty$, then $\|g\|_{p'}<\infty$ would imply that each $\{ x:|g(x)|>1/k\}$ has to be a set of finite measure.
If you do allow infinite measure support simple functions, then I believe your proof is right.
If you only allow simple functions with finite measure support, the result is still correct. In this case, you can show with the same method that for any finite measure $K$,
$$\left| \sup_{\substack{g \text{ simple}\\\operatorname{supp} g \subseteq K\\\|g\|_{\infty}\le 1}}\int_K fg\right |= \|f\|_{L^1(K)}$$
Replacing $K$ with a sequence of finite measure sets $K_n\uparrow X$ (where we used the $\sigma$-finite hypothesis) and using monotone convergence theorem,
$$ \|f\|_{L^1(X)} = \lim_{n\to\infty} \|f\|_{L^1(K_n)}=\lim_{n\to\infty} \left| \sup_{\substack{g \text{ simple}\\ \operatorname{supp} g \subseteq K_n\\\|g\|_{\infty}\le 1}}\int_{K_n} fg\right |$$
Pick a simple function $g_n$ supported in $K_n$ with $\|g_n\|_{L^\infty}\le 1$ such that $$\left|\int_{K_n} fg_n \right|\ge \left| \sup_{\substack{g \text{ simple}\\\operatorname{supp} g \subseteq K_n \\ \|g\|_{\infty}\le 1}}\int_{K_n} fg\right | - 1/n.$$
Then
$$\|f\|_{L^1(X)} \le \left|\lim_{n\to\infty} \int_{K_n} fg_n\right|.  $$
But Holder's inequality gives the reverse inequality, so LHS = RHS; we have just constructed a sequence of simple functions with finite measure support showing that
$$\left| \sup_{\substack{g \text{ simple}\\ |\operatorname{supp} g|<\infty \\|g\|_{\infty}\le 1}}\int_K fg\right |= \|f\|_{L^1(K)}.$$
PS: that you can find such simple functions $g_n$ converging to $g$ in $L^1$  with $\|g\|_{L^1}=1$ and $\|g_n\|_{L^1} \le 1$ is essentially by the definition of the Lebesgue integral. For other $p<\infty$, you can apply the $L^1$ result to $|g|^p\operatorname{sgn}(g)$ (or replace $\operatorname{sgn}$ with $e^{i \operatorname{Arg}g }$ if $g$ is complex valued). For $p=\infty$ you can use the direct construction
$$ g_n := \sum_{k=1}^{4^n} \frac{k}{2^n} \chi_{\{ k2^{-n}\le g < (k+1)2^{-n} \}} - \sum_{k=1}^{4^n} \frac{k}{2^n} \chi_{\{ -(k+1)2^{-n}\le g < -k2^{-n} \}}$$
You can of course truncate this to a finite measure set $K$ to obtain finite measure support simple functions with the same property on $K$.
