Is $\int fg$ equal to the supremum of $\int f_0g_0$, over simple functions $f_0\leq f, g_0\leq g$? Let $f, g$ be positive functions on a measure space $X$ with measure $\mu$. I have seen the definition that
$$\int f \,d\mu = \sup\limits_{0 \leq h \leq f} \int h \,d\mu$$
where the supremum is over simple functions $h$. Is it true that 
$$\int fg\, d\mu = \sup\limits_{\substack{0 \leq f_0 \leq f\\ 0 \leq g_0 \leq g}} \int f_0g_0 d\mu$$
with the supremum over simple functions $f_0, g_0$?
 A: Ok, let $h = fg$ and define $F_0 = \{f_0:X\to\Bbb R| f_0\leq f,f_0 \text{ is simple}\}$ and accordingly $G_0,H_0.$ Let us further denote 
$$
  I[f] = \int_Xf\mathrm d\mu:=\sup_{f_0\in F_0}\int_Xf_0\mathrm d\mu = \sup_{f_0\in F_0}I[f_0].
$$
Your question is whether
$$
  I[h] := \sup_{h_0\in H_0}I[h_0] \stackrel?=\sup_{f_0,\in F_0,g_0\in G_0} I[f_0g_0].
$$
First of all, note that $H_0 \supset \{f_0g_0|f_0\in F_0,g_0\in G_0\}$ since if
$$
  f_0 = \sum_i\alpha_i1_{A_i},\quad g_0 = \sum_j \beta_j1_{B_j}
$$
and $f_0\leq f$, $g_0\leq g$ then $h_0:=f_0g_0\leq fg$ and it is a simple function as well:
$$
  h_0 = \sum_{i,j}\alpha_i\beta_j1_{A_i\cap B_j}.
$$
As a restul, $I[h]\geq \sup_{F_0,G_0}I[f_0g_0]$.
For the converse, let $h_0 = \sum_k \gamma_k1_{C_k}$. If we construct $f_0\in F_0$ and $g_0\in G_0$ in such a way that $f_0g_0\geq h_0$, then from the monotonicity of $I$ on simple functions the desired result will follow. However, I am not sure how to do this construction, let me think of it.
A: Let $f_n$ be a sequence of simple functions converging pointwise to $f$ from below. Define $g_n$ similarly.
Then $f_n g_n$ converges pointwise to $fg$ from below. By the monotone convergence theorem we get that $\int f_n g_n \, d\mu$ converges towards $\int fg \, d\mu$.  
