Simple question about the supremum of lebesgue integrals Do we have the following equality?
Let $h$ be measurable and non-negative, $f$ another measurable function and $g$ a step-function, then:
$$\sup_g\left\{\int_X(fg)\,d\mu:0\leq g\leq h\right\}=\sup_g\left\{\int_Xg\,d\mu:0\leq g\leq fh\right\} $$
This seems obvious to me but I want to make sure that this is correct.
Any reasoning for an answer is welcome!
 A: Let $\mathcal{S}$ denote the set of simple finite measurable functions. That is, if $g \in \mathcal{S}$, then we have measurable sets $E_1, \cdots, E_n$ and positive real values $a_1, \cdots a_n$ such that 
$$
g=\sum_{j=1}^n a_j \chi_{E_j}
$$
Now, observe that one of your sets it is actually the integral of $fh$, since by definition 
$$
\int_X fh \ d\mu =\sup_{g \in \mathcal{S}} \left\{ \int_X g \ d\mu : 0 \leq g\leq fh\right\}
$$
$\bullet$ On one side, since $h$ is measurable and non-negative, it is a well known fact that there exist an increasing sequence on $\mathcal{S}$ converging to $h$, i.e. $\{\varphi_n\}\subset \mathcal{S}$ such that $\varphi_n \nearrow h$. Then we have one side of inequality between your sets because 
$$
\sup_{g \in \mathcal{S}} \left\{ \int_X fg \ d\mu : 0 \leq g\leq h\right\} \geq \lim_{n \to \infty} \int_X f\varphi_n \ d\mu = \int_X fh \ d\mu
$$
where the inequality follows by the definition of supremum and the equality by the Monotone Convergence Theorem (MCT).
$\bullet$ On the other hand, to get the other side of the inequality assuming that $f$ is also non-negative, we find $\{ \psi_n \}_n \subset \mathcal{S}$ such that $\psi_n \nearrow f$. Applying a couple of times the MCT, using that $\{ \psi_n\}$ is an increasing sequence (IS), and that  that $\psi_n \geq 0$ (since $\psi_n \in \mathcal{S}$) (+), we get
\begin{align}
\sup_{g \in \mathcal{S}} \left\{ \int_X fg \ d\mu : 0 \leq g\leq h\right\} & \overset{(MCT)}{=} \sup_{g \in \mathcal{S}} \left\{ \lim_{n \to \infty} \int_X \psi_n g \ d\mu : 0 \leq g\leq h\right\} \\
& \overset{(IS)}{=} \lim_{n \to \infty} \sup_{g \in \mathcal{S}} \left\{ \int_X \psi_n g \ d\mu : 0 \leq g\leq h\right\} \\
& \overset{(+)}{=} \lim_{n \to \infty} \sup_{g \in \mathcal{S}} \left\{ \int_X \psi_n g \ d\mu : 0 \leq \psi_n g\leq \psi_n h\right\} \ \ \ \ \ \text{note here that $\psi_ng \in \mathcal{S}$} \\
& \leq \lim_{n \to \infty} \sup_{p \in \mathcal{S}} \left\{ \int_X p \ d\mu : 0 \leq p \leq \psi_n h\right\} \\
& \overset{def}{=} \lim_{n \to \infty}  \int_X \psi_n h \ d\mu \\
& \overset{(TMC)}{=} \int_X fh \ d\mu
\end{align}
Then indeed the other inequality holds and hence your claims follows. 
