Suppose $f: E \rightarrow [0,\infty]$ is measurable. Prove the following: $\sup \{\int \psi \ | \ 0 \leq \psi \leq f, \ \psi \ \text{is a simple function}\}=\sup \{\int_F h \ | \ 0 \leq h \leq f, \ h \ \text{measurable and bounded}, \ F \subset E, \ m(F)<\infty \}$.
I don't have an idea o prove this preposition that is in a measure theory book and I want to know the result soon. Will really appreciate any help...
I would follow Michael Lee's instruction. To see the second part you can argue like this:
Assum $0 \leq h \leq f$ is measurable and bounded, and $F \subset E, m(F)< \infty$. We need to show that there exists a simple function $\psi$ with $ 0 \leq \psi \leq f$ and $\int \psi \, dm \geq \int_F h \, dm$. First observe, that you can construct a monotone sequence of simple functions $f_n$ with $f_n \to 1_F \cdot h $ and $ 1_F \cdot h \leq f_n \leq f $. Thus we get from the monotonic convergence theorem
$$ \lim \int f_n \, dm = \int_F h \, dm $$
Now from $m(F)< \infty$ and the fact that $h$ is bounded, we have $\int_F h \, dm < \infty$. Thus there exists an $n$ with $$\int f_n \, dm \geq \int_F h \, dm $$