# Are Lebesgue integral definitions equivalent for nonnegative bounded functions on finite support?

According to Royden's Real Analysis, if $$f$$ is a bounded measurable function, with $$m(E)<\infty$$, then the integral is defined as $$\int_E f = \sup \{\int_E\phi:\phi \text{ simple}, \phi \le f\}$$

On the next chapter, if $$f$$ is a nonnegative measurable function, the integral is defined as $$\int_E f = \sup \{\int_Eh: 0 \le h \le f \text{ is bounded, measurable, finite support}\}$$

Then if $$f$$ is a nonnegative, bounded function on a set of $$m(E)<\infty$$, it only makes sense for the two definitions to be equivalent. However, this is not obvious to me, so I want to show it.

This is what I have so far. $$$$\label{eq1} \begin{split} \int f & = \sup \{\int h: 0 \le h \le f, h \text{ is measurable}\} \\ & = \sup \{\sup \{ \int \phi: \phi \text{ simple}, \phi \le h\}: 0 \le h \le f, h \text{ is measurable}\} \end{split}$$$$

I feel like I'm almost done here. In fact, it looks equal to the first definition because intuitively, it's like the squeeze theorem. But I can't quite express it formally, so help would be appreciated.

If $$h$$ is non-negative and the set $$E$$ has a finite measure, then $$h$$ can be approximated uniformly on $$E$$ by non-negative simple functions $$\le h$$ (from below).
Suppose you define the integral using the second definition. Then, for any $$\varepsilon>0$$ there exists $$h$$, measurable with $$0\le h\le f$$ such that $$\int_E f\le \int_E h+\varepsilon$$ Now you can approximate $$h$$ uniformly by a simple functions $$g$$ on $$E$$ to conclude that $$\int_E h\le \int_E g+\varepsilon$$ (true because you can choose $$|g-h|=h-g\le\varepsilon/m(E))$$ and thus you have $$\int_E g\le \int_E f\le \int_E g+2\varepsilon$$ so your function is integrable according to your first definition.