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. \begin{equation} \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} \end{equation}

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.


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