Elaboration of a step in the proof of Proposition 9 on Page80 Royden “fourth edition”(2).

Question

The proposition and its proof are given below:

My question is:
In the proof of the backward direction, why $h$ has finite support? does this because our assumption was that $f = 0$ a.e. on $E$ so that the lower Lebesgue integral of $h$ (a bounded measurable function) over a set of finite measure can be defined and because by definition a function that vanishes outside a set of finite measure has a finite support?

Answer 1

Because that's how the authors define the integral of nonnegative measurable functions. See (8) in Definition on page 79 of Royden-Fitzpatrick:

For $f$ a nonnegative measurable function on $E$, we define the integral of $f$ over $E$ by $$ \int_Ef=\{\int_Eh\mid h \text{ bounded, measurable, of finite support and } 0\le h\le f \text{ on } E\} $$

By this definition, to show that $\int_Ef=0$, it suffices to show that $\int_Eh=0$ for all "such" $h$ in the definition above.

For any given such $h$, it is Lebesgue integrable by Theorem 3 (page 73 on Royden-Fitzpatrick). Thus to show that $\int_E h=0$, one can show that $$ \sup\{\int_E\varphi \mid \varphi \text{ simple and } \varphi\le h \text{ on } E\}=0. $$ And since $f$ is nonnegative, it suffices to show that $$ \sup\{\int_E\varphi \mid \varphi \text{ simple and } 0\le\varphi\le h \text{ on } E\}=0. $$