Equivalence between the two definitions of the Lebesgue Integral

Question

Royden defines the Lebesgue integral as :

$$\int_E f = \sup \{\int_E\phi_n:\phi_n \text{ simple}, \phi_n \le f\} \cdots (A)$$

On a later occasion, he defines the Lebesgue Integral as :

$\int_E f = \sup \{\int_Eh:h \text{ is bounded, measurable, of finite support, and } 0 \le h \le f, \phi_n \le f\}$

$\cdots (B)$

Also, the support set $E_0$ for a a measurable function $f$ on a set $E$ is defined as $E_0=\{x \in E | f(x) \ne 0\}$ and $m(E_0)$ is finite. The function f is then identically $0$ on $E−E_0$.

I am trying to prove the equivalence between the two definitions.

Using the Simple Approximation Lemma, we know that $f$ is any extended real valued measurable function $\iff$ it is possible to define a sequence of simple functions $\{\phi_n\}, ~\phi_n \le f$ that converge to $f$

However, using this definition, how does one introduce the set of finite support as required in definition $(B)$? The simple function in step $(A)$ might not even assume the value $0$ anywhere. So, how do we talk about the complement of the support set. Thanks!!

Answer 1

First of all, you have to assume $f\ge 0$. Moreover, if one of the $\sup$ is infinite, then so is the other. Thus let us assume w.o.l.g. that they are finite both. Then any simple function $\phi\le f$ clearly has finite support and is bounded. Thus the simple functions in $(A)$ are a subset of the functions $h$ in $(B)$. Now by your simple approximation lemma, any bounded, measurable $h$ may be approximated arbitrarily precisely by simple functions from below. This gives you the equivalence.