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Let $f,g$ be bounded measurable functions on a set $E$ of finite measure.
Show that:

If $f displaystyle overset{a.e.}= g $ then $displaystyleint_E f= int_E g$

I have this proof from Cupta book, but I'm confused about why $f$ and $g$ need to be bounded and where we benefit from this information in this proof?

What if $f, g$ are nonnegative, and unbounded? What will be exchange?

$quad$ (c) It is sufficient to show that
$$int_E (f-g)=0.$$
Since $f-g=0$ a.e., it follows that if $psigeq f-g$, then $psigeq 0$ a.e. Therefore, in view of Theorem 2.2 (b), we have
$$int_E psigeq 0$$
Thus
$$int_E (f-g)geq 0.$$
Similarly, one can prove that
$$int_E (f-g)leq 0.$$
This proves the result.

In many standard textbooks like Real Analysis by Royden and apparently this one, the Lebesgue integral is first defined for bounded measurable functions on a set of finite measure. With these restrictions the function is integrable when upper and lower Lebesgue integrals are equal and the integral is

$$int_E f = sup left{int_E phi , , | , ,phi text{ simple },,, phi leqslant f right} = inf left{int_E psi , ,| , ,psi text{ simple },,, psi geqslant f right}$$

The basic properties of the integral like linearity, monotonicity and $f overset{a.e.}= g implies int_E f = int_E g$ are proved often by exploiting the assumptions such as boundedness.

Later the integral is defined more generally by building on this foundation. For example, for a nonnegative function $f$ that may be unbounded on a set $E$ that may have infinite measure, the integral is defined as

$$int_E f = sup left{int_E h , , | , ,h text{ bounded, measurable, of finite support },,, 0 leqslant h leqslant f right}$$

This definition relies on the earlier development of the integral for the bounded function $h$. The basic properties are then shown to extend.

The proof you included is clearly introduced at a point where the focus is on the Lebesgue integral for bounded, measurable functions defined on a set of finite measure. It uses two results that exploit boundedness.

(1) Since $f-g$ is bounded and $E$ has finite measure , there exists a simple function $psi$ such that $psi geqslant f-g$ on $E$. This is clear since boundedness implies there exists a number $M$ such that $f - g leqslant M$ on $E$ and $psi = M chi_E$ is a simple function.

(2) For any simple function $psi geqslant 0$ a.e., we have $int_E psi geqslant 0$. This follows almost directly from the definition of the Lebesgue integral of a simple function.

What if f,g are nonnegative and unbounded?

The theorem carries through and follows from:

Let $f$ be a nonnegative, measureable function on a measureable set
$E$. Then $f = 0$ a.e. if and only if $displaystyle int_E f = 0$.

The only if part is easily shown by taking $E_0 = {x in E ,, | ,, f(x) = 0}$. We then have

$$int_E f = int_{E_0} f + int_{E setminus E_0} f$$

The first integral on the RHS is $0$ since the only bounded, measurable function $h$ of finite support on $E_0$ with $0 leqslant h leqslant f$ has $h(x) = 0$ for all $x in E_0$ and $sup_h int_{E_0} h = 0$. The second integral on the RHS is also $0$ by again applying the definitions and using the fact that the measure of $Esetminus E_0$ is $0$.