## 7.4. Lebesgue Integral

### Example 7.4.4(e): Lebesgue Integral for Simple Functions

We have seen before that the representation of a simple function is
not unique. Show that the Lebesgue integral of a simple function is independent
of its representation.

We have to show that if *s*is a simple function with two different representations, then the integrals using either representation agree.

First assume that there are two representations for *s*

such thatswith_{A}(x) = a_{j}X_{Aj}(x)disjoint andA_{j}anot zero_{j}

swith_{B}(x) = b_{k}X_{Bk}(x)disjoint andB_{k}bnot zero_{k}

*s*. Let's assume that the integral of

_{A}(x) = s_{B}(x)*s*using the first representation exists, i.e. the measure of all sets

*is finite.*

**A**_{j}
If *x B_{k}*
then

*x*must be contained in one of the

*'s because otherwise*

**A**_{j}*s*and

_{B}(x) # 0*s*. Therefore

_{A}(x) = 0*so that*

**B**_{k}**A**_{j}*m(*is finite for all

**B**_{k})*k*so that

*s*is integrable.

_{B}By the same reasoning we have

IfandB_{k}=_{j}(A_{j}B_{k})A_{j}=_{k}(B_{k}A_{j})

*is not empty, then for*

**A**_{j}**B**_{k}*x*we have

**A**_{j}**B**_{k}so thata_{j}= s_{A}(x) = s_{B}(x) = b_{k}

*a*in that case. Putting everything together gives:

_{j}= b_{k}so that the integrals agree regardless of the representation.

It remains to show that the integral of a simple function agrees with the integral over the canonical representation of that function, i.e. when the sets are disjoint and the coefficients are not zero.