## 7.4. Lebesgue Integral

### Example 7.4.2(b): Simple Functions

Are simple functions uniquely determined? In other words, if

No, two simple functions can be the same but have different representations.
For example, take the sets
*s*and_{1}*s*are two simple functions with_{2}*s*, do they have to have the same representation? If different representations are possible, which one is "the best"?_{1}(x) = s_{2}(x)whereand= [0, 1]A= (1, 2]B

and=A'AQ=A''AI

and=B'BQ=B''BI

*are the rationals and*

**Q***are the irrationals. Then all sets are measurable. Define two simple functions*

**I**Obviously the two simple functions have different representations, buts_{1}(x) = X_{A}(x) + X_{B}(x)

s_{2}(x) = X_{A' }(x) + X_{A'' }(x) + X_{B' }(x) + X_{B'' }(x)

*s*for all

_{1}(x) = s_{2}(x)*x*. Note that

*s*is a simple function as well as a step function, while

_{1}*s*is

_{2}*not*a step function.

Simple functions assume only finitely many values. If *s*
is a simple function and *{ a _{1}, ..., a_{n} }*
are the non-zero values of

*s*, define the sets

Then the setsA_{j}= { x: s(x) = a_{j}}

*{*are all disjoint and

**A**_{j}}Therefore, we can always assume without loss of generality that ifs(x) = a_{j}X_{Aj}(x)

*s(x) = c*is

_{j}X_{Ej}(x)*any*simple function, the sets

*are disjoint and the numbers*

**E**_{j}*c*are not zero. This representation (the sets are disjoint and the coefficients are non-zero) is the

_{j}*canonical representation of a simple function*.