Example 7.4.2(b): Simple Functions |
Are simple functions uniquely determined? In other words, if s1 and s2 are two simple functions with s1(x) = s2(x), do they have to have the same representation? If different representations are possible, which one is "the best"? |
A = [0, 1] and B = (1, 2]where Q are the rationals and I are the irrationals. Then all sets are measurable. Define two simple functions
A' = AQ and A'' = A
I
B' = BQ and B'' = B
I
s1(x) = XA(x) + XB(x)Obviously the two simple functions have different representations, but s1(x) = s2(x) for all x. Note that s1 is a simple function as well as a step function, while s2 is not a step function.
s2(x) = XA' (x) + XA'' (x) + XB' (x) + XB'' (x)
Simple functions assume only finitely many values. If s is a simple function and { a1, ..., an } are the non-zero values of s, define the sets
Aj = { x: s(x) = aj }Then the sets { Aj } are all disjoint and
s(x) =Therefore, we can always assume without loss of generality that if s(x) =aj XAj(x)
cj XEj(x)
is any simple function, the sets Ej are
disjoint and the numbers cj are not zero. This
representation (the sets are disjoint and the coefficients are non-zero) is
the canonical representation of a simple function.