# Integral of Positive Simple Function is Well-Defined

## Theorem

Let $\struct {X, \Sigma, \mu}$ be a measure space.

Let $f: X \to \R, f \in \EE^+$ be a positive simple function.

Then the $\mu$-integral of $f$, $\map {I_\mu} f$, is well-defined.

That is, for any two standard representations for $f$, say:

$\ds f = \sum_{i \mathop = 0}^n a_i \chi_{E_i} = \sum_{j \mathop = 0}^m b_j \chi_{F_j}$

it holds that:

$\ds \sum_{i \mathop = 0}^n a_i \map \mu {E_i} = \sum_{j \mathop = 0}^m b_j \map \mu {F_j}$

## Proof

The sets $F_0, \ldots, F_m$ are pairwise disjoint, and:

$X = \ds \bigcup_{j \mathop = 0}^m F_j$

From Characteristic Function of Disjoint Union, we have:

$\chi_X = \ds \sum_{j \mathop = 0}^m \chi_{F_j}$

Remark that $\map {\chi_X} x = 1$ for all $x \in X$, so that we have:

 $\ds f$ $=$ $\ds \sum_{i \mathop = 0}^n a_i \chi_{E_i} \cdot 1$ $\ds$ $=$ $\ds \sum_{i \mathop = 0}^n a_i \chi_{E_i} \paren {\sum_{j \mathop = 0}^m \chi_{F_j} }$ $\ds$ $=$ $\ds \sum_{i \mathop = 0}^n \sum_{j \mathop = 0}^m a_i \chi_{E_i} \chi_{F_j}$ $\ds$ $=$ $\ds \sum_{i \mathop = 0}^n \sum_{j \mathop = 0}^m a_i \chi_{E_i \cap F_j}$ Characteristic Function of Intersection: Variant 1

Repeating the argument with the $E_i$ and $F_j$ interchanged also yields:

$f = \ds \sum_{j \mathop = 0}^m \sum_{i \mathop = 0}^n b_j \chi_{F_j \cap E_i}$

Now whenever $x \in E_i \cap F_j$, for some $i, j$, then since the $E_i$, $F_j$ are disjoint, we find:

$x \in E_{i'} \cap F_{j'}$ implies $i = i'$ and $j = j'$

Thus, evaluating both expressions for $\map f x$ we find:

$a_i = \map f x = b_j$

In conclusion, we have:

$a_i = b_j$

if $E_i \cap F_j \ne \O$.

Furthermore, we have for all $i$ that:

$\ds E_i = E_i \cap X = E_i \cap \paren {\bigcup_{j \mathop = 0}^m F_j} = \bigcup_{j \mathop = 0}^m \paren {E_i \cap F_j}$

Similarly, we obtain for all $j$:

$\ds F_j = F_j \cap X = F_j \cap \paren {\bigcup_{i \mathop = 0}^n E_i} = \bigcup_{i \mathop = 0}^n \paren {F_j \cap E_i}$

With this knowledge, we compute:

 $\ds \sum_{i \mathop = 0}^n a_i \map \mu {E_i}$ $=$ $\ds \sum_{i \mathop = 0}^n a_i \map \mu {\bigcup_{j \mathop = 0}^m \paren {E_i \cap F_j} }$ $\ds$ $=$ $\ds \sum_{i \mathop = 0}^n a_i \sum_{j \mathop = 0}^m \map \mu {E_i \cap F_j}$ Measure is Finitely Additive Function $\ds$ $=$ $\ds \sum_{i \mathop = 0}^n \sum_{j \mathop = 0}^m a_i \map \mu {E_i \cap F_j}$ Summation is Linear $\ds$ $=$ $\ds \sum_{j \mathop = 0}^m \sum_{i \mathop = 0}^n b_j \map \mu {E_i \cap F_j}$ If $a_i \ne b_j$ then $E_i \cap F_j = \O$; $a_i \cdot 0 = b_j \cdot 0$ $\ds$ $=$ $\ds \sum_{j \mathop = 0}^m b_j \sum_{i \mathop = 0}^n \map \mu {E_i \cap F_j}$ Summation is Linear $\ds$ $=$ $\ds \sum_{j \mathop = 0}^m b_j \map \mu {\bigcup_{i \mathop = 0}^n \paren {E_i \cap F_j} }$ Measure is Finitely Additive Function $\ds$ $=$ $\ds \sum_{j \mathop = 0}^m b_j \map \mu {F_j}$

Hence the result.

$\blacksquare$