Integral of Positive Simple Function is Well-Defined

Theorem
Let $\left({X, \Sigma, \mu}\right)$ be a measure space.

Let $f: X \to \R, f \in \mathcal{E}^+$ be a positive simple function.

Then the $\mu$-integral of $f$, $I_\mu \left({f}\right)$, is well-defined.

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


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

it holds that:


 * $\displaystyle \sum_{i \mathop = 0}^n a_i \mu \left({E_i}\right) = \sum_{j \mathop = 0}^m b_j \mu \left({F_j}\right)$

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


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

From Characteristic Function of Disjoint Union, we have:


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

Remark that $\chi_X \left({x}\right) = 1$ for all $x \in X$, so that we have:

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


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

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 $f \left({x}\right)$ we find:


 * $a_i = f \left({x}\right) = b_j$

In conclusion, we have:


 * $a_i = b_j$

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

Furthermore, we have for all $i$ that:


 * $\displaystyle E_i = E_i \cap X = E_i \cap \left({\bigcup_{j \mathop = 0}^m F_j}\right) = \bigcup_{j \mathop = 0}^m \left({E_i \cap F_j}\right)$

by Intersection Distributes over Union: General Result.

Similarly, we obtain for all $j$:


 * $\displaystyle F_j = F_j \cap X = F_j \cap \left({\bigcup_{i \mathop = 0}^n E_i}\right) = \bigcup_{i \mathop = 0}^n \left({F_j \cap E_i}\right)$

With this knowledge, we compute:

Hence the result.