My textbook says the $\int s\, d\mu$ is unaffected by how $s$ is written.

Say a simple function is written in two ways $$s = \sum_{i=1}^{m} a_i\chi_{A_i} =\sum_{j=1}^{n} b_j\chi_{B_j} =t$$ then $\int s\, d\mu =\sum_{i=1}^{m} a_i\mu(A_i)= \sum_{j=1}^{n} b_j\mu(B_j)=\int t\, d\mu$

How is that true?

  • $\begingroup$ @user296602 It's not as obvious as you make it out to be with your rhetorical question. The fact that any simple function has many different representations requires us to justify the well defined-ness of our definition of the integral. $\endgroup$
    – Alex Ortiz
    Commented Mar 7, 2018 at 4:10

1 Answer 1


Here is one standard way to do this, following the presentation in Stein and Shakarchi's book on real analysis. Note that any simple function $\varphi$ can be written in a canonical form $\varphi = \sum_{j=1}^Na_j^*\chi_{E_j^*}$ where the $a_j^*$ are distinct and the $E_j^*$ are disjoint by taking $E_j^* = \varphi^{-1}(\{a_j^*\})$.

Lemma. Given a collection of sets $F_1, F_2, \ldots, F_n$ we can construct another collection $F_1^*, F_2^*, \ldots, F_N^*$ with $N=2^n-1$ so that $\bigcup_{k=1}^n F_k = \bigcup_{j=1}^N F_j^*$; the collection of $\{ F_j^*\}$ is disjoint; also $F_k = \bigcup_{F_j^*\subset F_k} F_j^*$, for every $k$.

Proof. For a fixed $k \in \{1,\dots,n\}$, let $F_k^0 = F_k^c$ and let $F_k^1 = F_k$. Consider the sets $F_1^{\epsilon_1}\cap\dots\cap F_n^{\epsilon_n}$ where $\epsilon_k$ is either $0$ or $1$. If we neglect the set $F_1^0\cap\dots\cap F_n^0= F_1^c\cap\dots\cap F_n^c$ which is not a subset of any of the $F_k$, then we get a collection of $2^n-1$ disjoint sets with the desired properties.

Proposition. $\int \varphi$ is well-defined.

Proof. First suppose that the sets $E_k$ in the definition of $\varphi$ are disjoint. Write $\varphi = \sum_{k=1}^na_k\chi_{E_k}$ in its canonical form $\varphi = \sum_{j=1}^Na_j^*\chi_{E_j^*}$ where the $a_j^*$ are distinct and the $E_j^*$ are disjoint. There are disjoint collections $J_1,\dots,J_N\subset\{1,\dots,n\}$ such that for each fixed $j\in\{1,\dots,N\}$, and every $k\in J_j$ we have that $a_k = a_j^*$ and since the $E_k$ are disjoint, that $\sum_{k\in J_j}\chi_{E_k} = \chi_{E_j^*}$. Thus, \begin{align*} \varphi = \sum_{k=1}^na_k\chi_{E_k} &= \sum_{j=1}^N\sum_{k\in J_j}a_k\chi_{E_k} \\ &= \sum_{j=1}^N a_j^* \sum_{k\in J_j}\chi_{E_k} \\ &= \sum_{j=1}^N a_j^* \chi_{E_j^*}, \end{align*} In this case it is clear that the integral is well defined. Now suppose that the sets $E_k$ in the definition of $\varphi = \sum_{k=1}^na_k\chi_{E_k}$ are not disjoint. By applying part a), we may construct disjoint sets $E_j^*$ such that $E_k = \bigcup_{E_j^*\subset E_k} E_j^*$ for every $k$. Now we apply the same procedure we used in the case where the sets $E_k$ in the definition of $\varphi$ were disjoint.


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