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I am reading "Measure, Integration & Real Analysis" by Sheldon Axler.

The following theorem is Theorem 3.7 on p.76 in Section 3A in this book.

3.7 integral of a simple function
Suppose $(X,\mathcal{S},\mu)$ is a measure space, $E_1,\dots,E_n$ are disjoint sets in $\mathcal{S}$, and $c_1,\dots,c_n\in [0,\infty]$. Then $$\int \left(\sum_{k=1}^n c_k\chi_{E_k}\right) d\mu=\sum_{k=1}^nc_k\mu(E_k).$$

The author wrote as follows on p.79 in this book.

The representation of a simple function $h:X\to [0,\infty]$ in the form $\sum_{k=1}^n c_k\chi_{E_k}$ is not unique. Requiring the numbers $c_1,\dots,c_n$ to be distinct and $E_1,\dots,E_n$ to be nonempty and disjoint with $E_1\cup\dots\cup E_n=X$ produces what is called the standard representation of a simple function [take $E_k=h^{-1}(\{c_k\})$, where $c_1,\dots,c_n$ are the distinct values of $h$]. The following lemma shows that all representations (including representations with sets that are not disjoint) of a simple measurable function give the same sum that we expect from integration.

3.13 integral-type sums for simple functions
Suppose $(X,\mathcal{S},\mu)$ is a measure space. Suppose $a_1,\dots,a_m,b_1,\dots,b_n\in [0,\infty]$ and $A_1,\dots,A_m,B_1,\dots,B_n\in\mathcal{S}$ are such that $\sum_{j=1}^m a_j\chi_{A_j}=\sum_{k=1}^n b_k\chi_{B_k}.$ Then $$\sum_{j=1}^m a_j\mu(A_j)=\sum_{k=1}^n b_k\mu(B_k).$$

Proof We assume $A_1\cup\dots\cup A_m=X$ (otherwise add the term $0\chi_{X\setminus (A_1\cup\dots\cup A_m)}$).
Suppose $A_1$ and $A_2$ are not disjoint. Then we can write $$a_1\chi_{A_1}+a_2\chi_{A_2}=a_1\chi_{A_1\setminus A_2}+a_2\chi_{A_2\setminus A_1}+(a_1+a_2)\chi_{A_1\cap A_2},\tag{3.14}\label{3.14}$$ where the three sets appearing on the right side of the equation above are disjoint.
Now $A_1=(A_1\setminus A_2)\cup (A_1\cap A_2)$ and $A_2=(A_2\setminus A_1)\cup (A_1\cap A_2);$ each of there unions is a disjoint union. Thus $\mu(A_1)=\mu(A_1\setminus A_2)+\mu(A_1\cap A_2)$ and $\mu(A_2)=\mu(A_2\setminus A_1)+\mu(A_1\cap A_2).$ Hence $$a_1\mu(A_1)+a_2\mu(A_2)=a_1\mu(A_1\setminus A_2)+a_2\mu(A_2\setminus A_1)+(a_1+a_2)\mu(A_1\cap A_2).$$ The equation above, in conjunction with $\ref{3.14}$, shows that if we replace the two sets $A_1,A_2$ by the three disjoint sets $A_1\setminus A_2,A_2\setminus A_1,A_1\cap A_2$ and make the appropriate adjustments to the coefficients $a_1,\dots,a_m$, then the value of the sum $\sum_{j=1}^m a_j\mu(A_j)$ is unchanged (although $m$ has increased by $1$).
Repeating this process with all pairs of subsets among $A_1,\dots,A_m$ that are not disjoint after each step, in a finite number of steps we can convert the initial list $A_1,\dots,A_m$ into a disjoint list of subsets without changing the value of $\sum_{j=1}^m a_j\mu(A_j)$.
The next step is to make the numbers $a_1,\dots,a_m$ distinct. This is done by replacing the sets corresponding to each $a_j$ by the union of those sets, and using finite additivity of the measure $\mu$ to show that the value of the sum $\sum_{j=1}^m a_j\mu(A_j)$ does not change.
Finally, drop any terms for which $A_j=\emptyset$, getting the standard representation for a simple function. We have now shown that the original value of $\sum_{j=1}^m a_j\mu(A_j)$ is equal to the value if we use the standard representation of the simple function $\sum_{j=1}^m a_j\chi_{A_j}$. The same procedure can be used with the representation $\sum_{k=1}^n b_k\chi_{B_k}$ to show that $\sum_{k=1}^n b_k\mu(B_k)$ equals what we would get with the standard representation. Thus the equality of the functions $\sum_{j=1}^m a_j\chi_{A_j}=\sum_{k=1}^n b_k\chi_{B_k}$ implies the equality $\sum_{j=1}^m a_j\mu(A_j)=\sum_{k=1}^n b_k\mu(B_k)$.

The author used the standard representation of a simple function to prove Theorem 3.13.
Why?

Suppose $a'_1,\dots,a'_{m'}\in [0,\infty]$.
Suppose $A'_1,\dots,A'_{m'}$ are disjoint elements of $\mathcal{S}.$
Suppose $\sum_{j=1}^m a_j\chi_{A_j}=\sum_{j=1}^{m'} a'_j\chi_{A'_j}$.
Suppose $\sum_{j=1}^m a_j\mu(A_j)=\sum_{j=1}^{m'} a'_j\mu(A'_j)$.
Suppose $b'_1,\dots,b'_{n'}\in [0,\infty]$.
Suppose $B'_1,\dots,B'_{n'}$ are disjoint elements of $\mathcal{S}.$
Suppose $\sum_{k=1}^n b_k\chi_{B_k}=\sum_{k=1}^{n'} b'_k\chi_{B'_k}$.
Suppose $\sum_{k=1}^n b_k\mu(B_k)=\sum_{k=1}^{n'} b'_k\mu(B'_k)$.
Then, by Theorem 3.7, $$\sum_{j=1}^m a_j\mu(A_j)=\sum_{j=1}^{m'} a'_j\mu(A'_j)=\int\left(\sum_{j=1}^{m'}a'_j\chi_{A'_j}\right)d\mu=\int\left(\sum_{k=1}^{n'}b'_k\chi_{B'_k}\right)d\mu=\sum_{k=1}^{n'} b'_k\mu(B'_k)=\sum_{k=1}^n b_k\mu(B_k).$$
So, I think we don't necessarily need to use the standard representation of a simple function to prove Theorem 3.13.
Am I wrong?


I modified the proof in this book as follows:

3.13 integral-type sums for simple functions
Suppose $(X,\mathcal{S},\mu)$ is a measure space. Suppose $a_1,\dots,a_m,b_1,\dots,b_n\in [0,\infty]$ and $A_1,\dots,A_m,B_1,\dots,B_n\in\mathcal{S}$ are such that $\sum_{j=1}^m a_j\chi_{A_j}=\sum_{k=1}^n b_k\chi_{B_k}.$ Then $$\sum_{j=1}^m a_j\mu(A_j)=\sum_{k=1}^n b_k\mu(B_k).$$

Proof without using a standard representation of a simple function:
Suppose $A_1$ and $A_2$ are not disjoint. Then we can write $$a_1\chi_{A_1}+a_2\chi_{A_2}=a_1\chi_{A_1\setminus A_2}+a_2\chi_{A_2\setminus A_1}+(a_1+a_2)\chi_{A_1\cap A_2},\tag{3.14'}\label{3.14'}$$ where the three sets appearing on the right side of the equation above are disjoint.
Now $A_1=(A_1\setminus A_2)\cup (A_1\cap A_2)$ and $A_2=(A_2\setminus A_1)\cup (A_1\cap A_2);$ each of there unions is a disjoint union. Thus $\mu(A_1)=\mu(A_1\setminus A_2)+\mu(A_1\cap A_2)$ and $\mu(A_2)=\mu(A_2\setminus A_1)+\mu(A_1\cap A_2).$ Hence $$a_1\mu(A_1)+a_2\mu(A_2)=a_1\mu(A_1\setminus A_2)+a_2\mu(A_2\setminus A_1)+(a_1+a_2)\mu(A_1\cap A_2).$$ The equation above, in conjunction with $\ref{3.14'}$, shows that if we replace the two sets $A_1,A_2$ by the three disjoint sets $A_1\setminus A_2,A_2\setminus A_1,A_1\cap A_2$ and make the appropriate adjustments to the coefficients $a_1,\dots,a_m$, then the value of the sum $\sum_{j=1}^m a_j\mu(A_j)$ is unchanged (although $m$ has increased by $1$).
Repeating this process with all pairs of subsets among $A_1,\dots,A_m$ that are not disjoint after each step, in a finite number of steps we can convert the initial list $A_1,\dots,A_m$ into a disjoint list $A'_1,\dots,A'_{m'}$ such that $\sum_{j=1}^m a_j\chi_{A_j}=\sum_{j=1}^{m'} a'_j\chi_{A'_j}$ and $\sum_{j=1}^m a_j\mu(A_j)=\sum_{j=1}^{m'} a'_j\mu(A'_j)$, where $a'_1,\dots,a'_{m'}\in [0,\infty]$.
Similary, repeating the above process with all pairs of subsets among $B_1,\dots,B_n$ that are not disjoint after each step, in a finite number of steps we can convert the initial list $B_1,\dots,B_n$ into a disjoint list $B'_1,\dots,B'_{n'}$ such that $\sum_{k=1}^n b_k\chi_{B_k}=\sum_{k=1}^{n'} b'_k\chi_{B'_k}$ and $\sum_{k=1}^n b_k\mu(B_k)=\sum_{k=1}^{n'} b'_k\mu(B'_k)$, where $b'_1,\dots,b'_{n'}\in [0,\infty]$.
By the hypothesis of Theorem 3.13, $\sum_{j=1}^{m'} a'_j\chi_{A'_j}=\sum_{j=1}^m a_j\chi_{A_j}=\sum_{k=1}^n b_k\chi_{B_k}=\sum_{k=1}^{n'} b'_k\chi_{B'_k}$.
By Theorem 3.7, $$\sum_{j=1}^m a_j\mu(A_j)=\sum_{j=1}^{m'} a'_j\mu(A'_j)=\int\left(\sum_{j=1}^{m'}a'_j\chi_{A'_j}\right)d\mu\\=\int\left(\sum_{k=1}^{n'}b'_k\chi_{B'_k}\right)d\mu=\sum_{k=1}^{n'} b'_k\mu(B'_k)=\sum_{k=1}^n b_k\mu(B_k).$$

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2 Answers 2

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But you haven't written a proof (of the same theorem).

(EDIT: the OP has since considerably changed the post, so what I say may appear misplaced, out of that context).

You've supposed many things but what if I don't suppose the things you suppose?

Let's agree to suppose we are given a measure space $(X,\mathcal{S},\mu)$ and $n,m\in\Bbb N$ and extended nonnegative reals $a_1,\cdots,a_m,b_1,\cdots,b_n$ and measurable $A_1,\cdots,A_m,B_1,\cdots,B_n$ which satisfy $\sum_{j=1}^ma_j\chi_{A_j}=\sum_{k=1}^nb_j\chi_{B_k}$.

I would now like to conclude $\sum_{j=1}^ma_j\mu(A_j)=\sum_{k=1}^nb_k\mu(B_k)$.

In your approach, you "suppose" also that there are disjoint $A'_1,\cdots,A'_{m'},B'_1,\cdots,B'_{n'}\in\mathcal{S}$ and coefficients $a'_1,\cdots,a'_m,b'_1,\cdots,b'_{n'}$ with $\sum_{j=1}^{m'}a'_j\mu(A'_j)=\sum_{j=1}^{m}a_j\mu(A_j)$, and similarly for $B,b$. But how do you know such $A'_\bullet$ and $a'_\bullet$ exist? It's not a hypothesis of the theorem, so the onus is on you to justify the use of these objects. Otherwise, because you have made extraneous suppositions, you're trying to prove a different theorem!

Similarly, you state $\sum_{j=1}^{m'}a'_j\chi_{A'_j}=\sum_{j=1}^ma_j\chi_{A_j}$... this is essentially the same thing as the "standard representation" because you supposed the $A'_\bullet$ to be disjoint. So your proof is essentially saying: "suppose the standard representation satisfies $\sum_{j=1}^{m'}a'_j\mu(A'_j)=\sum_{j=1}^ma_j\mu(A_j)$ and similarly for $B,b$. Then [conclude the result]". But this is no different to the idea of Axler's proof and makes the mistake of assuming something which you don't yet know - I challenge you to justify these "suppose"s without already proving theorem $3.13$.

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    $\begingroup$ @tchappyha The $A'$ have been plucked from thin air, was the main point of my response. But sure, I misspoke - it might not be exactly the standard representation. However, it is essentially the same thing, because if all the $A'_\bullet$ are disjoint it's completely trivial that the sum of $a'_\bullet\mu(A'_\bullet)$ equals the analogous sum for the standard representation. $\endgroup$
    – FShrike
    Commented May 27, 2023 at 16:56
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    $\begingroup$ Your new proof - which I haven't super carefully checked, since it looks about the same as Axler's - avoids explicit use of the standard representation, but breaking everything down into a representation where the sets are disjoint is basically the same thing as getting the standard representation, it is equally as difficult and the idea of the proof is the same. $\endgroup$
    – FShrike
    Commented May 27, 2023 at 16:58
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    $\begingroup$ @tchappyha The first sentence is not necessary at all, since the simple function will be zero outside of $\chi_{X\setminus(A_1\cup\cdots\cup A_m)}$ which means absolutely nothing is being affected by the change... I guess they just wanted to be thorough? But the other sentences are important. The proof idea is to compare all possible representations of the simple function to some "common" one, and for convenience the author chooses this "standard representation" as the baseline for comparison. The sentences about replacing the sets corresponding to each $a_j$, etc., are needed to get (cont.) $\endgroup$
    – FShrike
    Commented May 27, 2023 at 17:10
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    $\begingroup$ Get the standard form. Your proof doesn't use some central object for comparison (well, it uses the well-defined value of the integral as a basis for comparison), so it's a bit different and those sentences are not needed $\endgroup$
    – FShrike
    Commented May 27, 2023 at 17:11
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    $\begingroup$ FShrike, Thank you very much! $\endgroup$
    – tchappy ha
    Commented May 27, 2023 at 17:11
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I don't know if this is less convoluted than the explanation in the OP but here is an attempt to clarify things. First notice that as (real-valued) simple functions take only a finite number of values, any simple function can be expressed uniquely as $$\phi=\sum_{r\in\mathbb{R}\setminus\{0\}}\mathbb{1}_{\{\phi=r\}}$$ The sum above is reality a finite one. This is the canonical representation of a simple function.

Now, suppose there is a measure space $(X,\mathcal{S},\mu)$ in the background and let $\phi$ be a measurable nonnegative simple function. Let $\{a_1,\ldots,a_n\}$ be the set of all the different non--zero values that $\phi$ takes. Set $A_j:=\{\phi=a_j\}\in\mathcal{R}$, then the canonical representation of $\phi$ is $$\phi=\sum^n_{j=1}a_j\mathbb{1}_{A_j}$$ The integral of $\phi$ w.r.t. $\mu$ is then defined as $$\int g\,d\mu:=\sum_{r\in\mathbb{R}\setminus\{0\}}r\mu(\{\phi=r\})=\sum^n_{j=1}a_j\mu(A_j)$$ There is no ambiguity by virtue of the integral being defined on a canonical representation.

If $\phi=\sum^m_{k=1}b_k\mathbb{1}_{B_k}$, $B_k\in\mathcal{S}$, $b_k\geq0$, is a non-canonical representation of $\phi$ (as a linear combination of indicator functions) then the question is whether \begin{align} \sum^m_{k=1}b_k\mu(B_k)=\sum^n_{j=1}a_j\mu(A_j)\tag{0}\label{zero} \end{align}

I let the OP the convince himself/herself that it is enough to assume that the $\{B_1,\ldots,B_m\}\subset\mathcal{S}$ are pairwise disjoint and $b_k\neq0$ for all $k$ (I leave an argument to this effect at the end of the posting).

First, notice that $\bigcup^n_{j=1}A_j=\bigcup^m_{k=1}B_k$, and that if $A_j\cap B_k\neq \emptyset$, then $a_j=b_k$. Hence $a_j\mu(A_j\cap B_k)=b_k\mu(A_j\cap B_k)$ for all $1\leq j\leq n$ and $1\leq k\leq m$. This shows that \begin{align} \sum^n_{j=1}a_j\mu(A_j)&=\sum^n_{j=1}a_j\sum^m_{k=1}\mu(A_j\cap B_j)\\ &=\sum^m_{k=1}\sum^n_{j=1}b_k\mu(A_j\cap B_k)=\sum^m_{k=1}b_k\mu(B_k) \end{align}


If the sets $B_k$ in the (non-canonical) representation of $\phi$ are not disjoint, then we can decompose $\bigcup_kB_k$ in union of disjoint subsets as follows.

Lemma: For any finite collection $\mathcal{I}=\{B_1,\ldots,B_m\}$ of sets in a semiring $\mathcal{R}$, there exists another finite collection $\mathscr{C}=\{C_1,\ldots,C_M\}$ of pairwise disjoint sets in $\mathcal{R}$ such that

  1. For each $C_j\in\mathscr{C}$, there is $B_\ell\in\mathcal{I}$ with $C_j\subset B_\ell$.
  2. For each $B_\ell\in\mathcal{I}$, $B_\ell=\bigcup \{C_j\in\mathscr{C}: C_j\subset B_\ell\}$.

Proof: We proceed by induction on the number of elements of $\mathcal{I}$. For $m=1$ this is obvious. Suppose the result is true for $n$. Let $J=\{B_1,\ldots,B_n,B_{m+1}\}$ and $\mathcal{I}=\{B_1,\ldots,B_m\}$. Let $\mathcal{C}'=\{C_1,\ldots,C_M\}$ be a finite collection of sets in $\mathcal{R}$ for which (i) and (ii) hold for $\mathcal{I}$. Set \begin{align*} \mathcal{C}=\big\{C_j\cap B_{m+1}:1\leq j\leq M\big\}\cup\big\{C_j\setminus B_{m+1}:1\leq j\leq M\big\}\cup\Big\{B_{m+1}\setminus \bigcup^M_{j=1}C_j\Big\} \end{align*} As $\mathcal{R}$ is a semiring, $\mathcal{C}$ is a finite pairwise disjoint collection of sets in $\mathcal{R}$. It is easy to check that $\mathcal{C}$ satisfies (i) and(ii).

In the context of the OP, one can take $\mathbb{R}=\mathcal{S}$, since any $\sigma$-algebra is a semiring.


To complete the argument, notice that $$\phi=\sum^m_{j=1}b_k\mathbb{1}_{B_j}=\sum^M_{k=1}c_k\mathbb{1}_{C_k}$$ where $c_k=\sum \{b_j: C_k\subset B_j\}$. Then \begin{align} \sum^m_{j=1}b_j\mu(B_j)&=\sum^m_{j=1}b_j\sum\{\mu(C):C\in\mathscr{C},\,C\subset B_j\}\\ &=\sum_{C\in\mathscr{C}}\mu(C)\sum_{\substack{1\leq j\leq m\\ C\subset B_j}}b_j=\sum^M_{k=1}c_k\mu(C_k) \end{align}

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