Exercise 1.3 (Well-definedness of simple integral)

Question

Suppose that an unsigned simple function $f$ has two representations as the linear combination of indicator functions:
$$f = \sum_{i=1}^{n}a_i 1_{E_i} = \sum_{j=1}^{m}b_j 1_{F_j}$$
where $n,m\in\mathbb{N}$, $a_i, b_j$ lie in $[0,+\infty]$, and $E_i,E_j$ are measurable sets. Show that
$$\sum_{i=1}^{n} a_i \mu\left(E_i\right) = \sum_{j=1}^{m}b_j \mu\left(F_j\right)$$

Elu Thingol · Accepted Answer

We use a Venn diagram argument.

\begin{lemma} (Venn diagram argument) 
Let $E_1,\ldots,E_n \subseteq \Omega$ be sets. Then they partition $\mathbb{R}^d$ into $2^n$ disjoint sets, each of which is an intersection of $E_i$ or the complement $\Omega\setminus E_i$ for $i=1,\ldots,n$.
\end{lemma}

\begin{proof} We use induction.
\begin{itemize}
\item For the case $n=2$ we have the collection of $2^2 = 4$ sets: $$I_2 = \left\{E_1\cap E_2, E_1^c \cap E_2, E_1 \cap E_2^c, E_1^c \cap E_2^cight\}.$$ It is easy to verify that all of the above sets are pairwise disjoint and their union is $\Omega$ indeed.
\item Now suppose inductively that the theorem assertion is true for $n\in\mathbb{N}$, i.e, we have the set $I_{n}$ of cardinality $2^n$ consisting of all intersection of $E_1,E_1^c,\ldots,E_n,E_n^c$. Then it is easy to verify that the set $I_{n+1}$ can be written as $$I_{n+1} = \left\{A \cap B: A\in I_{n} 	ext{ and } B=E_{n+1},E_{n+1}^cight\}.$$ The elements of $I_{n+1}$ are disjoint since the elements of $I_n$ are, and it is a partition of $\Omega$ of cardinality $2^{n+1}$, as desired.
\end{itemize}
\end{proof}

Thus, consider the $k+k^{\prime}$ disjoint sets of the collection $I_{n+m}$ constructed as above. We throw away any sets that are empty, leaving us with a partition of $\Omega$ into $k$ non-empty disjoint sets $A_1,\ldots,A_k$ for some $0\leq k \leq 2^{n+m}$. As the $E_1,\ldots,E_n,F_1,\ldots,F_m$ and their complements are measurable, the $A_1,\ldots,A_k$ are too. By construction, each of the $E_1,\ldots,E_n,F_1,\ldots,F_m$ arise as unions of some of the $A_1,\ldots,A_k$, thus we can write
$$E_i = \bigcup_{j \in J_i} A_j, \qquad F_{i^{\prime}} = \bigcup_{j^{\prime} \in J_{i^{\prime}}^{\prime}} A_{j^{\prime}}$$
for all $i=1,\ldots,n$ and $i^{\prime}=1,\ldots,m$, and some subsets $J_i, J_{i^{\prime}}^{\prime} \subseteq \{1,\ldots,k\}$. By finite additivity of measure, we thus have
$$m(E_i) = \sum_{j \in J_i} m(A_j) \qquad m(F_{i^{\prime}}) = \sum_{j\in J_{i^{\prime}}^{\prime}}m(A_j).$$
Thus, our objective is now to show that
$$\sum_{i=1}^{n} a_i m(E_i) 
= \sum_{i=1}^{n} a_i \sum_{i \in J_i} m(A_j) 
\stackrel{!}{=}
\sum_{i^{\prime}=1}^{n} b_{i^{\prime}} \sum_{j \in J_{i^{\prime}}^{\prime}} m(A_j) 
= \sum_{i^{\prime}=1}^{n} b_{i^{\prime}} m(F_{i^{\prime}}).$$
To obtain this, we fix $1 \leq j \leq k$ and evaluate $a_1 1_{E_1} + \ldots + a_n 1_{E_n} = b 1_{F_1}+\ldots+b_m 1_{F_m}$. At such point, $1_{E_i}(x)$ is equal to $1_{J_i}(j)$, and similarly $1_{F_{i^{\prime}}}$ is equal to $1_{J_{i^{\prime}}^{\prime}}(j)$. From this, we conclude that
$$\sum_{i=1}^{n} a_i 1_{J_i}(j) = \sum_{i^{\prime}=1}^{n} b_{i^{\prime}} 1_{J_{i^{\prime}}^{\prime}}(j).$$
Multiplying this by $m(A_j)$ and then summing over all $j=1,\ldots,m$ we obtain the theorem assertion.

Exercise 1.3 (Well-definedness of simple integral)

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Exercise 1.3 (Well-definedness of simple integral)

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