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Background

Let $\chi_A$ be the characteristic function of the set $A$. A simple function $s$ is a function of the form $$s(x)=\sum_{i=1}^{n}a_i\chi_{E_i}(x),$$ where $a_i \in \mathbb{R}$ and $E_i$ are measurable sets. A simple function may have more than one representation (consider splitting $E_j$ into two disjoint sets).

If $(X,\mathcal{M},\mu)$ is a measure space and $s$ is as above, then the Lebesgue integral of $s$ is defined as $$\int s \ d\mu=\sum_{i=1}^{n}a_i\mu(E_i).$$

We are to show that the above definition of $\int s \ d\mu$ is consistent in light of the fact that $s$ may have multiple representations.

Attempt

Assume $s=\sum_{i=1}^{m}a_i\chi_{A_i}=\sum_{j=1}^{n}b_j\chi_{B_j}$ and further assume that $A_i$ are disjoint and $B_i$ are disjoint. Then \begin{align*} \sum_{i=1}^{m}a_i\chi_{A_i}&=\sum_{i=1}^{m}a_i\sum_{j=1}^n\chi_{A_i \cap B_j}\\ &=\sum_{j=1}^{n}b_j\sum_{i=1}^m\chi_{A_i \cap B_j}\\ &=\sum_{j=1}^{n}b_j\chi_{B_j}, \end{align*} since $a_i=b_j$ on $A_i \bigcap B_j \ \ ^{(*)}$.

Since $A_i \bigcap B_j$ form a partition of $\bigcup A_i=\bigcup B_i$, we have \begin{align*} \sum_{i=1}^{m}a_i\mu(A_i)&=\sum_{i=1}^{m}a_i\mu\left(\sum_{j=1}^n A_i \cap B_j\right)\\ &=\sum_{j=1}^{n}b_j\mu\left(\sum_{i=1}^m A_i \cap B_j\right)\\ &=\sum_{j=1}^{n}b_j\mu(B_j), \end{align*}

Question

Is this right? I am not too sure about $\ \ ^{(*)}$.

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The (*) formula is not correct. Precisely, it should be described as: If for some $i,j$, $A_i\cap B_j\neq \emptyset$, then a_i=b_j. The proof is as follows. Because $$\sum_{i=1}^ma_i\chi_{A_i}=\sum_{j=1}^nb_j\chi_{B_j},$$ if we take $x\in A_i\cap B_j$ for some $i,j$, then $a_i=b_j$.

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  • $\begingroup$ Is it wrong to assign a value to $A_i\cap B_j$ if the intersection is empty?..other than that, is the rest of my proof all right? $\endgroup$ – illysial Jun 15 '15 at 1:17
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    $\begingroup$ Yes, it is wrong. Other than that, the definition of the Lebesgue is not proper. We usually define the Lebesgue integral as $\int s d\mu=\sum_{i=1}^na_i\chi_{A_i}$ with $A_i$ are disjoint partition of $X$, that is, $\cup_iA_i=X$. You miss this condition. $\endgroup$ – Michael Jun 15 '15 at 2:30

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