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.


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*}


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


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 $a_i=b_j$.

  • $\begingroup$ 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
  • 1
    $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.