Let $(X,\mathcal{M},\mu)$ be a measure space. Let $s,t: X\to[0,\infty)$ two simple functions. If $E\in\mathcal{M}$, show that,

$$\int_E (s+t)\,d\mu=\int_E s\,d\mu+\int_E t\,d\mu$$


By definition of Lebesgue integral for simple functions, if $s(X)=\{a_i\}_{i=1}^n$ and $t(X)=\{b_i\}_{i=1}^m$ such that $a_i\neq a_j$ and $b_i\neq b_j$ if $i\neq j$ and $A_i=s^{-1}(a_i)$, $B_i=t^{-1}(b_i)$ so $A_i\cap A_j=\emptyset$ and $B_i\cap B_j=\emptyset$ if $i\neq j$, so we have

$$\int_E s\,d\mu+\int_E t\,d\mu=\sum_{k=1}^{n+m}c_k\mu(C_k\cap E)$$

where $c_j=a_j$ and $C_j=A_j$ if $j\in\{1,2,\ldots,n\}$. Also $c_j=b_{j-n}$ and $C_j=B_{j-n}$ if $j\in\{n+1,n+2,\ldots,n+m\}$.

Also $t+s=g$ is simple and $$g(x)=\sum_{k=1}^{n+m}c_j\mathcal{X}_{C_j}(x),$$ then $g$ is simple. But $C_j$ must be not disjoint. I don't know how to conclude.


1 Answer 1


Sorry for my stupidity, I noted that I can consider a colection $E_i$ of disjoint measurable subsets of $E$ such that $\bigcup E_i=E$ and $t,s$ assumes constant values in each $E_i$ (this since images of $t,s$ are finite). And I can conclude easily.


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