# Lebesgue integral is linear in simple functions.

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,

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

Attempt:

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.

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.