Show that the Lebesgue integral is well-defined for simple functions,
I am a bit confused and not sure if my proof is any good.
I want to show that the value of the expression $\sum_{k=1}^n\alpha_k\mu(A_k)$, does not depend on the choice of the simple-function representation of $f$.
Suppose that $f,g$ are simple functions and $f\leq g$, then they can be represented by:
$$f=\sum_{k=1}^m\alpha_k\chi_{A_k}$$
$$g=\sum_{j=1}^l\beta_j\chi_{B_j}$$
Can I say here that $\cup_{k=1}^mA_k=\mathbb R^n$? How may i show it?
$$\int fd\mu=\sum_{k=1}^m\alpha_k\mu(A_k)=\sum_{k=1}^m\sum_{j=1}^l\alpha_k\mu(A_k\cap B_j)$$
$$\int gd\mu=\sum_{j=1}^l\beta_j\mu(B_j)=\sum_{j=1}^l\sum_{k=1}^m\beta_j\mu(A_k\cap B_j)$$
If $\mu(A_k\cap B_j)\neq0$ then in particular $A_k\cap B_j \neq \phi$, so there exist $x\in A_k \cup B_j$ but then:
$$\alpha_k=f(x)\leq g(x)=\beta_j$$
And if $\mu(A_k\cap B_j)=0$ the the inequality is trivially valid and so:
$$\int fd\mu\leq \int gd\mu$$
and if $g=f$ we are simply dealing with representation of one simple function $f$.
Is this a good proof? If not please point out the correct way.
Best Answer
I don't see how you obtain $\alpha_k = f(x) \leq g(x) = \beta_j$.
I would show that if you have a simple function $\sum_{k = 1}^m \alpha_k \chi_{A_k}$, then a refinement doesn't change the value of the integral, i.e. if we take $B_{k,i} \subseteq A_k$ disjoint with $\bigcup_{i = 1}^{n_k} B_{k,i} = A_k$, then $$\int \sum_{k = 1}^m \alpha_k \chi_{A_k} = \int \sum_{k = 1}^m \sum_{i = 1}^{n_k} \beta_{k,i} \chi_{B_{k,i}}$$ with $\beta_{k,i} = \alpha_k$. Afterwards you show that if you have two representations of a simple function, then they have a common refinement.