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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.

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  • $\begingroup$ Hint: Induction over the number of possible images of $f$, which is finite. You may assume that $A_1,\dotsc, A_n$ are pairwise disjoint. $\endgroup$
    – user251257
    Commented Sep 29, 2016 at 20:41

2 Answers 2

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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.

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  • $\begingroup$ Can you kindly expand on the second part and demonstrate how to show they have a common refinement? $\endgroup$
    – havakok
    Commented Sep 29, 2016 at 20:59
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    $\begingroup$ If $\sum \alpha_k A_k = \sum \beta_i B_i$ just take all the intersections $A_k \cap B_i$ and the coefficients accordingly. It might be easier to assume that $A_i \cap A_j = \emptyset$ for $i \neq j$ and similarly for $B_i$. $\endgroup$
    – user301452
    Commented Sep 29, 2016 at 21:00
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A tidier proof: Assume $f:=\sum \alpha_k \chi_{A_k}=\sum\beta_j\chi_{B_j}=:g$ and wlog that the $A$'s form a disjoint collection and the $B$'s form a disjoint collection. Once you have arrived at $$\int fd\mu=\sum_{k=1}^m\alpha_k\mu(A_k)=\sum_{k=1}^m\sum_{j=1}^l\alpha_k\mu(A_k\cap B_j)$$ and $$\int gd\mu=\sum_{j=1}^l\beta_j\mu(B_j)=\sum_{j=1}^l\sum_{k=1}^m\beta_j\mu(A_k\cap B_j),$$ argue that for each $j,k$ we have $\alpha_k\mu(A_k\cap B_j)=\beta_j\mu(A_k\cap B_j)$:

  • If $\mu(A_k\cap B_j)=0$, there is nothing to prove.

  • Otherwise there exists $x\in A_k\cap B_j$. For this $x$ we have $\alpha_k=f(x)=g(x)=\beta_j$.

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