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Let $( X , \mathcal{A} , \mu)$ be measurable space, meaning that X are sets equipped with respective $\sigma$ -algebra $\mathcal{A}$ and $\mu:\mathcal{A} \rightarrow [0,\infty]$ with $\mu( \bigcup_{j=1}^\infty A_j)= \sum_{j=1}^{\infty} \mu(A_j) $ for $j \not= k: A_j \cap A_k = \emptyset $ and $\mu(\emptyset)=0$

A measurable function $s : X → \mathbb{R} $ is said to be a simple function if it´s range is finite, that means $s(x)= \sum_{j=1}^p \alpha_j 1_{A_j} (x)$ with $s(X)=\{\alpha_1,..,\alpha_p\}, A_j:= s^{-1} (\alpha_j) $

Definition: Be $s : X → \mathbb{R} $ a nonnegative simple function and $A \in \mathcal{A}$ so we define $\int_A s d\mu := \sum_{j=1}^p \alpha_j \mu (A_j \cap A) $ with $\alpha_j, A_j $ like above.

$s : X → \mathbb{R} $ are a nonnegative simple function. Show:

i) $ \int_A s d \mu= \int_X 1_A s d\mu$

ii) $ \int_{\bigcup_{j=1}^{\infty} A_j} s d\mu = \sum_{j=1}^{\infty} \int_{A_j} s d\mu $ for $ A_j \cap A_k = \emptyset$ for $ j \not=k$

Proof: i) Let $s(X):= \{\alpha_1,..,\alpha_p\}, (1_A s)^{-1} (\alpha_j)=:A_j,( s)^{-1} (\alpha_j)=:B_j$

$ \int_X s1_A d \mu= \sum_{\alpha_j \in s(A)} \alpha_j \mu (A_j \cap X)= \sum_{\alpha_j \in s(A)} \alpha_j \mu(A_j)=\sum_{\alpha_j \in s(A)} \alpha_j \mu (A \cap A_j)= \sum_{i=1}^p \alpha_j \mu (A \cap B_j)= \int_A s d \mu$

The second to last equation because if $ \alpha_i \in s(X) \setminus s(A)$ so $ B_i=s(\alpha_i) \not\in A$ and so $ \mu(B_i \cap A)= \mu(\emptyset)= 0$

ii) I guess that the $A_j´s$ in the exercise and the $A_j´s$ in my definition are different?

$s(X)= \{\alpha_1,.., \alpha_p\}, B_i := s^{-1}(\alpha_j)$

$ \int_{\bigcup_{j=1}^{\infty} A_j} s d\mu= \sum_{i=1}^p \alpha_i \mu (B_i \cap \bigcup_{j=1}^{\infty} A_j) = \sum_{i=1}^p \alpha_i \mu (\bigcup_{j=1}^{\infty} A_j \cap B_i) = \sum_{i=1}^p \alpha_i \sum_{j=1}^{\infty} \mu (A_j \cap B_i) = \sum_{i=1}^p \sum_{j=1}^{\infty} \alpha_i \mu (A_j \cap B_i)$

Can i change the order of the sum and the series?

EDIT:
I would like to prove it without using the linearity of the integral.

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    $\begingroup$ 1) $s:X\to\mathbb R$ is a simple function if (firstly) $s$ is measurable and (secondly) its range is finite. You forgot the first part. 2) It is kind of weird to define $\int_Asd\mu$ the way you did here. It is better to switch: take i) as definition and based on that prove that $\int_Asd\mu$ satisfies the equation mentioned in your definition. $\endgroup$
    – drhab
    Commented Aug 17, 2016 at 10:52
  • $\begingroup$ I editet some things, also your first point. To 2:In my lecture we define it like that so i want to stay by this definition.If i can proof the equality of i) and my definitions it doesn´t matter from which definition i come.. $\endgroup$ Commented Aug 17, 2016 at 11:08
  • $\begingroup$ I understand that you stick to the definition of your lecture. Profit of my definition: it also serves as definition if $s$ is not simple. Secondly it is generally accepted as definition. I hope that once you will be converted :-). $\endgroup$
    – drhab
    Commented Aug 17, 2016 at 11:44

2 Answers 2

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If $s=\sum_{i=1}^{p}\alpha_{i}1_{B_{i}}$ where the $B_{i}$ are disjoint measurable sets and the $\alpha_{i}$ are elements of $\left[0,\infty\right)$ then for measurable set $A$ we find:

$$\int_{A}sd\mu=\int1_{A}\sum_{i=1}^{p}\alpha_{i}1_{B_{i}}d\mu=\int\sum_{i=1}^{p}\alpha_{i}1_{B_{i}\cap A}d\mu=\sum_{i=1}^{p}\alpha_{i}\mu\left(B_{i}\cap A\right)$$

"My" definition (see my comment on the question) takes care of the first equality. "Yours" takes care of the last.

If $A:=\bigcup_{j=1}^{\infty}A_{j}$ where the $A_{j}$ are measurable and mutually disjoint then: $$\mu\left(B_{i}\cap A\right)=\sum_{j=1}^{\infty}\mu\left(B_{i}\cap A_{j}\right)$$

So we arrive at:

$$\int_{A}sd\mu=\sum_{i=1}^{p}\alpha_{i}\sum_{j=1}^{\infty}\mu\left(B_{i}\cap A_{j}\right)=\sum_{j=1}^{\infty}\sum_{i=1}^{p}\alpha_{i}\mu\left(B_{i}\cap A_{j}\right)=\sum_{j=1}^{\infty}\int_{A_{j}}sd\mu$$

The switch of summation is allowed because all terms are nonnegative.


addendum:

Let $a_{i,j}\in\left[0,\infty\right]$ for $i,j\in\left\{ 1,2,\dots\right\}$ and let $t\in\left(0,\infty\right)$

If $\sum_{i=1}^{\infty}\sum_{j=1}^{\infty}a_{i,j}>t$ then $\sum_{i=1}^{n}\sum_{j=1}^{\infty}a_{i,j}>t$ for some $n$.

Then also $\sum_{i=1}^{n}\sum_{j=1}^{m}a_{i,j}>t$ for $m$ large enough, so that $$\sum_{j=1}^{\infty}\sum_{i=1}^{\infty}a_{i,j}\geq\sum_{j=1}^{m}\sum_{i=1}^{n}a_{i,j}=\sum_{i=1}^{n}\sum_{j=1}^{m}a_{i,j}>t$$

Analogously it can be shownt that: $$\sum_{j=1}^{\infty}\sum_{i=1}^{\infty}a_{i,j}>t\implies\sum_{i=1}^{\infty}\sum_{j=1}^{\infty}a_{i,j}>t$$

So for every $t\in(0,\infty)$ we have: $$\sum_{j=1}^{\infty}\sum_{i=1}^{\infty}a_{i,j}>t\iff\sum_{i=1}^{\infty}\sum_{j=1}^{\infty}a_{i,j}>t$$

This can only be true if: $$\sum_{j=1}^{\infty}\sum_{i=1}^{\infty}a_{i,j}=\sum_{i=1}^{\infty}\sum_{j=1}^{\infty}a_{i,j}$$

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  • $\begingroup$ thanks, as beginner in this topic it is really helpful to see this calculation! But why i can switch the summation? I don´t get it that it follows from the nonnegative terms? $\endgroup$ Commented Aug 17, 2016 at 13:59
  • $\begingroup$ I have added something that might help. $\endgroup$
    – drhab
    Commented Aug 17, 2016 at 18:39
  • $\begingroup$ @drhab I'm working on a similar problem, but I am not getting anywhere with it. Would it be possible for you to take a look? math.stackexchange.com/questions/2453786/… Thank you! (Matt 25:21) $\endgroup$
    – user100463
    Commented Oct 2, 2017 at 13:13
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To prove $i)$ with your definition, just observe that

$$(s \, 1_A) (x) = \sum_{j=1}^p \alpha_j 1_{A \cap A_j} (x).$$

By the definition of the Lebesgue integral for simple function you get immediately

$$\int_X s \, 1_A \, d\mu = \sum_{j=1}^p \alpha_j \mu( A \cap A_j)$$

which proves the claim.

For ii) you can use drhab explanation.

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