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.

My exercise: Let $s,t: X\to[0,\infty)$ two simple functions. If $A\in\mathcal{A}$, show that,

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

Proof: $s(X)= \{\alpha_{i}\}_{i=1}^n, t(X)= \{\beta_{i}\}_{i=1}^m, A_i:= s^{-1} (\alpha_i), B_i:= s^{-1} (\beta_i)$

$s= \sum_{i=1}^n \alpha_i 1_{A_i}, t= \sum_{i=1}^m \beta_i 1_{B_i}$

$ \int_A s\,d\mu + \int_A t\,d\mu= \sum_{j=1}^n \alpha_j \mu(A_j \cap A) + \sum_{j=1}^m \beta_j \mu(B_j \cap A)$

$=\sum_{j=1}^n \alpha_j \mu(A_j \cap A \cap X) + \sum_{j=1}^m \beta_j \mu(B_j \cap A \cap X)$

$=\sum_{j=1}^n \alpha_j \mu(A_j \cap A \cap (\bigcup_{i=1}^m B_i)) + \sum_{j=1}^m \beta_j \mu(B_j \cap A \cap (\bigcup_{i=1}^n A_i))$

$= \sum_{j=1}^n \alpha_j \sum_{i=1}^m \mu (A_j \cap A \cap B_i) + \sum_{j=1}^m \beta_j \sum_{i=1}^n \mu(B_j \cap A \cap A_i)$

$=\sum_{j=1}^n\sum_{i=1}^m (\alpha_j + \beta_j) \mu (A_j \cap B_i \cap A)$

I don´t know if this is helpful for my problem..

  • 1
    $\begingroup$ I think you need to use different indices at the beginning. See my answer below. $\endgroup$ Commented Aug 17, 2016 at 15:33

1 Answer 1


Note that $s+t=\sum_{j=1}^n\sum_{k=1}^m\left ( \alpha_j+\beta_k \right )\chi_{(A_j\cap B_k\cap A)}$

and compute:

$\int_A s\,d\mu + \int_A t\,d\mu= \sum_{j=1}^n \alpha_j \mu(A_j \cap A) + \sum_{k=1}^m \beta_k \mu(B_k \cap A)=$

$\sum_{j=1}^n \alpha_j \mu(A_j \cap A\cap \bigcup^{m}_{k=1}B_k)+ \sum_{k=1}^m \beta_k \mu(B_k \cap A\cap \bigcup^{n}_{j=1}A_j)=$

$\sum_{j=1}^n \alpha_j \sum_{k=1}^m \mu(A_j \cap B_k\cap A)+\sum_{k=1}^m \beta_k \sum_{j=1}^n \mu(B_k \cap A_j\cap A)=$

$\sum_{j=1}^n\sum_{k=1}^m\left ( \alpha_j+\beta_k \right )\mu (A_j\cap B_k\cap A)=$



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