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Question:

Proposition 2.13 - Let $\phi$ and $\psi$ be simple functions in $L^+$.

a.) If $c\geq 0$, $\int c\phi = c\int \phi$.

b.) $\int(\phi + \psi) = \int \phi + \int \psi$.

c.) If $\phi\leq \psi$, then $\int \phi\leq \int \psi$.

d.) The map $A\rightarrow \int_{A}d\mu$ is a measure on $M$.

Attempted proof a.) - If $c\geq 0$ then $$\int c\phi = \int c\sum_{j}a_j \chi_{E_j} = c\int \sum_{j}a_j \chi_{E_j} = c\int \phi$$

Attempted proof b.) Let $\phi = \int_{j}a_j\chi_{E_j}$ and $\psi = \sum_{k}b_k\chi_{f_k}$, then $$\int \phi + \int \psi = \int \sum_{j}a_j\chi_{E_j} + \int \sum_{k}b_k\chi_{F_k} = \sum_{j}a_j\mu(E_j) + \sum_{j}b_k\mu(F_k)$$

Before I go on with c and d I saw that after this last step we can say $$\sum_{j}a_j\mu(E_j) + \sum_{j}b_k\mu(F_k) = \sum_{j}a_k\sum_{k}\mu(E_j\cap F_k) + \sum_{k}b_k\sum_{j}\mu(E_j\cap F_k)$$ I don't understand how they are equal. So this is a more of an algebra question I guess more than a measure theory question. Any suggestions is greatly appreciated.

Background Information:

We fix a measure space $(X,M,\mu)$, and we define $$L^+ = \ \ \text{the space of all measurable functions from} \ X \ \text{to} \ [0,\infty]$$ If $\phi$ is a simple function in $L^+$ with standard representation $\phi = \sum_{1}^{n}a_j\chi_{E_j}$, we define the integral of $\phi$ with respect to $\mu$ by $$\int \phi d\mu = \sum_{1}^{n}a_j\mu(E_j)$$

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  • $\begingroup$ In a) you are practically using the proposition that you want to prove. At this stage, we do not know if integrals satisfy scalar linearity (to move out a constant). $\endgroup$
    – user305860
    Jun 23, 2016 at 14:38

1 Answer 1

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Let $(X,M, \mu)$ be a measure space.

Question:

Proposition 2.13 - Let $\phi$ and $\psi$ be simple functions in $L^+$.

a.) If $c\geq 0$, $\int c\phi = c\int \phi$.

b.) $\int(\phi + \psi) = \int \phi + \int \psi$.

c.) If $\phi\leq \psi$, then $\int \phi\leq \int \psi$.

d.) The map $A\rightarrow \int_{A}d\mu$ is a measure on $M$.

Proof a.) - If $c\geq 0$ then \begin{align*} \int c\phi &= \int c\sum_{j}a_j \chi_{E_j} = \int \sum_{j}ca_j \chi_{E_j} = \sum_{j}ca_j \mu(E_j)= \\ &= c\left(\sum_{j}a_j \mu(E_j) \right)= c\int \sum_{j}a_j \chi_{E_j} = c\int \phi \end{align*}

Proof b.) Let $\phi = \sum_{j}a_j\chi_{E_j}$, $\psi = \sum_{k}b_k\chi_{F_k}$ and $\phi +\psi = \sum_{i}c_i\chi_{H_i}$ be the standard representations of $\phi$, $\psi$ and $\phi + \psi$.

Then $\{E_j\}_j$ is a finite family of disjoint measurable sets, such that $X= \bigcup_jE_j$. We also have that $\{F_k\}_k$ is a finite family of disjoint measurable sets, such that $X= \bigcup_kF_k$. And $\{H_i\}_i$ is a finite family of disjoint measurable sets, such that $X= \bigcup_kH_i$.

So we have that $\{E_j\cap F_k \cap H_i\}_{j,k,i}$ is a finite family of disjoint measurable sets, such that, for all $j$, $E_j =\bigcup_{k,i}(E_j\cap F_k \cap H_i)$; for all $k$, $F_k =\bigcup_{j,i}(E_j\cap F_k\cap H_i)$ and $H_i =\bigcup_{j,k}(E_j\cap F_k \cap H_i)$. So we have $$ \mu(E_j) =\sum_{k,i}\mu(E_j\cap F_k \cap H_i)$$ $$ \mu(F_k) =\sum_{j,i}\mu(E_j\cap F_k\cap H_i)$$ and $$\mu(H_i) = \sum_{j,k}\mu(E_j\cap F_k \cap H_i)$$

Moreover, for all $j$, $k$, $i$, if $E_j\cap F_k \cap H_i \neq \emptyset$, then $a_j+b_k=c_i$.

\begin{align*} \int(\phi + \psi) &= \sum_{i}c_i\mu(H_i)= \sum_{i,j,k}c_i\mu(E_j\cap F_k \cap H_i)= \sum_{\substack{i,j,k \\ E_j\cap F_k \cap H_i \neq \emptyset}}c_i\mu(E_j\cap F_k \cap H_i)= \\ & = \sum_{\substack{i,j,k \\ E_j\cap F_k \cap H_i \neq \emptyset}}(a_j+b_k)\mu(E_j\cap F_k \cap H_i)= \sum_{i,j,k }(a_j+b_k)\mu(E_j\cap F_k \cap H_i)=\\ &= \sum_{i,j,k }a_j\mu(E_j\cap F_k \cap H_i)+ \sum_{i,j,k }b_k\mu(E_j\cap F_k \cap H_i)= \\ &= \sum_j a_j\sum_{i,k }\mu(E_j\cap F_k \cap H_i)+ \sum_k b_k\sum_{i,j }\mu(E_j\cap F_k \cap H_i)= \\ &= \sum_j a_j\mu(E_j)+ \sum_k b_k\mu( F_k )= \\ &=\int \phi + \int \psi \end{align*}

Proof c.) Let $\phi = \sum_{j}a_j\chi_{E_j}$ and$\psi = \sum_{k}b_k\chi_{F_k}$ be the standard representations of $\phi$ and $\psi$.

Then $\{E_j\}_j$ is a finite family of disjoint measurable sets, such that $X= \bigcup_jE_j$. And $\{F_k\}_k$ is a finite family of disjoint measurable sets, such that $X= \bigcup_kF_k$.

So we have that $\{E_j\cap F_k \}_{j,k}$ is a finite family of disjoint measurable sets, such that, for all $j$, $E_j =\bigcup_k(E_j\cap F_k)$ and for all $k$, $F_k =\bigcup_j(E_j\cap F_k)$. So we have $$ \mu(E_j) =\sum_{k,i}\mu(E_j\cap F_k \cap H_i)$$ and $$ \mu(F_k) =\sum_{j,k}\mu(E_j\cap F_k\cap H_i)$$

Moreover, for all $j$, $k$, if $E_j\cap F_k \neq \emptyset$, then $a_j\leq b_k$.

\begin{align*} \int\phi &= \sum_{j}a_j\mu(E_j)= \sum_{j,k}aj\mu(E_j\cap F_k)= \sum_{\substack{j,k \\ E_j\cap F_k \neq \emptyset}}a_j\mu(E_j\cap F_k) \leq \\ & \leq \sum_{\substack{j,k \\ E_j\cap F_k \neq \emptyset}}b_k\mu(E_j\cap F_k)= \sum_{j,k }b_k\mu(E_j\cap F_k)= \\ &= \sum_k b_k\sum_{j }\mu(E_j\cap F_k )= \sum_k b_k\mu( F_k )= \int \psi \end{align*}

Proof d.) Since for all $A \in M$, $\int_{A}d\mu=\int\chi_{A}d\mu=\mu(A)$, we have that the map $A\rightarrow \int_{A}d\mu$ is just the map $A \rightarrow \mu(A)$ a measure on $M$.

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  • $\begingroup$ Really well done, you made that so easy to grasp for me. Thank you! $\endgroup$
    – Wolfy
    Jun 23, 2016 at 20:25
  • $\begingroup$ Folland's proof of part d suggests to me that the statement of the proposition contains an error. I think he meant to write $A \mapsto \int_A \phi d\mu$. $\endgroup$ Sep 12, 2018 at 19:23

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