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integration with respect to non-canonical simple functions

Integration of non-negative functionsIntegral of simple functions in standard and non-standard representationRepresentations of Simple FunctionsSubclasses of simple functions dense in $L^2$If $mu$ is $sigma$-finite, then there exist increasing simple functions $s_n rightarrow f$ with $mu({x:s_n neq 0})< infty$Real Analysis, Folland Proposition 2.13 Integration of Nonnegative Functionsprove Lesbesgue Integration definition for simple function is well definedFinite sigma-algebra; integrating a non-simple function.Simple question on simple functions: interpreting the sum signA simple function and its canonical form.

0

$begingroup$

I am currently reading Rudin. Rudin defines integration of simple functions as follows:

Let $phi : X rightarrow mathbb{R}$ be a nonnegative simple function with the canonical representation:

$$phi = Sigma_{j = 0}^n c_j chi_{E_j}$$

Recall given a simple function $phi : X rightarrow mathbb{R}$. We have the following canonical representation:

$$phi = Sigma_{j = 0}^n c_j chi_{E_j}$$

$$c_j: text{ are distinct}$$
$$c_0 = 0, text{ and }, E_j = phi^{-1}({c_j})$$

Define integral of $phi$ as follows:

$$int{ phi } = int{ phi d mu} := Sigma_{j = 0}^n c_j mu(E_j)$$

What I am trying to prove is integration with respect to non-canonical representation:

For a nonnegative simple function $phi = Sigma_{j = 1}^m a_k chi_{A_k}$, then

$$int{ phi mu} = Sigma_{k = 1}^m a_k dmu(A_k)$$

My work:

I will first do it for the case when $A_m$ are disjoint. Set $A_0 = (A_1 cup ldots cup A_k)^{c}$ together with $a_0 := 0 $.

$int phi dmu = Sigma c_j chi_{E_j} = Sigma c_j mu(E_j) = Sigma c_j Sigma mu(E_j cap A_k) = Sigma c_j mu(A_k)$. The last step is by interchanging the sum as we are dealing with finite summation.

If we write $phi = Sigma a_k chi_{A_k}$. If we apply the same trick as above we arrive at the following formula:

$$Sigma a_i u(A_k) = Sigma a_i u(E_k).$$

By uniqueness of the $c_j$ we must have that $c_j = a_i$. Completing the proof.

Now we prove it for general $A^{prime}_m$. We can form disjoint set $A_m$ such that:

$$bigcup A_m = bigcup A^{prime}_m$$

By the result we proved $int(phi dmu) = Sigma a_k mu(A_k) = Sigma a_k mu(A_k)$ by invoking the following properties of measure $mu(A_1 cup A_2) = mu(A_1) +mu(A_2)$.

measure-theory

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asked 2 days ago

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0

$begingroup$

I am currently reading Rudin. Rudin defines integration of simple functions as follows:

Let $phi : X rightarrow mathbb{R}$ be a nonnegative simple function with the canonical representation:

$$phi = Sigma_{j = 0}^n c_j chi_{E_j}$$

Recall given a simple function $phi : X rightarrow mathbb{R}$. We have the following canonical representation:

$$phi = Sigma_{j = 0}^n c_j chi_{E_j}$$

$$c_j: text{ are distinct}$$
$$c_0 = 0, text{ and }, E_j = phi^{-1}({c_j})$$

Define integral of $phi$ as follows:

$$int{ phi } = int{ phi d mu} := Sigma_{j = 0}^n c_j mu(E_j)$$

What I am trying to prove is integration with respect to non-canonical representation:

For a nonnegative simple function $phi = Sigma_{j = 1}^m a_k chi_{A_k}$, then

$$int{ phi mu} = Sigma_{k = 1}^m a_k dmu(A_k)$$

My work:

I will first do it for the case when $A_m$ are disjoint. Set $A_0 = (A_1 cup ldots cup A_k)^{c}$ together with $a_0 := 0 $.

$int phi dmu = Sigma c_j chi_{E_j} = Sigma c_j mu(E_j) = Sigma c_j Sigma mu(E_j cap A_k) = Sigma c_j mu(A_k)$. The last step is by interchanging the sum as we are dealing with finite summation.

If we write $phi = Sigma a_k chi_{A_k}$. If we apply the same trick as above we arrive at the following formula:

$$Sigma a_i u(A_k) = Sigma a_i u(E_k).$$

By uniqueness of the $c_j$ we must have that $c_j = a_i$. Completing the proof.

Now we prove it for general $A^{prime}_m$. We can form disjoint set $A_m$ such that:

$$bigcup A_m = bigcup A^{prime}_m$$

By the result we proved $int(phi dmu) = Sigma a_k mu(A_k) = Sigma a_k mu(A_k)$ by invoking the following properties of measure $mu(A_1 cup A_2) = mu(A_1) +mu(A_2)$.

measure-theory

share|cite|improve this question

asked 2 days ago

NewbieNewbie

449312

$endgroup$

add a comment | 

$begingroup$

I am currently reading Rudin. Rudin defines integration of simple functions as follows:

Let $phi : X rightarrow mathbb{R}$ be a nonnegative simple function with the canonical representation:

$$phi = Sigma_{j = 0}^n c_j chi_{E_j}$$

Recall given a simple function $phi : X rightarrow mathbb{R}$. We have the following canonical representation:

$$phi = Sigma_{j = 0}^n c_j chi_{E_j}$$

$$c_j: text{ are distinct}$$
$$c_0 = 0, text{ and }, E_j = phi^{-1}({c_j})$$

Define integral of $phi$ as follows:

$$int{ phi } = int{ phi d mu} := Sigma_{j = 0}^n c_j mu(E_j)$$

What I am trying to prove is integration with respect to non-canonical representation:

For a nonnegative simple function $phi = Sigma_{j = 1}^m a_k chi_{A_k}$, then

$$int{ phi mu} = Sigma_{k = 1}^m a_k dmu(A_k)$$

My work:

I will first do it for the case when $A_m$ are disjoint. Set $A_0 = (A_1 cup ldots cup A_k)^{c}$ together with $a_0 := 0 $.

$int phi dmu = Sigma c_j chi_{E_j} = Sigma c_j mu(E_j) = Sigma c_j Sigma mu(E_j cap A_k) = Sigma c_j mu(A_k)$. The last step is by interchanging the sum as we are dealing with finite summation.

If we write $phi = Sigma a_k chi_{A_k}$. If we apply the same trick as above we arrive at the following formula:

$$Sigma a_i u(A_k) = Sigma a_i u(E_k).$$

By uniqueness of the $c_j$ we must have that $c_j = a_i$. Completing the proof.

Now we prove it for general $A^{prime}_m$. We can form disjoint set $A_m$ such that:

$$bigcup A_m = bigcup A^{prime}_m$$

By the result we proved $int(phi dmu) = Sigma a_k mu(A_k) = Sigma a_k mu(A_k)$ by invoking the following properties of measure $mu(A_1 cup A_2) = mu(A_1) +mu(A_2)$.

measure-theory

share|cite|improve this question

asked 2 days ago

NewbieNewbie

449312

$endgroup$

share|cite|improve this question

asked 2 days ago

NewbieNewbie

449312

share|cite|improve this question

asked 2 days ago

NewbieNewbie

449312

asked 2 days ago

NewbieNewbie

449312

asked 2 days ago

NewbieNewbie

449312