Simple Function: Difference between revisions
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* the function <math>A \mapsto \int_A f</math> is a measure on <math>\mathcal{M}</math>. | * the function <math>A \mapsto \int_A f</math> is a measure on <math>\mathcal{M}</math>. | ||
===Proof<ref>Craig, Katy. ''MATH 201A | ===Proof<ref>Craig, Katy. ''MATH 201A Lectures 12-13''. UC Santa Barbara, Fall 2020.</ref>=== | ||
Let <math> f = \sum_{i=1}^n a_i 1_{E_i}</math> and <math> g = \sum_{j=1}^m b_j 1_{F_j}</math> be simple functions with their corresponding standard representations. | Let <math> f = \sum_{i=1}^n a_i 1_{E_i}</math> and <math> g = \sum_{j=1}^m b_j 1_{F_j}</math> be simple functions with their corresponding standard representations. | ||
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<math>c \int f = c \sum_{i=1}^n a_i 1_{E_i} = \sum_{i=1}^n ca_i 1_{E_i} = \int cf</math>. | <math>c \int f = c \sum_{i=1}^n a_i 1_{E_i} = \sum_{i=1}^n ca_i 1_{E_i} = \int cf</math>. | ||
Next, we show the second statement. Notice that | |||
<math>E_i = \cup_{j=1}^m (E_i \cap F_j)<\math> and <math>F_j = \cup_{i=1}^n (F_j \cap E_i).<\math> | |||
Then | |||
<math>\int f + \int g = \sum_{i=1}^n a_i \mu(E_i) + \sum_{j=1}^m b_j \mu(F_j)<\math> | |||
<math><\math> | |||
==References== | ==References== |
Revision as of 05:32, 11 December 2020
The simplest functions you will ever integrate, hence the name.
Definition
Let be a measure space. A measurable function is a simple function[1] if is a finite subset of . The standard representation[1] for a simple function is given by
,
where is the indicator function on the disjoint sets that partition , where .
Integration of Simple Functions
These functions earn their name from the simplicity in which their integrals are defined[2]. Let be the space of all measurable functions from to Then
where by convention, we let . Note that is equivalent to and that some arguments may be omitted when there is no confusion.
Furthermore, for any , we define
Properties of Simple Functions
Given simple functions , the following are true[2]:
- if ;
- ;
- if , then ;
- the function is a measure on .
Proof[3]
Let and be simple functions with their corresponding standard representations.
We show the first claim. Suppose . Then , implying . Similarly, . Thus, the first statement holds for this case.
Suppose . Then
.
Next, we show the second statement. Notice that
<math>E_i = \cup_{j=1}^m (E_i \cap F_j)<\math> and <math>F_j = \cup_{i=1}^n (F_j \cap E_i).<\math>
Then <math>\int f + \int g = \sum_{i=1}^n a_i \mu(E_i) + \sum_{j=1}^m b_j \mu(F_j)<\math>
<math><\math>