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 May 1st, 2018, 04:06 AM #1 Senior Member   Joined: Nov 2011 Posts: 197 Thanks: 2 Integration Axioms What are the axioms of Integration?
 May 1st, 2018, 04:34 AM #2 Senior Member   Joined: Sep 2016 From: USA Posts: 383 Thanks: 207 Math Focus: Dynamical systems, analytic function theory, numerics 1. The real numbers are a closed ordered field. 2. Thou shalt not covet thy neighbor's wife. 3. He who smelt it dealt it. Note: If in addition you accept the axiom of choice, then 3 is equivalent to: 4. He who said the rhyme, did the crime.
 May 1st, 2018, 11:05 AM #3 Math Team   Joined: Jan 2015 From: Alabama Posts: 3,194 Thanks: 871 Typically, the term "axiom" refers to a "mathematical structure", such as the "integers", "real numbers", and "functions of real numbers". Integration however is a operation applied to the mathematical structure of functions on the real numbers. "Integration", itself, is not a mathematical structure so the word "axiom" does not apply.
May 1st, 2018, 11:08 AM   #4
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Quote:
 Originally Posted by Country Boy Typically, the term "axiom" refers to a "mathematical structure", such as the "integers", "real numbers", and "functions of real numbers". Integration however is a operation applied to the mathematical structure of functions on the real numbers. "Integration", itself, is not a mathematical structure so the word "axiom" does not apply.
One could perhaps interpret OP's question as pertaining to the axioms of measure theory, in which case the axioms are described here.

 May 1st, 2018, 11:27 AM #5 Senior Member   Joined: Oct 2009 Posts: 406 Thanks: 141 It is in fact possible to give axioms for integration. So called Daniell integration theory does this and is an alternative way to construct the Lebesgue integral and Lebesgue measure. Given are a set $X$, a set $\mathcal{U}$ consisting of function $X\rightarrow \mathbb{R}$, and a function $I:\mathcal{U}\rightarrow \mathbb{R}$. The following axioms are considered: - $\mathcal{U}$ is a vector space for the usual operations on functions $X\rightarrow \mathbb{R}$ - For any $f,g\in \mathcal{U}$, we have that $\sup(f,g)$ and $\inf(f,g)$ are also in $\mathcal{U}$. - $I$ is a linear function - If $f\geq 0$, then $I(f)\geq 0$ - If $f_n\in \mathcal{U}$ for every $n\in \mathbb{N}$ is such that $$f_1\geq f_2 \geq f_3 \geq ...\geq 0$$ and if $f_n\rightarrow 0$ pointswise, then $I(f_n)\rightarrow 0$.
 May 1st, 2018, 11:27 AM #6 Senior Member   Joined: Oct 2009 Posts: 406 Thanks: 141 It is in fact possible to give axioms for integration. So called Daniell integration theory does this and is an alternative way to construct the Lebesgue integral and Lebesgue measure. Given are a set $X$, a set $\mathcal{U}$ consisting of function $X\rightarrow \mathbb{R}$, and a function $I:\mathcal{U}\rightarrow \mathbb{R}$. The following axioms are considered: - $\mathcal{U}$ is a vector space for the usual operations on functions $X\rightarrow \mathbb{R}$ - For any $f,g\in \mathcal{U}$, we have that $\sup(f,g)$ and $\inf(f,g)$ are also in $\mathcal{U}$. - $I$ is a linear function - If $f\geq 0$, then $I(f)\geq 0$ - If $f_n\in \mathcal{U}$ for every $n\in \mathbb{N}$ is such that $$f_1\geq f_2 \geq f_3 \geq ...\geq 0$$ and if $f_n\rightarrow 0$ pointswise, then $I(f_n)\rightarrow 0$.

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