Editing Absolute convergence (section)

==Absolute convergence over sets==

A generalization of the absolute convergence of a series, is the absolute convergence of a sum of a function over a set.  We can first consider a countable set <math>X</math> and a function <math>f : X \to \R.</math>  We will give a definition below of the sum of <math>f</math> over <math>X,</math> written as <math display=inline>\sum_{x \in X} f(x).</math>  

First note that because no particular enumeration (or "indexing") of <math>X</math> has yet been specified, the series <math display=inline>\sum_{x \in X}f(x)</math> cannot be understood by the more basic definition of a series.  In fact, for certain examples of <math>X</math> and <math>f,</math> the sum of <math>f</math> over <math>X</math> may not be defined at all, since some indexing may produce a conditionally convergent series.  

Therefore we define <math display=inline>\sum_{x \in X} f(x)</math> only in the case where there exists some bijection <math>g : \Z^+ \to X</math> such that <math display=inline>\sum_{n=1}^\infty f(g(n))</math> is absolutely convergent.  Note that here, "absolutely convergent" uses the more basic definition, applied to an indexed series.  In this case, the value of the '''sum of <math>f</math> over <math>X</math>'''<ref>{{Cite book|title=Analysis I|last=Tao|first=Terrance|publisher=Hindustan Book Agency|year=2016|isbn=978-9380250649|location=New Delhi|pages=188–191}}</ref> is defined by 
<math display=block>\sum_{x \in X}f(x) := \sum_{n=1}^\infty f(g(n))</math>

Note that because the series is absolutely convergent, then every rearrangement is identical to a different choice of bijection <math>g.</math>  Since all of these sums have the same value, then the sum of <math>f</math> over <math>X</math> is well-defined.  

Even more generally we may define the sum of <math>f</math> over <math>X</math> when <math>X</math> is uncountable.  But first we define what it means for the sum to be convergent.

Let <math>X</math> be any set, countable or uncountable, and <math>f : X \to \R</math> a function.  We say that '''the sum of <math>f</math> over <math>X</math> converges absolutely''' if 
<math display=block>\sup\left\{\sum_{x \in A} |f(x)|: A\subseteq X, A \text{ is finite }\right\} < \infty.</math>

There is a theorem which states that, if the sum of <math>f</math> over <math>X</math> is absolutely convergent, then <math>f</math> takes non-zero values on a set that is at most countable.  Therefore, the following is a consistent definition of the sum of <math>f</math> over <math>X</math> when the sum is absolutely convergent.  
<math display=block>\sum_{x \in X} f(x) := \sum_{x \in X : f(x) \neq 0} f(x).</math>

Note that the final series uses the definition of a series over a countable set.

Some authors define an iterated sum <math display=inline>\sum_{m=1}^\infty \sum_{n=1}^\infty a_{m,n}</math> to be absolutely convergent if the iterated series <math display=inline>\sum_{m=1}^\infty \sum_{n=1}^\infty |a_{m,n}| < \infty.</math><ref>{{Cite book|title=The Way of Analysis|last=Strichartz|first=Robert|publisher=Jones & Bartlett Learning|year=2000|isbn=978-0763714970|pages=259–260}}</ref>  This is in fact equivalent to the absolute convergence of <math display=inline>\sum_{(m,n) \in \N \times \N} a_{m,n}.</math>  That is to say, if the sum of <math>f</math> over <math>X,</math> <math display=inline>\sum_{(m,n) \in \N \times \N} a_{m,n},</math> converges absolutely, as defined above, then the iterated sum <math display=inline>\sum_{m=1}^\infty \sum_{n=1}^\infty a_{m,n}</math> converges absolutely, and vice versa.