Difference between revisions of "User:Boris Tsirelson/sandbox2"
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* in this case $\int_{\{0,1,2,\dots\}} a(n)\,\mu(\rd n) = \sum_{n=0}^\infty a_n$. | * in this case $\int_{\{0,1,2,\dots\}} a(n)\,\mu(\rd n) = \sum_{n=0}^\infty a_n$. | ||
The same holds when $a_n$ are complex numbers or elements of a Banach space. | The same holds when $a_n$ are complex numbers or elements of a Banach space. | ||
| − | In this formulation, the order of terms is evidently irrelevant, since the notion of a measure space does not stipulate any ordering of points. | + | In this formulation, the order of terms is evidently irrelevant, since the notion of a measure space does not stipulate any ordering of points. The claims about multiple and iterated series become special cases of [[Fubini theorem]]. |
Revision as of 18:09, 21 July 2012
Relations to Lebesgue integral
It is possible to treat absolutely convergent series as a special case of Lebesgue integrals. To this end, the countable set $\{0,1,2,\dots\}$ is treated as a measure space; all subsets are measurable, and the counting measure $\mu$ is used: $\mu(A)$ is the number of points in $A$ ($\infty$ if $A$ is infinite). A sequence $(a_n)$ of real numbers is just a function $a:\{0,1,2,\dots\}\to\R$ (measurable, since everything is measurable on this discrete space). It is easy to see that
- the function $a$ is integrable if and only if the series $\sum a_n$ converges absolutely, and
- in this case $\int_{\{0,1,2,\dots\}} a(n)\,\mu(\rd n) = \sum_{n=0}^\infty a_n$.
The same holds when $a_n$ are complex numbers or elements of a Banach space. In this formulation, the order of terms is evidently irrelevant, since the notion of a measure space does not stipulate any ordering of points. The claims about multiple and iterated series become special cases of Fubini theorem.
Boris Tsirelson/sandbox2. Encyclopedia of Mathematics. URL: http://encyclopediaofmath.org/index.php?title=Boris_Tsirelson/sandbox2&oldid=27158