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Difference between revisions of "User:Boris Tsirelson/sandbox2"

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==Relations to Lebesgue integral==
 
  
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It is possible to treat absolutely convergent series as a special case of [[Lebesgue integral]]s. 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
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\begin{equation}\label{ab}
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* a function $a$ is integrable if and only if the series $\sum a_n$ converges absolutely, and
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E=mc^2
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* in this case $\int_{\{0,1,2,\dots\}} a(n)\,\mu(\rd n) = \sum_{n=0}^\infty a_n$.
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\end{equation}
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The same holds when $a_n$ are complex numbers or elements of a Banach space.
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By \eqref{ab}, it is possible. But see \eqref{ba} below:
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In this formulation, the order of terms is evidently irrelevant, since the notion of a measure space does not stipulate any order between points. The claims about multiple and iterated series become special cases of [[Fubini theorem]].
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\begin{equation}\label{ba}
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E\ne mc^3,
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On the other hand, absolutely convergent series may be used [[Lebesgue integral#series|when constructing]] Lebesgue integral.
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\end{equation}
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which is a pity.

Revision as of 20:22, 8 July 2013

\begin{equation}\label{ab} E=mc^2 \end{equation} By \eqref{ab}, it is possible. But see \eqref{ba} below: \begin{equation}\label{ba} E\ne mc^3, \end{equation} which is a pity.

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