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showing measures are equal

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Gerald

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Feb 28, 2014
2
between point (6) and (7) on page 323 of [this pdf file][1]. [1]: http://59clc.files.wordpress.com/2012/08/functional-analysis-_-rudin-2th.pdf

rudin claims that since the integrals (w.r.t two different measures) of real valued continuous functions are equal, then the measures are equal.

I think he concludes the integrals are equal for continuous functions by using real and imaginary parts, but even so how does it then follow that the measures are equal.

Many thanks
 

ThePerfectHacker

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Jan 26, 2012
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I think he concludes the integrals are equal for continuous functions by using real and imaginary parts, but even so how does it then follow that the measures are equal.
Integration is a positive linear functional on the space of continuous functions. It can be shown, as in the representation theorems, that such linear functionals induce measures for which the functional is integration along that induced measure.

Riesz?Markov?Kakutani representation theorem - Wikipedia, the free encyclopedia
 
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Gerald

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Feb 28, 2014
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ThePerfectHacker

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Jan 26, 2012
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How does that show the measures are equal?
Because more general functions can be approximated by continuous functions. If the measures are equal on the continuous functions they are forced to be equal everywhere.