Are Measures Really Equal If Integrals of Continuous Functions Match?

[Gerald1]

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]

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

 

[Gerald1]

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

How does that show the measures are equal?

 

[ThePerfectHacker]

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.

 

[blue_raver22]

for your question. I agree that Rudin's claim may not be entirely clear and may require further explanation. However, I believe that Rudin's statement is based on the fundamental theorem of calculus, which states that if two functions have the same derivative, then they must be equal (up to a constant).

In this case, since the integrals of two different measures are equal for all continuous functions, this implies that the measures must be equal (up to a constant) in order for the integrals to be equal. This is because the integral of a function is essentially a measure of its area under the curve, and if two measures have the same integral for all continuous functions, then they must be equal.

Furthermore, Rudin also mentions that the measures are equal "almost everywhere," which means that they may differ at a few isolated points, but overall they are equal. This is a common concept in mathematics, and it is often used to describe situations where two things are equal except for a few exceptions.

In conclusion, while Rudin's statement may require further explanation, I believe that it is based on the fundamental theorem of calculus and the concept of measures being equal almost everywhere. I hope this helps clarify the concept for you. Thank you for your interest in this topic.