Why Did Radon Choose Riemann-Stieltjes Over Lebesgue Integration?

[jsr9119]

Hey guys,

I'm doing a paper on the Radon transform and several sources I've come across cite the Lebesgue integral as a necessary tool to handle measures in higher order transforms.
But, Radon's original paper employs the Riemann-Stieltjes integral in its place.

I read that Lebesgue is more general and so Radon could have used it in place of RSI. Is this the case?

Thanks,

Jeff

 

[micromass]

The Lebesgue integral is indeed more general than the Riemann integral.
Using measure theory, we can also develop the Lebesgue-Stieltjes integral, and this is a generalization of the Riemann-Stieltjes integral.

So yes, the paper could probably be written with Lebesgue instead of Riemann. But there may be technical differences between the two.

 

[Stephen Tashi]

Is there a distinction between "Lebesgue Integration" and "integration with respect to Lebesgue measure"? My impressions is that "Lebesgue measure" on the real number line is a particular measure that implements the usual notion of length, so the measure of a single point would be zero. On the other hand, is "Lebesgue integration" defined with respect to an arbitrary measure?

For Lebesgue Integration to include Riemann-Stieljes integration as a special case, is it necessary to use measures other than Legesgue measure? (I'm thinking of the specific example of defining an integration that can integrate a discrete probability density function by the method of assigning non-zero measure to certain isolated points and turning "integration" into summation.)

 

[mathman]

Lebesgue-Stieljes integral is best described as Lebesgue integration with respect to a given measure.

 

[pwsnafu]

Is there a distinction between "Lebesgue Integration" and "integration with respect to Lebesgue measure"? My impressions is that "Lebesgue measure" on the real number line is a particular measure that implements the usual notion of length, so the measure of a single point would be zero. On the other hand, is "Lebesgue integration" defined with respect to an arbitrary measure?

Yes. Lebesgue integration is defined with respect to a measure. The procedure is the same, but different measures give different integrals.

For Lebesgue Integration to include Riemann-Stieljes integration as a special case, is it necessary to use measures other than Legesgue measure?

The Lebesgue measure is derived from the set function $m((a,b])=b-a$.
The Stieljes measure is derived from the set function $m(a,b]) = g(b)-g(a)$ for some monotonically increasing function g.

(I'm thinking of the specific example of defining an integration that can integrate a discrete probability density function by the method of assigning non-zero measure to certain isolated points and turning "integration" into summation.)

Summation is a special case of Lebesgue integration, using the counting measure over Z, or Dirac measure over R.

 

[micromass]

In my experience, it's a bit ambiguous. When talking about Lebesgue integration, sometimes people talk about general integration wrt a measure and sometimes they talk about integration wrt Lebesgue measure. It's usually clear from the context though.