Talk:Lebesgue–Stieltjes integration

Is there a good reason why some math pages have an annoying message in front about what to do if you don't understand the article? It seems rather condescending and insulting to me. Gene Ward Smith 04:32, 7 Feb 2005 (UTC)


 * I think it's because many people would be quite confused if they arrived at such pages through, say, the Random page feature. The message suggests places they could look to gain some background to the topic, as a compromise between including high-level mathematical content and an audience that is (presumably) not so mathematically-minded.  I think that it is a good idea, but this is not to say that it couldn't be improved.  Ben Cairns 07:40, 9 Feb 2005 (UTC).


 * I am also for removing that message "an audience that is (presumably) not so mathematically-minded" -> that would apply to every particular topic in science. Diego Torquemada 09:09, 31 March 2006 (UTC)

example
could you please show me some simple example of Lebesgue-Stieltjes integration?

Added example Oded (talk) 21:17, 15 April 2008 (UTC)

Lebesgue-Stieltjes additive set function
where is that?

Right. This is really unexplained. I will try to put this in sometime this week. Oded (talk) 15:17, 15 April 2008 (UTC)

Rewrote definition Oded (talk) 19:09, 15 April 2008 (UTC)

Definition wrong?
The definition seems to be wrong: \mu_g[s,t]=g(t)-g(s) would imply that the measure is always continuous, which it isn't. On the other hand, this is a direct consequence of the sup-minus-inf thing, which leads me to suspect that this is also wrong (but I can't verify it because I've never seen this particular formulation of it before). The usual way of defining it is \mu_g(s,t]=g(t)-g(s), or swapping the open and closed sides; you have to pick the left or right endpoint to leave out in order to make intervals add up without double counting. I'm editing it without consultation because the article seems to be rarely updated. Pirsq (talk) 04:41, 13 May 2009 (UTC)

Problem at lower bound
The way the measure \mu is defined on (s,t] intervals is insensitive to the value of w({a}). Therefore there is no way the Caratheodory extension for \mu on [a,b] is unique, because the value of \mu({a}) is unconstrained.  I therefore added take w({a})=0.  I think sometimes the convention w({a})=g(a) is used, so you could change it to this if you feel strongly.  It is equivalent to essentially assuming that g was 0 'just before' a.  Obviously this will affect the value of the integral, e.g. \int_a^a f(x)dg(x) = f(a)g(a).  I'm not sure if people find this desirable or not. --Wstrong (talk) 05:19, 22 August 2010 (UTC)

Integration by parts
If a function is "regular" at a discontinuity, then it isn't right continuous there. So the integral in the "Integration by parts" section isn't defined in these cases. I think the correct formulation is that we should be integrating the 'regularized' U against dV, and the 'regularized' V against dU, but I'm not sure. —Preceding unsigned comment added by 132.216.229.73 (talk) 02:17, 7 October 2010 (UTC)


 * Or rather, the measure corresponds to the right continuous 'de-regularization' of the function. I'll try to clarify it.
 * Also, the phrase "Under a slight generalization of this formula, the extra conditions on U and V can be dropped" is unclear. Where are 'the extra conditions'? In the source I see generalizations to (a) infinite a or/and b, and (b) complex-valued functions (which also includes non-increasing real-valued functions, while the theorem in the source is formulated for increasing real-valued functions). I'll correct it, too. Boris Tsirelson (talk) 09:11, 3 November 2019 (UTC)
 * I did. Boris Tsirelson (talk) 19:48, 3 November 2019 (UTC)

Lebesgue–Stieltjes measure is a regular Borel measure,
The statement "The Lebesgue–Stieltjes measure is a regular Borel measure, and conversely every regular Borel measure on the real line is of this kind." is wrong. regular Borel measure is defined on the sigma-algebra of Borel subsets while the Lebesgue–Stieltjes measure is the unique extension of the Borel measure in question onto the sigma algebra of the corresponding Lebesgue-measurable subsets due to the standard Lebesgue expansion procedure. Thus, they are defined on differents algebras and they cannot be the same. 193.136.33.223 (talk) 15:50, 12 August 2023 (UTC)