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This question is a follow-up to the post "Calculating a Lebesgue integral involving the Cantor Function."

Let $\varphi: [0,1] \rightarrow [0,1]$ be the Cantor (ternary) function, and let $m_\varphi$ be the Lebesgue-Stieltjes measure associated to it.

Attached is an excerpt from the answer received on this post:

Notice that $\varphi(1-x) = 1 - \varphi(x)$. From this, it is easy to show that the Cantor measure $m_\varphi$ is invariant under the transformation $x \mapsto 1-x$.

I know very little about this type of measure, and hence have had trouble seeing why the above is easy to show. I tried to ask the responder about this point, but to no avail. Hence, I am posting this to see if anyone visiting would be up for either explaining this point about invariance, or giving me a push in the right direction to proving it.

Also: the responder changed the notation in the integral from $dm_\varphi$ to $m_\varphi dx$, and I would be grateful for any clarification as to the relationship of these two notations.

Thank you in advance for any response!

Community
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Vulcan
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  • Concerning the notation, see http://math.stackexchange.com/questions/5230/notation-question-integrating-against-a-measure/5237#5237 –  Feb 12 '12 at 18:59

1 Answers1

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It should just be a consequence of

$\int_a^bf(x)dg(x) = f(b)g(b)-f(a)g(a)-\int_a^b g(x)df(x) $

which can be seen as integration by parts for the Stieltjes integral.

Not sure about the Lebesgue-Stieltjes integral myself, but it's not too hard to show that this holds for the Riemann-Stieltjes integral: you just have to rollback to the definition and rearrange the terms of the sum in order to get something like

$\sum f(c_i)(g(x_{i+1})-g(x_i)) = f(a)g(a)-f(b)g(b)- \sum g(x_i)(f(c_{i+1})-f(c_i))$

From this you should be able to give sense of the answer you got:

$\int_a^b \varphi(x)d\varphi(x) = \varphi(b)\varphi(b)-\varphi(a)\varphi(a)-\int_a^b \varphi(x)d\varphi(x) = 1 - \int_a^b \varphi(x)d\varphi(x)$

Manlio
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