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I came across the following challenging problem that concerns evaluating a Lebesgue integral rather than being asked to prove something about it:

Let $\varphi: [0,1] \rightarrow [0,1]$ be the Cantor (ternary) function, and let $m_\varphi$ be the Lebesgue-Stieltjess measure associated to it. Let $f(x) = x$. Evaluate $$\int_{[0,1]} f \; dm_\varphi.$$

A friend of mine suggested that if you study this integral in Mathematica, its value is fairly large, but I am having trouble thinking of how to proceed in computing the value without a computer algebra package. I post this question in hopes that anyone visiting will find the problem curious too, and to see if anyone visiting had some suggestions on how to proceed in computation.

Michael Hardy
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Vulcan
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    Another interesting approach is to come up with some natural random variable, for which $m_\phi$ is the CDF. Your problem is to find the mean of that random variable. – GEdgar Feb 09 '12 at 15:08
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    We've seen this question before, where $f(x)=x^n$ for various values of $n$. – Michael Hardy Feb 09 '12 at 16:18

2 Answers2

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The value of the integral is $\frac{1}{2}$, by symmetry.

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 $x \mapsto 1-x$. Thus $$\int x\, m_\varphi(dx) = \int (1-x)\, m_\varphi(dx) = 1 - \int x\,m_\varphi(dx).$$

Nate Eldredge
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  • I really like this approach and it leads to a very clean solution (as opposed to something more messy looking). Just to leave no stone unturned: what is the relationship between $dm_\varphi$ and $m_\varphi(dx)$? Also, I am having trouble thinking about how the invariance argument should go, so any help would be appreciated! – Vulcan Feb 11 '12 at 22:01
  • @Vulcan: Just different notation. For a measure $\mu$, the Lebesgue integral of $f$ with respect to $\mu$ can be written as $\int f,d\mu$ or $\int f(x) \mu(dx)$ or $\int f(x) d\mu(x)$. They all mean the same. – Nate Eldredge Feb 28 '21 at 20:20
  • The start for the invariance argument would be to note that for an interval $(a,b]$ we have $$m_\varphi((a,b]) = \varphi(b) - \varphi(a) = (1-\varphi(a)) - (1-\varphi(b)) = \varphi(1-a) - \varphi(1-b) = m_\varphi(1-b, 1-a]).$$ Since $\varphi$ is continuous and thus $m_\varphi$ puts no mass at points, this is the same as $m_\varphi([1-b, 1-a)$ which is the image of $(a,b]$ under the map $x \mapsto 1-x$. – Nate Eldredge Feb 28 '21 at 20:25
  • Since a Lebesgue-Stieltjes measure is defined in terms of its action on intervals, you can then go on to show that $m_{\varphi}(A) = m_\varphi(1-A)$ for all Borel sets $A$, and conclude $\int f(x) m_\varphi(dx) = \int f(1-x) m_\varphi(dx)$ for all integrable Borel functions $f$. – Nate Eldredge Feb 28 '21 at 20:26
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Here's a geometric way of understanding the solution. Since Lebesgue integration begins by building up a function out of simple functions, construct a sequence of simple functions that approaches the Cantor function from below. This wouldn't be very hard to do, although the notation is a little cumbersome. Here is the basic idea:

$\varphi_1(x) = \frac121_{[1/3,2/3]}(x)$

$\varphi_2(x) = \varphi_1(x) + \frac141_{[1/9,2/9]}(x) + \frac341_{[7/9,8/9]}(x)$

$\varphi_3(x) = \varphi_2(x) + \frac181_{[1/27,2/27]}(x) + \frac381_{[7/27,8/27]}(x)+\frac581_{[19/27,20/27]}(x) + \frac781_{[25/27,26/27]}(x)$

The first $\varphi_1$ is simply the very middle portion of the function, and then each subsequent $\varphi_n$ adds in the parts directly between $\varphi_{n-1}$. Notice that each of the functions is sort of averaging to $\frac12$. Their integrals are

$\int \varphi_1 \ dm = \frac12\cdot\frac13$

$\int \varphi_2 \ dm = \frac12\cdot\frac13 + \frac19(\frac14+\frac34) = \frac12\cdot\frac13+\frac19$

$\int \varphi_3 \ dm = \frac12\cdot\frac13+\frac19 + \frac1{27}(\frac18+\frac38+\frac58+\frac78) = \frac12\cdot\frac13+\frac19+2\cdot\frac1{27}$

Then it follows that $\int\varphi_n\ dm = \frac16\sum_{j=1}^n \left(\frac23\right)^j \to \frac16\cdot\frac{1}{1-\frac23}=\frac12$

Greg Zitelli
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