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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.

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    $\begingroup$ 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. $\endgroup$
    – GEdgar
    Feb 9 '12 at 15:08
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    $\begingroup$ We've seen this question before, where $f(x)=x^n$ for various values of $n$. $\endgroup$ Feb 9 '12 at 16:18
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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 $$\int x\, m_\varphi(dx) = \int (1-x)\, m_\varphi(dx) = 1 - \int x\,m_\varphi(dx).$$

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  • $\begingroup$ 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! $\endgroup$
    – Vulcan
    Feb 11 '12 at 22:01
  • $\begingroup$ @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. $\endgroup$ Feb 28 at 20:20
  • $\begingroup$ 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 $x \mapsto 1-x$. $\endgroup$ Feb 28 at 20:25
  • $\begingroup$ 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 $f$. $\endgroup$ Feb 28 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 $\int\varphi_n\ dm = \frac16\sum_{j=1}^n \left(\frac23\right)^j \to \frac16\cdot\frac{1}{1-\frac23}=\frac12$

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