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I am not an expert on measure theory. I am sorry if this question is too simple for some experts here.

Suppose the measure $\mu$ is singular continuous on $\mathbb{R}$, such as the cantor measure. Are there good references about the calculus with respect this measure?

Here is one concrete question. Let $\mu$ be the Cantor measure, what is a general procedure to evaluate

$$ \int_0^1 f(x)\mu(d x) =? $$

for some continuous function $f(x)$. For some specific functions, such as $f(x)=x$, $f(x)=1-x$, $f(x)=1$, one can use integration by parts and symmetry to solve this problem. How about for general continuous function $f$?

Thanks a lot!

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  • $\begingroup$ What on earth is the stochastic-calculus tag doing here? $\endgroup$
    – Yemon Choi
    Sep 16, 2014 at 23:34
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    $\begingroup$ @YemonChoi: I removed it. I also voted to migrate to Math.SE since I think it is a better fit there. $\endgroup$ Sep 16, 2014 at 23:43
  • $\begingroup$ Yemon Choi, people working on stochastic calculus might have this knowledge as well. $\endgroup$
    – Anand
    Sep 17, 2014 at 0:51
  • $\begingroup$ Nate Eldredge, may I know what is Math.SE? Thanks. $\endgroup$
    – Anand
    Sep 17, 2014 at 1:02
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    $\begingroup$ Anand, just because people working on X might know how to answer your question, it doesn't mean your question belongs under that tag. $\endgroup$
    – Yemon Choi
    Sep 17, 2014 at 1:24

1 Answer 1

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A typical way to integrate continuous functions against complicated measures is to approximate them (in the weak topology) by simpler measures.

One approach for the Cantor measure is to follow the middle-thirds construction, where you write the Cantor set as a decreasing intersections of sets $C_n$ each of which is a finite union of closed intervals. If $\mu_n$ is the Lebesgue measure on $C_n$, normalized so that $\mu_n(C_n)=1$, then $\mu$ is the weak limit of the $\mu_n$. So you can say $$\int f\,d\mu = \lim_{n \to \infty} \frac{1}{m(C_n)} \int_{C_n} f\,dm.$$ You could also approximate by discrete measures: let $B_n$ be the finite set consisting of the endpoints of the $2^n$ intervals remaining after the $n$th stage of the construction (so $|B| = 2^{n+1}$). Normalized counting measure on $B_n$ converges weakly to $\mu$, i.e. $$\int f\,d\mu = \lim_{n \to \infty} 2^{-(n+1)} \sum_{x \in B_n} f(x).$$

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  • $\begingroup$ Thanks Nate Eldredge for your help! Do you have a reference for this so that I can learn it more systematically? In my problem, I have a general singular continuous measure, not necessary the Cantor measure. Thanks a lot! $\endgroup$
    – Anand
    Sep 17, 2014 at 1:06
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    $\begingroup$ @Anand: Weak convergence of measures is covered in most probability texts, but the canonical introduction is Billingsley's Convergence of Probability Measures. $\endgroup$ Sep 17, 2014 at 1:08
  • $\begingroup$ thanks a lot! I agree. But all information that I know about my measure $\mu$ is that it is a singular continuous (nonnegative) measure. In this setting, do you have any idea to construct a sequence of measures $\mu_n$ that converge to $\mu$ in the weak sense? Is there a general construction procedure? Thank you very much! $\endgroup$
    – Anand
    Sep 17, 2014 at 1:12
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    $\begingroup$ To get an effective approximation strategy, you need to know something about the measure you're using and its properties. Asking "how do I integrate against a general measure" is like asking "how do I find the cardinality of a set" - strategies for finding the cardinality of $\{1,2\} \cup \{2,3,4\}$ will be much different from those for finding the cardinality of $\aleph_{\aleph_5}^{\aleph_6}$. It would help to know what kind of results you're trying to prove. $\endgroup$ Sep 17, 2014 at 1:15
  • $\begingroup$ I agree. According to the Lebesgue's decomposition theorem, $\mu$ has three parts. I study them separately. The absolutely continuous part is easy; the pure jump part reduces to the study of the Dirac delta measure. The problem arises if the singular continuous part is not vanishing, I need to prove something. Thanks a lot! $\endgroup$
    – Anand
    Sep 17, 2014 at 1:19

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