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consider the ternary cantor set C, and the asscoiated cantor function f, and the associated Lebesgue-Stieltjes measure u.

what is the integral of f over all of R with respect to u?

my attempt:

i know that under the Lebesgue measure, the integral of the cantor function is 1/2 using a symmetry argument.

but under this measure, u, is the integral still the same?

i was thinking about breaking up the integral into 3 parts: (-infty, 0), (0, 1), and (1, infty).

since the cantor function is constant on (-infty, 0) and (1, infty), it follows that the integral is 0 on those intervals...

now im kind of stuck here because im not sure how to handle (0, 1) would be appreciated

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Try using the iterative definition of the Cantor function, which gives a sequence of functions that converge uniformly to the Cantor function; then integrate each of those (or try a few and see if you can spot a pattern). – Arturo Magidin Feb 9 '11 at 5:23
thanks Arturo. this may sound like a stupid question but how do i integrate y=x under this measure, say on [0,1]. by definition, the integral is equal to the sup of the integral of all simple functions that sit underneath not sure where to go from here – jack Feb 9 '11 at 15:40
@jack: Simple functions are easy to integrate: $$\int \sum\alpha_i\chi_{E_i} du = \sum\alpha_iu(E_i) = \sum\alpha_i\int_{E_i}fd\lambda.$$So integrating simple functions mounts to integrating $f$ over measurable sets contained in $[0,1]$. – Arturo Magidin Feb 9 '11 at 19:06
@Arturo: y=x is not a simple function – jack Feb 13 '11 at 0:39
@jack: Yes, I know $y=x$ is not a simple function; my point was that since you know the integral is the sup of the simple functions, and simple functions are easy to integrate, you could have tried to go from there. At this point, though, you have Shai Covo's answer below, so you don't have to work it out from first principles. – Arturo Magidin Feb 13 '11 at 3:58
up vote 5 down vote accepted

You should get that the answer is $1/2$, immediately from integration by parts. See this, in particular the sections "Integration by parts" and "Related concepts". Further, see this for a proof of the integration by parts formula.

EDIT: Explicitly, since $f$ is continuous and non-decreasing, and is constant on $\mathbb{R}-[0,1]$, $\int_{\mathbb R} {f(x)du(x)} = \int_0^1 {f(x)df(x)} $, and it holds $$ \int_0^1 {f(x)df(x)} = f(1)f(1) - f(0)f(0) - \int_0^1 {f(x)df(x)} $$ (integration by parts). Since $f(0)=0$ and $f(1)=1$, it thus follows that $\int_{\mathbb R} {f(x)du(x)} = 1/2$.

EDIT: More generally, if $F$ is any continuous distribution function (the Cantor function $f$ is a particular example), then $\int_{\mathbb R} {F(x)dF(x)} = 1/2$. As before, this can be proved using integration by parts, which is allowed since $F$ is continuous and non-decreasing. (Under certain conditions, this also follows from a change of variable.) Indeed, for any $a < b$, $$ \int_a^b {F(x)dF(x)} = F(b)F(b) - F(a)F(a) - \int_a^b {F(x)dF(x)}. $$ Hence the result follows by letting $a \to -\infty$ and $b \to \infty$.

Another way to obtain the general result is as follows. Let $X$ be an arbitrary random variable with continuous distribution function $F$. Then, $\int_{\mathbb R} {F(x)dF(x)}$ expresses the expectation of the random variable $F(X)$. As is well known, in this case $F(X) \sim {\rm uniform}(0,1)$. Hence, $\int_{\mathbb R} {F(x)dF(x)} = 1/2$.

Remark. With $f$ as above, integration by parts gives $$ \int_0^1 {xdf(x)} = xf(x)\big|_0^1 - \int_0^1 {f(x)dx} = 1 - 1/2 = 1/2. $$ The left-hand side, $\int_0^1 {xdf(x)}$, expresses the expectation of the Cantor distribution.

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Let us first observe the following restricted change-of-variables type formula for Stieltjes integrals: $$\int_{f(I)} g(x)\,dx = \int_I g(f(y))\,df(y)$$ for every Borel $I\subseteq\mathbb{R}$ and every non-decreasing continuous $f:I\rightarrow\mathbb{R}$:
compare the corresponding Darboux sums and see also this question.

In particular for $f:[0;1]\rightarrow[0;1]$ the Cantor function and $g(x):=x$, $1/2=\int_0^1 x\,dx = \int_0^1 f(y)\,df(y)$.

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