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Let $f$ be a function such that agrees with the cantor function on $[0,1]$, vanishes on $(-\infty,0)$, and is identically $1$ on $(1,+\infty)$ and let $\mu_f$ the Borel measure associated to $f$. Show that

(a) each of the $2^n$ intervals remaining after the n-th step in the construction of the Cantor set has measure $2^{-n}$ under $\mu_f$,

(b) the Cantor set $C$ has measure $1$ under $\mu_f$,

(c) each $x$ in $\mathbb{R}$, satisfies $\mu_f(\{x\})=0$

$(c)$ follows immediately because $f$ is continuous and

\begin{align}\mu_f(\{x\}) &=\lim\mu_f\bigl((x-2^{-n},x]\bigr)\\ &=\lim \bigl(f(x)-f(x-2^{-n})\bigr)\\ &=f(x)-f(x)=0 \end{align}

$(b)$, Now $\mu_f([0,1])=\mu_f((0,1])=f(1)-f(0)=1$, and also the Cantor function is constant on each open interval which is removed to make up the Cantor set. Since the set $[0,1]-C$ is a countable disjoint union of open intervals and on each the Cantor function $f$ is constant, thus $$\mu_f([0,1]\setminus C)=\sum_n \mu_f((a_n,b_n))=\sum_n \bigl(F(b_n)-F(a_n)\bigr)=0$$ Hence $\mu_f(C)=\mu_f([0,1])=1$.

(a) For the first point I'm not sure.

The definition of my book of the Cantor singular function is the following: on the various intervals removed from the n−1 step for the construction of the nth step the $f(x)$ be $1/2^n,3/2^n,5/2^n\ldots$ for each such interval. After all the steps $f$ is defined on $[0,1]-C$, and for the points in the Cantor set we have

$$f(x):=\sup\{f(t):t<x \text{ and } t\in [0,1]-C\}$$

I'd appreciate any help for the first point.

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If the interval is $[a,b]$, $\mu([a,b]) = f(b) - f(a) = 2^{-n}$.

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  • $\begingroup$ Sorry but I can't see why the difference is always $2^{-n}$. $\endgroup$ – Jose Antonio Jul 29 '15 at 15:42
  • $\begingroup$ By construction of the Cantor function: as you said, at the $n$'th step you remove an interval of length $2^{-n}$ from the middle of each interval of length $2^{1-n}$, leaving intervals of length $2^{-n}$. $\endgroup$ – Robert Israel Jul 29 '15 at 20:30

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