What measure does Lebesgue measure induce on the fat Cantor set?

I know that the fat Cantor set under the subspace topology is homeomorphic to Cantor space $\{0,1\}^{\mathbb N}$ under the product topology induced by the discrete topology on $\{0,1\}$. Call the natural homeomorphism $f$.

What about the measure induced by Lebesgue measure on Cantor space via $f$? Is it the same (up to a constant) as the usual product measure, i.e. the probability measure used to describe an infinite sequence of coin tosses? A reference would be helpful.

• All the uncountable Polish spaces have the same Borel sets, so their measure theory is fairly the same. Apr 30 '13 at 1:11
• Are you saying that the Borel measure on the fat Cantor set (probably) induces the product measure on Cantor space? Or that Lebesgue measure does? Apr 30 '13 at 1:21
• I guess you're saying that the Borel measure induces the product measure and Lebesgue measure induces its completion. The product measure on Cantor space isn't complete, right? Apr 30 '13 at 1:50
• I don't remember all the gory details, in fact I don't even remember the exact proof that the Borel sigma algebras of all the uncountable Polish spaces are isomorphic. But the point is that the result might not be a scalar multiplication of the usual product measure, but it should be some very nice measure with respect to the product measure. Apr 30 '13 at 1:56
• @AsafKaragila: What do you mean by nice? It could very well be singular. Apr 30 '13 at 19:24

Yes, it is.

Note that the Cantor set is in fact a compact group, and the product measure $\mu$ is the Haar measure, that is, the invariant finite, regular Borel measure, which is by Haar's theorem unique up to a scaling factor.

That Lebesgue measure restricted to the fat Cantor set is invariant is an easy exercise (it follows easily from the fact that the Lebesgue measure is translation invariant).

It is also regular, because it is a restriction of a regular measure, so by Haar's theorem it is a scalar multiple of the product measure.

Let $C$ be the fat Cantor set; $\mu$ is the restriction of the Lebesgue measure to $C$, normalized so that $\mu(C)=1$. For every $n\ge 1$, the pre-Cantor set of generation $n$ consists of $2^{-n}$ intervals $I_{n, k}$ of equal length, and the sets $C\cap I_{n,k}$ are congruent to one another. Therefore, $\mu(C\cap I_{n,k})=2^{-n}$. Observe that this is the same measure that $I_{n,k}$ gets from coin-tossing. Also, the intervals $I_{n,k}$ generate the Borel $\sigma$-algebra because every open subset of $C$ is a union of such intervals. Conclusion: yes, the measures are the same up to normalization.

• Looks good, except $\mu(I_{n,k})$ doesn't really make sense, you should write $\mu(C\cap I_{n,k})$ instead. Apr 30 '13 at 19:54
• @tomasz Thanks, I made the corrections. Apr 30 '13 at 21:04
• This answer is also helpful, thanks. Apr 30 '13 at 22:18

Caution: Thanks to tomasz, I realize my "answer" is almost surely irrelevant to OP's question. Yet I decide not to delete it in case it might interest some others. Perhaps more importantly, it's kind of difficult for me to delete something that took me quite a while to finish.

This is an elaboration of Asaf Karagila's comment, "all the uncountable Polish spaces have the same Borel sets".All the stuff is borrowed from Probability measures on metric spaces, Parthasarathy(1967).

Main result: Two Borel sets of Polish spaces are isomorphic, iff they have the same cardinality.

Defination: Given Polish spaces $$X_1$$ and $$X_2$$, Borel sets, $$B_1$$ and $$B_2$$ and $$B_1 \subseteq X_1$$, $$B_2 \subseteq X_2$$, $$B_1$$ is isomorphic to $$B_2$$, which is denoted as $$B_1 \sim B_2$$, iff there is a bijection $$\phi : B_1 \to B_2$$, and both $$\phi$$ and its inverse, $$\phi^{-1}$$ are measurable functions.

Theorem $$1$$: Let $$M = 2^{\omega}$$ endowed with product topology, there exists a Borel subset $$E \subseteq M$$ such that $$E \sim [0,1]$$.

Proof: Consider the well-known continuous surjective function $$\tau$$ from the Cantor space onto the interval $$[0,1]$$: $$(x_n)\mapsto\sum_{n\in\omega}\frac{x_n}{2^{n+1}}$$ The problem for this function is that it's not injective, which means, we need to make a choice between eventual constant sequences that ends with $$1$$ and those ending with $$0$$. One such set $$E$$ is the set of sequences in which $$0$$ occurs infinite times and the sequence that only comsists of $$1$$. It's not difficult to see the function $$\tau$$ restricted on $$E$$ and its inverse are both continuous which implies that they're also measurable.

Theorem $$2$$: There exists a Borel set $$E_1 \subseteq M$$,$$(M = 2^{\omega})$$ such that $$E_1 \sim [0,1]^{\omega}$$.

Proof: With the help of the bijection defined in Theorem $$1$$, we have the induced bijection from $$2^{E}$$ to $$[0,1]^{\omega}$$.Since $$2^{\omega} = (2^{\omega})^{\omega}= 2^{\omega \times \omega}$$, $$M$$ is homeomorphic to $$2^{M}$$ by $$\iota$$, hence $$M \sim 2^{M}$$. Thus there's a subset of $$M$$,$$E_1 = \iota^{-1}(E)$$. So we have $$\tau \circ \iota$$ that guarentee the isomorphism between $$E_1$$ and $$[0,1]^{\omega}$$.

Theorem 3: Let $$X$$ be a Polish space, and $$G \subseteq X$$ be a Borel set. Then there exists a Borel set $$E' \subseteq M$$ such that $$G \sim E'$$.

Proof:Theorem 2 makes our life easier by enabling us to turn our attention away from the Cantor space $$M$$ to Hilbert cube $$[0,1]^{\omega}$$. One important result is that every Polish Space is homeomorphic to a $$G_\delta$$ subspace of the Hilbert Cube (see here). We are done.

• Remark:Thanks to Theorem $$1$$ to $$3$$, we don't need to consider an arbitary Polish space. We only have to prove the theorem for the Cantor space.

Theorem 4: Let $$\mathscr{N} = \omega^{\omega}$$ denote the Baire space. For any Polish space $$X$$, there's a continuous surjection $$\psi$$ from $$\mathscr{N}$$ onto $$X$$.

Proof: This theorem is omnipresent in various textbooks, notes, et cetera. e.g. See here.

Theorem 5: Let $$X$$ be a Polish space and $$E \subseteq X$$ be a Borel set, then there exist a continuous function $$\psi$$ that maps $$\mathscr{N}$$ into $$X$$ such that $$\psi(\mathscr{N}) = E$$.

Proof: We show this by induction. It's obvious that it's true for closed sets, since they're complete with the same metric in $$X$$. Suppose it holds for a denumerable set of subsets of $$X$$, which is $$\{A_n\}_{n \in \omega}$$ with $$\psi_n(\mathscr{N}) = A_n$$. Define a continuous function as a cartesan product of $$\{\psi_n\}_{n \in \omega}$$, i.e. $$(\psi_0, \psi_1, \psi_2, \ldots) : \mathscr{N}^{\omega} \to X^{\omega}$$. Also define a subset of $$\mathscr{N}^{\omega}$$ as $$\mathscr{N}' = \{ x \in \mathscr{N}^{\omega} : \forall m, n \in \omega \psi_m(x) = \psi_n(x)\}$$. As a continuous preimage of a diagonal set, $$\mathscr{N}'$$ is closed in $$\mathscr{N}^{\omega}$$. Now define $$\psi :$$\mathscr{N}'$$\to X$$, for any $$x \in \mathscr{N}'$$ and let $$\pi_1$$ projects $$x$$ to its first component. $$x \mapsto \psi_1(\pi_1(x))$$ Now we have $$\psi(\mathscr{N}') = \bigcap_{n \in \omega}A_n$$. Notice that $$\mathscr{N}$$ is homeomorphic to $$\mathscr{N}^{\omega}$$. So all we need is a composition of two isomophism. To complete the proof, we also have to prove it holds for $$\bigcup_{n \in \omega}A_n$$. Let $$\mathscr{N}_j = \{ x \in \mathscr{N}: \pi_1(x) = j\}$$ for each $$j$$. $$\mathscr{N}_j$$ is isomorphic to $$\mathscr{N}$$, so there a continuous surjection $$\phi_j$$such that $$\phi_j(\mathscr{N}_j) = A_j$$. Define $$\phi : \mathscr{N} \to \bigcup_{n \in \omega}A_n$$, for each $$x$$ in $$\mathscr{N}$$: $$x \mapsto \phi_j(x) \text{iff } \pi_1(x) = j$$.

Theorem $$6$$: Given $$X$$ an uncountable seperable metric space, then there is a partition of $$X$$ with two atoms. One atom is countable, and the other, which is dense in itself, consists of all condensation points.

Proof: This is essentially Cantor–Bendixson theorem.

Theorem $$7$$: Let $$X$$ be a Polish space, $$Y$$ be an uncountable seperable metric space. Assume that there exists a continuous surjection $$\phi$$ from $$X$$ onto $$Y$$. Then there exists a subset $$C \subseteq X$$, such that $$\phi : C \to Y$$ is a homeomorphism. Moreover, $$C$$ is homeomorphic to the Cantor space, and $$\phi(C)$$ is compact.

Proof: With $$\bf{AC}$$, we can construct a subset $$A \subseteq X$$ such that $$\phi$$ ristricted to $$A$$ is a bijection. By Theorem $$6$$, we have a subset $$D \subseteq A$$ such that $$D$$ is an uncountable perfect set. Notice that $$D$$, as a Hausdorff uncountable perfect set contains a copy $$C$$ of Cantor set, which is the same as the construction in the proof of $$|\mathbb{R}| = 2^{\aleph_0}$$. $$\phi(C)$$ is compact follows from the fact it's homeomorphic to the Cantor space, which is compact.

Corrollary: Let $$X$$ be a Polish space, $$E$$ be an uncountable Borel set in $$X$$. Then there exists a compact subset $$C \subseteq E$$ such that $$C$$ is homeomorphic to the Cantor set. In particular, $$|E| = \mathfrak{c}$$.

Lemma: Given $$\{E_m\}_{m \in \omega}$$ a set of disjoint Borel sets, and $$\{F_n\}_{n \in \omega}$$ the same, then $$E_i \sim F_i$$ for all $$i \in \omega$$ implies that $$\bigcup_{i \in \omega} E_i \sim \bigcup_{i \in \omega} F_i$$

Theorem 8: Let $$E_1 \subseteq F \subseteq E_0$$ be three Borel subsets of Cantor space. Then $$E_0 \sim E_1$$ implies that $$E_0 \sim F$$.

Proof: Let $$\psi : E_0 \to E_1$$ be the mapping that guarentee the isomorphism. Define recursively, $$E_{n+1} = \psi(E_n)$$, $$A_{n+1} = E_n \setminus F$$, $$B_{n} = F \setminus E_n$$, $$D_{n+1} = E_n \setminus E_{n+1}$$, $$E_{\infty} = \bigcap_{n \in \omega} E_n$$. Via $$\psi^{m-n}$$ and $$\psi^{i-j}$$ , we have $$D_m \sim D_n$$ for all $$m,n \in \omega$$,$$E_i \sim E_j$$ for all $$i,j \in \omega$$. Similarly, it also holds for the class $$\{A_n\}$$ and $$\{B_m\}$$. Thus we have: $$E = E_{\infty} \cup \bigcup_{i \in \omega}A_i \cup \bigcup_{j \in \omega}B_j \sim E_{\infty} \cup \bigcup_{i \in \omega \setminus \{0\}}A_i \cup \bigcup_{j \in \omega}B_j = F$$

Theorem $$9$$:Two Borel sets $$E_1$$ and $$E_2$$ of Polish spaces $$X_1$$ and $$X_2$$ are isomorphic, if they have the same cardinality.

Proof: As shown, WLOG, Let $$X_1 = X_2$$ be the Cantor spaces. The only interesting case is when $$E_1$$ and $$E_2$$ are uncountable. To prove the theorem, it suffice to show $$E_1 \sim E_2 \sim 2^{\omega}$$. Since both $$E_1$$ and $$E_2$$ contains a copy of the Cantor space, by appealing to Theorem $$8$$, we have $$E_1 \sim 2^{\omega}$$ and $$E_2 \sim 2^{\omega}$$. Moreover $$\sim$$ is an equivalence relation, we have $$E_1 \sim E_2$$.

• How does this answer the question? May 1 '13 at 3:15
• @tomasz: I'm sorry I don't follow your comment. There's nothing special about "Lebesgue measure induce on the fat Cantor set". All uncountable Polish spaces are isomorphic. May 1 '13 at 3:21
• The question was about two particular measures, not Borel algebras. Any two uncountable Polish spaces are Borel isomorphic. May 1 '13 at 3:27
• @tomasz: Thank you for pointing that out. To be frank, I didn't read the question carefully and haste to give an answer, which is in effect a supplement on Asaf's comment. Hope you don't mind asking. What can we say about two measures, if we've already known the isomorphism between two Borel algebras. May 1 '13 at 4:01
• As far as I can tell, nothing really; that the Borel algebras are isomorphic is really trivial in this case, because the fat Cantor set is homeomorphic to the usual Cantor set, which is much stronger than mere Borel isomorphism. What the theorem really gives us is that when considering an abstract Borel measure on a Polish space, we can very well assume that the Polish space is any uncountable Polish space we want. But two measures on a Polish space can be rather different (singular), e.g. the Lebesgue measure on reals and the image of the Cantor measure by the usual embedding. May 1 '13 at 8:02