I got stuck on exercise 11 in chapter 6 of Michael Taylor's Measure Theory and Integration. Let $Z=\prod_{k=1}^{+\infty} \Big\{ 0,\,2\Big\}$ and $G: Z \to [0,1]$ given by $$ G(a)=\sum_{k=1}^{+\infty}a_{k}\,3^{-k}.$$
I'm stuck on the second part of the exercise. Define $K$ to be the cantor set and equip $Z$ with the product measure $\mu = \prod_{n=1}^{+\infty} P_{n}$ where $P_{n}$ is the uniform distribution on $\{ 0,\, 2 \}.$ Define $\mu_{K}(S)= \mu(G^{-1}(S)).$ Show that $\mu_{K}$ is the Lebesgue-Stijeles(https://en.wikipedia.org/wiki/Lebesgue%E2%80%93Stieltjes_integration) measure induced by the Cantor function(https://en.wikipedia.org/wiki/Cantor_function) $\varphi.$ I've included what I've done so far, including the solution to the first part, in case it contains any useful elements. The first question consisted in proving that $G$ is a homeomorphism onto Cantor set; the function $G$ is obviously continuous and closed (Z has the discrete topology, and G goes from a compact set to a T2 space) then I showed G is 1-1 and finally that $G(Z)=K.$ To show that $G(Z)=K,$ I observed that $$ K=\cap_{n=1}^{+\infty} I_{n},$$ where $$ I_{n}:= [0,\frac{1}{3}]\cap \frac{1}{3}\,I_{n-1} \cup [\frac{2}{3},1]\cap \Big( \frac{1}{3}I_{n-1} + \frac{2}{3} \Big). $$ Finally, I pointed out that $$ I_{n}=\Big\{ x \in [0,\,1]: a_{i} \in \{0,\,2\} \quad i = 1,\dots,n \Big\}. $$ As for the second part of the exercise, I intended to prove the proposition about $\mu_{K}$ on all the intervals of the form $$[ \frac{k}{3^{n}},\,\frac{k+1}{3^{n}}] $$ with $k \in \mathbb{N}, \, k \leq 3^{n}-1$, and then extend the result with a standard 'density' argument. I proved that when $k$ is odd then $$ \mu_{K}\Big(\,[\frac{k}{3^{n}},\,\frac{k+1}{3^{n}}]\Big)= 0 = \frac{1}{2^{n}}-\frac{1}{2^{n}}\\ =\varphi(\frac{k+1}{3^{n}})-\varphi(\frac{k}{3^{n}})= \int_{\frac{k}{3^{n}}}^{\frac{k+1}{3^{n}}}d\varphi. $$ I then proved that $$ \mu_{K}((0,\frac{1}{3^{n}}))=\mu_{K}\Big(\{ 0 \} \times \dots \times \{ 0 \} \times \prod_{k=n+1}^{+\infty} \Big\{ 0,\,2 \Big\}\Big)= \frac{1}{2^{n}}. $$ What I need to conclude is a sort of invariance by translation property of the two measures: if I translate the interval of $\frac{2}{3^{n}}$ to the right the result does not change. I tried to show that $$ \varphi(x+\frac{2}{3^{n}})= \frac{1}{2^{n}}+\varphi(x), $$ but I did not succeed. Any suggestion would be welcomed.