Hausdorﬀ dimension of the product of two generalized cantor sets It is Exercise 9, Chapter 7 from Stein and Shakarchi's Real Analysis.

Consider the set $C_{\xi_1} \times C_{\xi_2}$ in $R^2$, with $C_{\xi}$ as in the previous exercise. Show that $C_{\xi_1} \times C_{\xi_2}$ has strict Hausdorff dimension $\dim(C_{\xi_1}) + \dim(C_{\xi_2})$.

$C_{\xi}$ is defined similarly to the standard Cantor set. We remove the middle part of length $\xi$ instead of length $1/3$.
It is trivial when $\xi_1 = \xi_2$. I have so far failed to prove the general case.
Below is a list of relevant theorems provided in the book.
Lemma 2.2 Suppose a function $f$ deﬁned on a compact set $E$ satisﬁes a Lipschitz condition with exponent $\gamma$. Then

*

*$m_\beta(f(E)) \le M^\beta m_\alpha(E)$ if $\beta=\alpha/\gamma$.

*$\dim f(E) \le \frac{1}{\gamma} \dim E$
Theorem 2.9 Suppose $S_1$, $S_2$, ..., $S_m$ are $m$ similarities, each with the same ratio $r$ that satisﬁes $0 < r < 1$. Then there exists a unique nonempty compact set $F$ such that $$F=S_1(F) \cup \cdots \cup S_m(F)$$
Theorem 2.12 Suppose $S_1$, $S_2$, ..., $S_m$ are $m$ separated similarities with the common ratio $r$ that satisﬁes $0 < r < 1$. Then the set $F$ has Hausdorff dimension equal to $\log m / \log(1/r)$.
 A: Let $r_i=(1-\xi_i)/2$ so $\alpha_i=\dim(C_{\xi_i})=\frac{\log 2}{|\log r_i|}$. Write $\alpha=\alpha_1+\alpha_2$. To bound the $\alpha$ dimensional measure $m_\alpha$ of $\Lambda:=C_{\xi_1} \times C_{\xi_2}$ from above, fix a large $n$ and find $k=k(n)$ such that $r_2^{k+1} \le r_1^n <r_2^k$. Then $\Lambda$ can be covered by $2^{n+k}$ rectangles of width $r_1^n$ and height $r_2^k$, so
$$m_\alpha(\Lambda) \le \lim_n 2^{n+k} (2r_2^k)^\alpha \le 
2^\alpha \lim_n \Bigl[(2r_2^{\alpha_2})^k \cdot (2 r_1^{\alpha_1})^n \Bigr] \cdot r_2^{-\alpha_1} =2^\alpha r_2^{-\alpha_1}\,.$$
On the other hand, for each $i=1,2$, the set $C_{\xi_1}$ supports a Borel probability measure $\mu_i$ such that $\mu_i(J) \le c_i |J|^{\alpha_i}$
for all intervals $J$ and some constants $c_i$. Then $\mu=\mu_1 \times \mu_2$ satisfies $\mu(Q) \le c_1 c_2 {\rm diam}(Q)^\alpha$ for all squares $Q$. A uniform lower bound for $m_\alpha(\Lambda)$ now follows from the mass distribution principle, see near page 55 in [1] or Lemma 1.2.8 in [2].
[1]  Falconer, Kenneth. Fractal geometry: mathematical foundations and applications. John Wiley & Sons, 2004.
[2] https://www.yuval-peres-books.com/fractals-in-probability-and-analysis/
