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Im stuck with this exercise

For $A\subset\Bbb R^n$ and $B\subset \Bbb R^m$ show that $\dim_H(A\times B)=\dim_H(A)+\dim_H(B)$

where $\dim_H$ is the Hausdorff dimension. I know that when $A$ and $B$ are open the above holds. However Im unable to generalize the result.

I tried to relate the following inequalities and identities

$$\operatorname{diam}(A)\lor\operatorname{diam}(B)\le\operatorname{diam}(A\times B)\le\operatorname{diam}(A)+\operatorname{diam}(B)\tag1$$

$$\mathcal H_*^r(A)<\infty\implies\mathcal H_*^s(A)=0,\quad\forall s>r\\ \mathcal H_*^r(A)>0\implies\mathcal H_*^s(A)=\infty,\quad\forall s\in[0,r)\tag2$$

$$x,y\in[0,1]\implies xy<x+y\,\text{ and }\, x^{r+s}<x^r,\quad\forall r,s>0\tag3$$

$$\inf A+\inf B=\inf(A+B)\text{ and }\sup A+\sup B=\sup(A+B)\tag4$$

where $\mathcal H_*^s$ is the $s$-dimensional Hausdorff outer measure and $\rm diam$ is the diameter of a set. By example I find that

$$\dim_H(A)+\dim_H(B)=\inf\{r>0:\exists \alpha\in[0,r]\text{ such that }\mathcal H_*^{r-\alpha}(A)+\mathcal H_*^\alpha(B)=0\}\tag5$$

so a line of action is try to relate $(2)$ and $(5)$ in something like

$$\mathcal H_*^{r-\alpha}(A)+\mathcal H_*^\alpha(B)=0\implies\mathcal H_*^r(A\times B)=0\tag6$$

using the definition of $\mathcal H_*^s$, $(1)$ and maybe $(3)$. However I found nothing. Some help will be appreciated.


EDIT: to clarify some things: from the definitions of Hausdorff dimension the statement to be proved can be stated as $$ \begin{align}\dim_H(A)+\dim_H(B)&=\inf\left\{s+t>0:\sup_\epsilon\inf\left\{\sum_{k=0}^\infty a_k^r+b_k^s: a_k,b_k<\epsilon\right\}=0\right\}\\ &=\inf\left\{s+t>0:\sup_\epsilon\inf\left\{\sum_{k=0}^\infty c_k^{r+s}: c_k<\epsilon\right\}=0\right\}\\ &=\sup\left\{s+t\ge0:\sup_\epsilon\inf\left\{\sum_{k=0}^\infty a_k^r+b_k^s: a_k,b_k<\epsilon\right\}=\infty\right\}\\ &=\sup\left\{s+t\ge0:\sup_\epsilon\inf\left\{\sum_{k=0}^\infty c_k^{r+s}: c_k<\epsilon\right\}=\infty\right\}\\ &=\dim_H(A\times B)\end{align}\tag{*} $$ where $a_k,b_k,c_k$ are the diameters of sequences of covers $(A_k),(B_k),(C_k)$ of $A\subset\Bbb R^n$, $B\subset\Bbb R^m$ and $A\times B\subset\Bbb R^{n+m}$ respectively.

This together with $(2)$ seems the way to go, however I can't found appropriate bounds because I can't relate covers $A$ and $B$ with covers of $A\times B$ such that it make possible to find these bounds.

At my disposal, in the context where this exercise appear, there is not too many theorems to solve this exercise, by example I dont know the Frostman's lemma that @DavidUlrich state in the comment. I need to solve it from elementary theorems as the stated above.

P.S.: I know some more identities related to the Hausdorff dimension, by example that is an increasing function or that $\dim_H(f(A))\le\dim_H(A)$ for $f$ Lipschitz or that $\dim_H(\bigcup_k A_k)=\sup_k\dim_H(A_k)$.

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    $\begingroup$ The fact that $\dim(A\times B)\le\dim(A)+\dim(B)$ is easy from the definition. For the other inequality apply Frostman's Lemma on $A$ and on $B$, then consider the measure $\mu_A\times\mu_B$ on $A\times B$. $\endgroup$ – David C. Ullrich Apr 28 '18 at 16:35
  • $\begingroup$ @David Im unable to find such solution (I refer to the "easy" inequality you says) and the Frostman lemma is not known in the context where this exercise appear. $\endgroup$ – Masacroso Apr 29 '18 at 16:36
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    $\begingroup$ The assertion is false. Indeed, there are sets $A$ and $B$ in $\mathbb R$ of Hausdorff dimension zero whose Cartesian product exceeds dimension one. This appears as example 7.8 on page 97 of the second edition of Falconer's Fractal Geometry. If, however, at least one of those sets has Hausdorff dimension equal to it's upper box-counting dimension, then the equality can be proved (as it's done in that text). $\endgroup$ – Mark McClure Apr 30 '18 at 14:39
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    $\begingroup$ @MarkMcClure thank you very much, you saved my life! $\endgroup$ – Masacroso Apr 30 '18 at 14:49
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    $\begingroup$ @Masacroso WOO - I'm a hero! (A slight overstatement, perhaps, but I'm glad it helps. :) $\endgroup$ – Mark McClure Apr 30 '18 at 14:50

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