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I wanted to know if the problems mentionned in this book are solved. More specifically, at some places, the author says that he doesn't know the answer, for example :"I do not know whether this equations are always true" p.361 4.1.8, or "I do not know..." p.189 2.10.26. Are there counterexamples or proofs of these asumptions? I add some details : in theorem 2.10.25, if $f:X\rightarrow Y$ is a lipschitzian map of metric spaces, $A\subset X$, $0\leq k<\infty$, and $0\leq m<\infty$, then $$ \int_Y^*\mathscr{H}^k(A\cap f^{-1}\{y\})d\mathscr{H}^m(y)\leq (\mathrm{Lip} f)^m \frac{\alpha(k)\alpha(m)}{\alpha(k+m)}\mathscr{H}^{k+m}(A). $$ (where $\mathscr{H}^n$ is the Hausdorff measure associated to the metrics on $X$ and $Y$, and $\int^*$ is the upper integral) provided either $\{y\in Y, \mathscr{H}^k(A\cap f^{-1})>0\}$ is the union of countable family of sets with finite $\mathscr{H}^m$ measure, or $Y$ is boundedly compact (each close bounded subset is compact). The question Federer asks is to determine if everything after "provided either..." is necessary.

The other question is about currents : let $S$, $T$ be two currents on open subsets $A$,$B$ of euclidean spaces, of degree $i$ and $j$ : if $S$, $T$ are representable by integration, do we always have $\Vert S\times T\Vert=\Vert S\Vert\times\Vert T\Vert$, and $\overrightarrow{S\times T}(a,b)=(\wedge_i p)\overrightarrow{S}(a)\wedge(\wedge_j q)\overrightarrow{T}(b)$ for $\Vert S\times T\Vert$ almost all $(a,b)\in A\times B$, where $p:A\rightarrow A\times B$ and $q:B\rightarrow A\times B$ are the canonical injections (it is indeed the case if either $\overrightarrow{S}$ or $\overrightarrow{T}$ is simple $\Vert S\Vert\times\Vert T\Vert$ almost everywhere).

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    $\begingroup$ I wonder if you might get better input from math overflow on this one. $\endgroup$ – James S. Cook Jul 18 '14 at 15:09
  • $\begingroup$ Thank you Mr Cook, I asked the question on math overflow and added the same détails here. $\endgroup$ – Paul-Benjamin Jul 20 '14 at 9:01
  • $\begingroup$ I'll put a bounty on your question here once it's possible. $\endgroup$ – James S. Cook Jul 20 '14 at 13:49
  • $\begingroup$ well, I tried. Sorry. $\endgroup$ – James S. Cook Jul 29 '14 at 16:03
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    $\begingroup$ It seems it has been proved in Measure theory and fine properties of functions, revised edition, by Evans. $\endgroup$ – Hua Apr 25 '17 at 9:33

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