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Let $K$ be a compact set, $K \subset \mathbb{R}^n \times [a,b]$ and, for each $t \in [a,b]$ define $K_t = \{ x\in \mathbb{R}^n $ ; $(x,t) \in K\}$. If $\forall t \ K_t$ has measure zero in $\mathbb{R}^n$, then $K$ has measure zero in $\mathbb{R}^{n+1}$(In the sense os Lebesgue.)

Its a problem from an analysis book, and it should be true. (not a true x false question.)

I've been strugling to prove this assertion unsuccessfully. I don't need this result in this generality, one could restrict to the case $K = [0,1]^2$

I'm sorry for the abuse, but I really need an answer for this question. I feel i should use the fact that $[a,b]$ is both connected and compact. but how? I know that the set $\{t \in [a,b] ; K_t \neq \emptyset \}$ is compact in $[a,b]$

The book is Elon Lages lima, Curso de Análise II. A book in portuguese.

Grateful, Henrique.

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    Something doesn't fit well here, imo: you wrote in the very first line "...s.t., if we define $\,K_t=...\,$ . If..." . So ***what*** if we define that?? What is "t" in the definition of $\,K_t\,$? What is the book, page...?2012-06-30
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    I've edited it so it reads better, @DonAntonio. It's possible I changed the meaning of the question (though I don't think this is likely) so feel free to revert my edit if this is the case.2012-06-30
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    As for solving the problem itself, could you use Fubini's theorem? If all of the t-slices are 0, surely the measure of set is zero (this follows from integrating along the t-direciton).2012-06-30
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    The only way I can make sense of all this (perhaps it's only me) is: if $\,K=\prod_iK_i\,\,,\,\text{each} \,\,K_i\in\mathbb R^n\,$ and a nullset, and $\,I\in\mathbb R\,$ is any compact set (for example, aclosed finite interval), then $\,K\times I\in\mathbb R^{n+1}\,$ is a nullset. Is this what the OP meant?2012-06-30
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    @DonAntonio Yes, basically. Although $K_i \subset \mathbb{R}^n$2012-06-30

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