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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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    @DonAntonio Yes, basically. Although $K_i \subset \mathbb{R}^n$2012-06-30

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As defined in Rudin's book Real and complex analysis, $ \mathcal{L}^{n+1}(K) = \int_a^b \mathcal{L}^n(K_t)\, dt, $ since $K_t$ is the $t$-section of a measurable set. Your "theorem" follows immediately.

Comment: the solution of your problem depends heavily on the definition of product measures. Rudin defines a product measure by means of sections, and you problem is almost a definition. In $\mathbb{R}^{n+1}$ there are different constructions of Lebesgue's measure.

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    A possible hint: fix \epsilon>0. For each $t$, you can construct a block $\{B_k(t)\}_k$ such that $\sum_{k} |B_k(t)|$ is very small and $K_t \subset \bigcup_k B_k(t)$. Each $B_k(t)$ is a bounded interval, and you may slightly modify it and take it open. But then you should probably be in a position to pick up $t_1,\ldots, t_N$ such that $K$ is covered by $\{B_k(t_1),\ldots,B_k(t_N)\}_k$. I guess this may be converted into a complete proof.2012-06-30