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.