$K_t=\{x\in\mathbb{R}^m;(x,t)\in K\}$. If $\forall t\in[a,b],$ $\operatorname{med}K_t=0$, then $\operatorname{med}K=0$ in $\mathbb{R}^{m+1}.$ 
Let $K\subset \mathbb{R}^m\times [a,b]$ be compact. For each $t\in
 [a,b]$, let $K_t = \{x\in \mathbb{R}^m; (x,t) \in K\}$. If, for all
  $t\in [a,b]$ we have $\operatorname{med} K_t = 0$, then $\operatorname{med} K = 0$ in
  $\mathbb{R}^{m+1}$

ps: this proof Prove that a product of nullsets in $\mathbb{R}^n$ by a compact set in $\mathbb{R}$ is a nullset in $\mathbb{R}^{n+1}$ is too complicated for me
If $\operatorname{med} K_t = 0$, then we can cover $K_t$ by open blocks $B_i$ such that $\sum \operatorname{vol} B_i <\varepsilon$. Now I need to cover just $K$ by open blocks and prove their sum is $< \varepsilon$. Since we covered $K_t$, I'm thinking about using these same blocks for $K$. If we cover a finite surface in $\mathbb{R}^2$ by blocks in $\mathbb{R}^3$, then these same blocks cover a line of this surface, right? Shouldn't it be simple to prove?
I did not involve compactness yet. If the cartesian product is compact, then for an open cover of the cartesian, I can find a finite open subcover. This idea is certainly related to the volume, but I don't know how, exactly.
 A: HINT: 


*

*Show that if $K_t\subset U$, where $U$ open subset of $\mathbb{R}^n$, then there exists $\delta > 0$ so that $K_{s}\subset U$ for all $s \in (t-\delta, t+ \delta)$. 


Indeed, assume the contrary. There exists a sequence $(k_n,t_n) \in K$ so that $t_n \to t$ and $k_n \in \mathbb{R}^n \backslash U$. Pass to a convergent subsequence, so may assume $(k_n, t_n) \to (k, t) \in K_t$. But then $k \in \mathbb{R}^n \backslash U$ ( a closed subset), contradiction. 


*Let $\epsilon > 0$. For every $t$ consider $U_t$ a finite union of open boxes so that $K_t \subset U_t$ and $m(U_t) < \epsilon$. Let now $\delta_t>0$ ( from 1.) so that $K_s \subset U_t$ for all $s \in V_t = (t-\delta_t, t+ \delta_t)$.

*Let $I$ a closed interval containing the projection of $K_t$ on the last component $\mathbb{R}$. Take a sufficiently fine division of $I$ into closed intervals $I = I_1 \cup \ldots \cup I_m$ so that every $I_l$ is contained in some $V_{t_l} = ( t_l - \delta_{t_l}, t_l + \delta_{t_l})$. We conclude that $K$ is contained in 
$$\cup_{l=1}^m I_l \times U_{t_l}$$ a union of boxes of size at most $m(I) \times \epsilon$.  
