Measure zero of Cross section of a compact implies that compact has measure zero Let $K \subset \mathbb{R}^n\times [a,b]$ a compact subset. For each $t \in [a,b]$, let $K_t= \{x \in \mathbb{R}^n : (x,t) \in K\}$. Suppose that, for all $t \in [a,b]$, $K_t$ has measure zero in $\mathbb{R}^n$. Thus, $K$ has measure zero in $\mathbb{R}^{n+1}$.
The definition of measure zero in $\mathbb{R}^n$ here is:
$A$ has measure zero if given $\epsilon > 0$, there is $\{Q_i\}_{i \in \mathbb{N}}$ rectangles in $\mathbb{R}^n$ such that
$$A \subset \bigcup_{i \in \mathbb{N}} Q_i,$$
and,
$$\sum_{i \in \mathbb{N}}v(Q_i) < \epsilon,$$
where $v(Q) = (b_1 - a_1) \cdots (b_n - a_n)$, if $Q = [a_1,b_1]\times \cdots \times [a_n, b_n]$. The previous summation is called total volume of the cover.
There are some interesting points here:

*

*If $K_t$ has measure zero in $\mathbb{R}^n$, then $K_t \times \{t\}$ has measure zero in $\mathbb{R}^{n+1}$.

*For every $t \in [a,b]$, $K_t$ is compact. Then, $K_t \times \{t\}$ is compact.

*$K = \bigcup_{t \in [a,b]}K_t \times \{t\}$.

Then, I tried to take an finite open cover (of rectangles) of $K_t \times \{t\}$ which total volume is less then $\epsilon '$. Then, use it to construct an open cover of $K$, and using Lindelöf theorem to get a countable cover of $K$. But I'm stuck because I can't prove that total volume is less then $\epsilon$.
 A: By Tonelli's theorem,
$$m(K) = \int_{\mathbb{R}}\int_{\mathbb{R}^n}1_K(x, t)\,dx\,dt = \int_{\mathbb{R}}m(K_t)\,dt.$$
Hence $m(K) = 0$ if and only if $m(K_t) = 0$ for almost every $t \in \mathbb{R}$.
A: The use of the Lebesgue measure would simplify things a lot. But it's possible to use only the definition of the total volume of countable unions of rectangles (not necessarily open). So, let $v_p$ denote the total volume of a countable union of rectangles $V \subseteq \mathbb{R}^p$.
The solution below is based on this article.
It relies on the following lemma, which says that a cover of a $K_t$ is actually a cover of the sets $K_u$ that are "close enough".
Lemma. Let $K \subseteq \mathbb{R}^n \times [a,b]$ be compact, let $t \in \mathbb{R}$ and let $T$ be an open set in $\mathbb{R}^n$ with $K_t \subseteq T$. Then there is an interval $(c,d) \ni t$ such that $K_u \subseteq T$ for every $u \in (c,d)$.
The proof goes as follows:
Let $(a_k)$ be a strictly increasing sequence converging to $t$ and $(b_k)$ a strictly decreasing sequence converging to $t$. Define the sequence of sets $(F_k)$ in the following manner:
$$F_k = K \cap ((\mathbb{R}^n \setminus T) \times [a_k,b_k]).$$
It is clear that this sequence is decreasing and all the $F_k$ are compact sets. It follows that
$$\bigcap_{k = 1}^{\infty} F_k = K \cap \left( \bigcap_{k = 1}^{\infty} (\mathbb{R}^n \setminus T) \times [a_k, b_k] \right) = K \cap ((\mathbb{R}^n \setminus T) \times \{ t \}) = \varnothing,$$
because $K_t \cap (\mathbb{R}^n \setminus T) = \varnothing$. From compactness, it follows that $F_{k_0} = \varnothing$ for some $k_0 \in \mathbb{N}$. Let $c = a_{k_0}$ and $d = b_{k_0}$. Then,
$$K \cap ((\mathbb{R}^n \setminus T) \times [c,d]) = \varnothing \Rightarrow K_u \cap (\mathbb{R}^n \setminus T) = \varnothing \ \ \ \forall u \in [c,d],$$
and then $K_u \subseteq T \ \ \ \forall u \in (c,d)$, as desired.
This result allows to prove the initial proposition.
Consider an arbitrary $\delta > 0$. As the projection $P \, \colon \mathbb{R}^n \times \mathbb{R} \to \mathbb{R}$, $(x,t) \mapsto t$, is a continuous function, $K$ is compact implies that so is the set
$$L = P(K) = \{ t \in \mathbb{R} : (x,t) \in K \ \text{ for some } \ x \in \mathbb{R}^n \}.$$
Therefore, there is a bounded open interval $Q$ such that $L \subseteq Q$. Choose $\varepsilon > 0$ such that $\varepsilon < \delta/v_1(Q)$.
Call a set $U \subseteq \mathbb{R}$ $\varepsilon$-good if there is a countable union of open rectangles $V \subseteq \mathbb{R}^n$ with total volume less than $\varepsilon$ and $K_u \subseteq V$ for every $u \in U$. Let $\mathcal{F}$ be the set of open $\varepsilon$-good sets contained in $Q$. It follows that
$$L \subseteq \bigcup_{U \in \mathcal{F}} U \subseteq Q.$$
Indeed, if $t \in L$, then $(y,t) \in K$ for some $y \in \mathbb{R}^n$.  As $K_t$ has measure zero, there is an open cover of rectangles $\bigcup_{j = 1}^{\infty} I_{j,t}$ of $K_t$ such that $\sum_{j = 1}^{\infty} v_n(I_{j,t}) < \varepsilon$. Call this cover $V$ and see that, according to the previous lemma, there exists an interval $(c,d)$ that contains $t$ and satisfies $K_u \subseteq V$ for every $u \in (c,d)$. As $Q$ is an open interval, $Q \cap (c,d) \ni t$ also satisfies this property and is, therefore, $\varepsilon$-good. This implies that $t \in \bigcup_{U \in \mathcal{F}} U$.
As each open set in $\mathbb{R}$ is a countable union of open intervals, it follows that $\bigcup_{U \in \mathcal{F}} U$ can be written as a countable union of open intervals $\bigcup_{i = 1}^{\infty} G_i$ that covers $L$. Using the fact that $L$ is compact, there is a finite subcover $\bigcup_{i = 1}^{m} G_i$. As each $G_i$ is a subset of an $\varepsilon$-good set, the $G_i$ themselves are $\varepsilon$-good.
To make this union disjoint, let
$$S_j = G_j \setminus \bigcup_{i < j} G_i$$
for every $1 \leq j \leq m$. It is evident that the $S_j$ are finite unions of not necessarily open intervals (here, I consider a point an interval of length zero, to avoid any posssible complications - this is one of the reasons this proof is simpler when using the Lebesgue measure). As $S_j \subseteq G_j$ for every $j$, i.e., every $S_j$ is a subset of an $\varepsilon$-good set, they are $\varepsilon$-good sets themselves.  Therefore, for every $j$, there is a countable union of open rectangles $V_j = \bigcup_{l = 1}^{\infty} T_{j,l}$ with total volume less than $\varepsilon$ and satisfying $K_s \subseteq V_j$ for every $s \in S_j$.
It follows that
$$K \subseteq \bigcup_{j = 1}^{m} V_j \times S_j = \bigcup_{j = 1}^{m} \bigcup_{l = 1}^{\infty} T_{j,l} \times S_j.$$
Indeed, if $(x,t) \in K$, then $t \in S_k$ for some $1 \leq k \leq m$, as the union of the $S_j$ covers $L$. Then, $x \in K_t \subseteq V_k$, by construction of the $V_j$, and $(x,t) \in \bigcup_{j = 1}^{m} V_j \times S_j$. The expression on the right indicates that this cover is a countable cover of rectangles of total volume equal to
$$\sum_{j = 1}^{m} \sum_{l = 1}^{\infty} v_n(T_{j,l}) \cdot v_1(S_j) = \sum_{j = 1}^{m} v_1(S_j) \sum_{l = 1}^{\infty} v_n(T_{j,l}) < \varepsilon \sum_{j = 1}^{m} v_1(S_j) = \varepsilon \cdot v_1 \left( \bigcup_{j = 1}^{m} S_j \right).$$
However, $\bigcup_{j = 1}^{m} S_j \subseteq Q$, hence
$$\varepsilon \cdot v_1 \left( \bigcup_{j = 1}^{m} S_j \right) \leq \varepsilon \cdot v_1(Q) < \delta.$$
As the choice of $\delta$ is arbitrary, it follows by definition that $K$ has measure zero.
