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 $\epsilon$.
Best Answer
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 $t \in \mathbb{R}$.