# Prove that a product of nullsets in $\mathbb{R}^n$ by a compact set in $\mathbb{R}$ is a nullset in $\mathbb{R}^{n+1}$

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.

-
Something doesn't fit well here, imo: you wrote in the very first line "...s.t., if we define $\,K_t=...\,$ . If..." . So what if we define that?? What is "t" in the definition of $\,K_t\,$? What is the book, page...? – DonAntonio Jun 30 '12 at 4:44
I've edited it so it reads better, @DonAntonio. It's possible I changed the meaning of the question (though I don't think this is likely) so feel free to revert my edit if this is the case. – Potato Jun 30 '12 at 4:46
As for solving the problem itself, could you use Fubini's theorem? If all of the t-slices are 0, surely the measure of set is zero (this follows from integrating along the t-direciton). – Potato Jun 30 '12 at 4:47
The only way I can make sense of all this (perhaps it's only me) is: if $\,K=\prod_iK_i\,\,,\,\text{each} \,\,K_i\in\mathbb R^n\,$ and a nullset, and $\,I\in\mathbb R\,$ is any compact set (for example, aclosed finite interval), then $\,K\times I\in\mathbb R^{n+1}\,$ is a nullset. Is this what the OP meant? – DonAntonio Jun 30 '12 at 4:51
@DonAntonio Yes, basically. Although $K_i \subset \mathbb{R}^n$ – Henrique Tyrrell Jun 30 '12 at 13:04

## 1 Answer

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.

-
I'm sorry, I can't understand what is it you're integrating. As for measures, it's a book in analysis so that no heavy measure theory should be involved. The only thing defined about lebesgue measure on the text is the notion of a set of measure zero. I don't think you can integrate measures, or anything like that, yet. Also: Fubini's theorem hasn't been proved yet so I guess it shouldn't be used. – Henrique Tyrrell Jun 30 '12 at 13:42
As in the book: A set $X$ is said to have measure zero if $\forall \epsilon$ there is a covering of X by n-dimensional blocks $\{B_k\}_{k \in \mathbb{N}}$ such that $\sum_{k \in \mathbb{N}} |B_k| \lt \epsilon$ where $|B|$ denotes the volume of the block B and a block in $\mathbb{R^n}$ is an n-fold product of compact intervals in $\mathbb{R}$ – Henrique Tyrrell Jun 30 '12 at 14:06
I'm sorry, but I can't help you immediately. As I said, I learned measure theory from an abstract viewpoint (Lebesgue measure is, for me, a consequence of the Riesz representation theory). You should try to work out the case in which $K \subset \mathbb{R}$, since you can easily draw blocks. – Siminore Jun 30 '12 at 17:23
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. – Siminore Jun 30 '12 at 17:28