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Show that if $m <n$ then $\mathbb{R}^m$ seen as subset of $\mathbb{R}^n$ has zero measure.

A set $X \subset \mathbb{R}^m$ have zero measure if for all $\epsilon > 0$ it is possible to obtain a sequence of open cubes $m$-dimensional $C_1 , C_2 , ..., C_i , ...$ such that $X \subset \bigcup_{i=1}^{\infty} C_i$ and $\sum_{i=1}^{\infty} vol(C_i) < \epsilon$

I have tried to use the following theorem:

If $m<n$ and $f:U \rightarrow \mathbb{R}^n$ of class $C^1$ in a open set $U \subset \mathbb{R}^m$ then $f(U)$ has null measure in $\mathbb{R}^n$.

I use the function $f: \mathbb{R}^m \rightarrow \mathbb{R}^n$, $f(x_1,...,x_m)=(x_1,...,x_m,0,...0)$.

My doubt is if this function satisfies the hypotheses of the theorem and the question.

Mrcrg
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  • What've you tried so far? – scoopfaze Jul 01 '19 at 17:59
  • What is your definition of zero measure? – Thomas Browning Jul 01 '19 at 17:59
  • @scoopfaze I added my definition in the question – Mrcrg Jul 01 '19 at 18:08
  • It's good practice on this forum to tell us what you've tried so far; we don't want to be answering homework questions for people here. What did you consider already? – Vedvart1 Jul 01 '19 at 18:11
  • @ThomasBrowning I used a theorem, which says that given a function of class $C^1$ defined in an open set $U \subset \mathbb{R}^m$, $f:U \rightarrow \mathbb{R}^n$ then $f(U)$ has null measure in $\mathbb{R}^n$. but I'm not sure it's right – Mrcrg Jul 01 '19 at 18:12
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    If you have such theorem, then it's easy as heck. Just choose $U = R^m$ and any non-singular linear map as $f$. – Jakobian Jul 01 '19 at 18:19
  • @Jakobian Unfortunately, as he's stated the theorem, that isn't possible - it seems you need $U$ to be a proper subset of $\mathbb{R}^m$, so you cannot take equality. – Vedvart1 Jul 01 '19 at 18:55
  • @Vedvart1 why not? There's no proper/improper subset notation involved. – Jakobian Jul 01 '19 at 19:08
  • @Jakobian Correct me if I'm wrong, but doesn't $U \subset \mathbb{R}^m$ mean U is a proper subset of $\mathbb{R}^m$, with only $U \subseteq \mathbb{R}^m$ allowing equality? – Vedvart1 Jul 01 '19 at 19:31
  • @Vedvart1 It depends on the book/author. Usually it's used to mean just subset, proper or not. – Jakobian Jul 01 '19 at 19:41
  • @Jakobian Ah okay, never seen it used ambiguously like that, thanks for letting me know. – Vedvart1 Jul 01 '19 at 20:29

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