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I have read in my teacher’s book the following statement.

  • Every subset A of a subspace C of $\mathbb{R}^n$ with $dim(C)<n$ has measure 0

And I’m having issues proving it and mainly understanding why. Is it because measure space $(\mathbb{R}^n , M( \mathbb{R}^n), \lambda_n)$ where $M( \mathbb{R}^n)$ is the $\sigma$-algebra of the measurable sets of $\mathbb{R}^n$, and $\lambda_n$ is Lebesgue Measure is a complete measure space?

MNM
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  • I assume "measure $0$" means "Lebesgue measure $0$". But what kind of dimension do you use? If you look at the definition, the question should answer itself. Anyway, you should include it in the question. – Michał Miśkiewicz Nov 16 '18 at 19:40
  • Actually there are $0$-dimensional subsets of $\mathbb R^1$ with positive Lebesgue measure. – bof Jul 08 '21 at 03:07

1 Answers1

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If you recall from calculus, if we integrated over a region, say $\displaystyle\int_{[a,b]}dx$, we would get the area of a rectangle with base length (b-a) if we were in $\mathbb{R}^2$ or $\displaystyle\int_{[a,b] \times [c,d]}dx$, the volume of a parallelepiped whose base has length (b-a) and width (d-c).

In $\mathbb{R}^3$, the measure of A is the volume of the set A. If dim C $<$n, then one of the dimensions of the higher dimensional rectangles is 0.

So if the set A is a square, it has volume 0. Thus the square would have measure 0. Now generalize this idea to higher dimensions.

Joel Pereira
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