Every subset of a subspace of $\mathbb{R}^n$ of dim $<n$ has measure 0

Question

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?

Answer 1

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.