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

Solution 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.

Solution 2:

I am not sure if the below approach is correct. Rudin Real and Complex Analysis book states the result asked in the question within the proof of Theorem 2.20 regarding the Lebesgue measure on ${\boldsymbol{R^{k}}}$. Hence trying to use only results upto this portion of the book.

Part (a) of Theorem 2.20 states that $m\left(W\right)=\text{vol}\left(W\right)$ for every $k$-cell $W$. Using this, we can wrap or cover every compact (closed and bounded) set, $E$, in the lower subspace, $Y$, with dimension, $m<k$, with a suitable $k$-cell from the higher dimensional space. The additional parameters for this $k$-cell in the higher dimension can be $-\frac{\epsilon}{2},\frac{\epsilon}{2}$ with $\epsilon>0$. This gives $m\left(E\right)=\text{vol}\left(E\right)\leq\epsilon^{k-m}\text{volm}$. Here, volm is the volume of the wrapper of $E$, in the $m$-dimensional sub space, $Y$.

Since $\epsilon$ was arbitraty, $m\left(E\right)=\text{vol}\left(E\right)=0$. The entire space (and hence the subspace) is $\sigma-$compact so it can be written as the countable union of compact sets (which are closed and bounded and hence covered like the set $E$ above) all of which have measure zero, so the measure of the entire subspace is zero.

Related Questions: Sub-dimensional linear subspaces of $\mathbb{R}^{n}$ have measure zero.

Lebesgue measure of a subspace of lower dimension is 0

Lebesgue measure of a subspace of lower dimension

Any linear subspace has measure zero

proof that the lebesgue measure of a subspace of lower dimension is 0.