1

I am reading an old lecture notes in measure theory and I encountered the folllowing:

Assume $f: \mathbb{R}^n \to \mathbb{R}^n$ is a linear transformation. If $f$ is not invertible, then $f$ maps $\mathbb{R}^n$ onto a linear subspace of lower dimension, and hence, onto a set of Lebesgue measure $0$.

However, I could not find something before that statement about showing that a set has measure $0$ except the fact that by monotonicity, a subset of a null set is also a null set. I believe that the Lebesgue measure acts like a volume. So intuitively, an object with dimension less than $n$ in $\mathbb{R}^n$ has $0$ volume. I'm not sure about this interpretation, but how do I prove the quoted statement?

  • Hint: the Lebesgue measure associated to linear mappings has something to do with the determinant. – Daan Oct 22 '24 at 08:54

2 Answers2

0

This is overkill, but Sard's Theorem is a great tool for this sort of thing. Copying from Wikipedia, the theorem asserts this:

let $f: \mathbb {R} ^{n}\rightarrow \mathbb {R} ^{m}$ be $C^k$ (that is, $k$ times continuously differentiable), where $k \ge \max > \{n-m+1, 1\}$. Let $X \subset \mathbb {R} ^{n}$ denote the critical set of $f$ ,which is the set of points $ x\in \mathbb {R} ^{n}$ at which the Jacobian matrix of $f$ has rank $< m$. Then the image $f(X)$ has Lebesgue measure 0 in $\mathbb {R} ^{m}$.

In your case, $f$ is linear, hence very smooth (it's $C^\infty$). And because it's not invertible, it's rank is everywhere less than $n = m$. So the set $X$ is all of $\mathbb R^n$. And then the theorem tells you that the image of $f$ has measure $0$ in the codomain.

The proof given here seems fairly straightforward and readable, and the 1D case gives a bunch of the main ideas.

John Hughes
  • 100,827
  • 4
  • 86
  • 159
0

The claim follows from this theorem:

Theorem. Let $T:\mathbb{R}^n\to\mathbb{R}^m$ be a linear transformation and take measurable $E\subset\mathbb{R}^n$, then \begin{equation} \mu(TE)=\mathrm{det}|T|\:\mu(E). \end{equation}

Here, we see that if $m<n$, then $\mathrm{det}|T|=0$, by which the claim follows.

The proof of the complete theorem decomposes $T$ using a polar decomposition. That is, take $T=QU$ for a positive definite symmetric $U=(T^\top T)^{1/2}$ and an orthogonal $Q$. We can diagonalise $Q=O\Lambda^\top O$ with $\Lambda:(x_1,...,x_n)\mapsto(\lambda_1x_1,...,\lambda_nx_n)$ for positive $\lambda_i>0$. Clearly $\Lambda$ does not change the orientation of $E$ and hence $\mu(\Lambda E)=\mathrm{det}|\Lambda|\mu(E)$. Additionally, the Lebesgue measure is translation invariant, hence $\mu(TE)=\mu(\Lambda E)$ and $\mathrm{det}|\Lambda|=\mathrm{det}|T|$.

The result is taken from these lecture notes, specifically theorem 2.33, if you are interested. Of course, it goes without saying that the credits of the proof go to the author of the lecture notes.

Daan
  • 646
  • Thanks for the reference! Actually, the next theorem in my lecture notes after the quote in my post is the theorem you linked. The proof is significantly more readable I think. But how the proof of this theorem proves that the image of noninvertible map has measure $0$? –  Oct 22 '24 at 10:17
  • You are almost right. The theorem states that the measure $\nu$ of a linear map that is not invertable (applied to $E$) is zero if that measure is equivalent w.r.t. the Lebesgue measure ($\nu(E)=0$ if and only if $\mu(E)=0$). The two become equivalent if $E$ has positive measure, since $\nu(TE)$ is then zero if and only if $\mu(TE)=0$ which only happens if $T$ has determinant zero. – Daan Oct 22 '24 at 10:25
  • Keep in mind that, if $\mu^n$ denotes the Lebesgue measure in $\mathbb{R}^n$, then $TE$ might still have positive measure for $\mu^m$ if $m<n$ even if $\mu^n(TE)=0$, an easy example is the unit square $A={(x,y):0\leq x\leq 1,0\leq y\leq1}$ and the linear map $T:(x,y)\mapsto(x,0)$, then $\mu^2(TA)=0$ whereas $\mu^1(A)=1$. – Daan Oct 22 '24 at 10:28
  • What is the reasoning behind the first sentence: If $T$ is singular, then its range is a lower-dimensional subspace of $\mathbb{R}^n$, which has Lebesgue measure zero? –  Oct 22 '24 at 14:03
  • The theorem holds for any linear map $T$, and in particular also for singular matrices. Recall that a linear map $T$ is uniquely associated to a matrix $A_T$ such that $T(v)=A_Tv$ for any $v\in\mathbb{R}^n$. Now the dimension of the image of $T$ added to the dimension of the kernel equals $n$, hence if $T:\mathbb{R}^n\to\mathbb{R}^m$ with $m<n$, then $\mathrm{dim ker}T=n-m>0$ which means that there are $v\in\mathbb{R}^n$ with $v\neq 0$ and $T(v)=0$ (which is the definition of singular). But this implies that $\mathrm{det}T=0$. Hence $T$ being singular implies $\mu(TE)=0$. – Daan Oct 22 '24 at 14:30
  • I'm sorry if I could not fully understand your point. I think your reasoning is circular. I am asking about a part in the proof so I should not use the theorem because that is what I am trying to show –  Oct 22 '24 at 14:56
  • In that case, you could approach it by approximating $TE$ using $\mathbb{R}^n$ covering balls. Because the space spanned by $TE$ is a subspace, you can make the union of these balls arbitrarily small, resulting in the claim. Take a look at this post for example. – Daan Oct 22 '24 at 15:08