Singular Maps Collapse Dimension
For an invertible linear map on \(\mathbb{R}^n\), determinant scaling describes how Lebesgue measure changes. A singular linear map is different: its range lies in a subspace of dimension strictly less than \(n\). Thus, even if the original set has positive \(n\)-dimensional measure, its image has zero \(n\)-dimensional measure. The image can still contain a line segment or a planar region when viewed inside its lower-dimensional subspace; the zero measure here is specifically the ambient Lebesgue measure on \(\mathbb{R}^n\).
Throughout, \(n\ge1\), and \(T:\mathbb{R}^n\to\mathbb{R}^n\) is a linear map. A set is bounded if it is contained in some ball of finite radius. We use Lebesgue measure \(m\) on \((\mathbb{R}^n,\mathcal{L})\), which is complete: every subset of a Lebesgue measurable null set is Lebesgue measurable. A singular map means a linear map whose matrix has determinant zero, equivalently, by the rank-nullity theorem of finite-dimensional linear algebra, one whose range has dimension less than \(n\).
The key measure fact is that a lower-dimensional subspace has zero \(n\)-dimensional Lebesgue measure. The proof first handles bounded pieces, then uses a countable union for the entire subspace. An orthogonal change of coordinates sends any \(r\)-dimensional subspace to the coordinate subspace \(\mathbb{R}^r\times\{0\}^{n-r}\). Orthogonal maps preserve Lebesgue measure by the determinant-scaling result for invertible linear maps, since the absolute value of their determinant is \(1\).
Proof. Choose an orthogonal change of coordinates sending \(H\) to \(\mathbb{R}^r\times\{0\}^{n-r}\). It suffices to show that this coordinate subspace has measure zero, because the change of coordinates preserves Lebesgue measure. For each positive integer \(k\), let \(C_k=[-k,k]^r\times\{0\}^{n-r}\). For every \(\varepsilon>0\), \(C_k\) is contained in the rectangle \([-k,k]^r\times[-\varepsilon,\varepsilon]^{n-r}\), whose measure is \((2k)^r(2\varepsilon)^{n-r}\). Since \(n-r\ge1\), this bound tends to zero as \(\varepsilon\) tends to zero. Therefore \(m^*(C_k)=0\), where \(m^*\) is Lebesgue outer measure. The coordinate subspace is \(\bigcup_{k=1}^{\infty}C_k\), so countable subadditivity gives it outer measure zero. A set of outer measure zero is Lebesgue measurable and has measure zero. The same is therefore true of \(H\). When \(r=0\), the factor \((2k)^r\) is \(1\), and the bound \((2\varepsilon)^n\) still tends to zero, so the argument includes the zero subspace. \(\square\)
The Image Theorem
Proof. Write \(H=T(\mathbb{R}^n)\). Singularity gives \(\dim H<n\), so the lemma gives \(m(H)=0\). Since \(T(A)\subseteq H\), completeness of Lebesgue measure implies that \(T(A)\) is measurable and has measure zero. This argument uses neither boundedness nor measurability of \(A\); it applies to any set \(A\subseteq\mathbb{R}^n\). \(\square\)
The theorem is an ambient-measure statement, not a claim that the image has no lower-dimensional size. For example, a line segment has zero area in \(\mathbb{R}^2\), although it has positive length. Nor does the theorem say that the image fills its range: the shape and extent of \(T(A)\) still depend on both \(A\) and \(T\).
Worked Example: A Square Collapsed onto a Line
Let \(A=[0,1]^2\subset\mathbb{R}^2\), and define \(T(x,y)=(x+y,2x+2y)\). Find the exact image and its two-dimensional Lebesgue measure.
Set \(t=x+y\). For \((x,y)\in A\), \(0\le t\le2\), and \(T(x,y)=(t,2t)\). Conversely, every \(t\in[0,2]\) occurs: if \(0\le t\le1\), take \((x,y)=(t,0)\); if \(1<t\le2\), take \((x,y)=(1,t-1)\). Each selected point belongs to \(A\), since both its coordinates lie in \([0,1]\) in the relevant case. Thus $$ T(A)=\{(t,2t):0\le t\le2\}. $$
The matrix of \(T\) is \(\begin{pmatrix}1&1\\2&2\end{pmatrix}\), whose determinant is \(1\cdot2-1\cdot2=0\); its range is the line \(\{(t,2t):t\in\mathbb{R}\}\). The theorem therefore gives \(m(T(A))=0\). As a direct check, divide \([0,2]\) into \(N\) intervals of length \(2/N\). Above each interval, the corresponding part of the segment lies in a rectangle of width \(2/N\) and height \(4/N\), because the second coordinate is twice the first. The \(N\) rectangles have total area at most \(N(2/N)(4/N)=8/N\), which tends to zero. This independently verifies that the segment has zero area.
Worked Example: Projecting a Disk onto a Diameter
Let \(A=\{(x,y)\in\mathbb{R}^2:x^2+y^2\le1\}\), the closed unit disk, and define \(T(x,y)=(x,0)\). Determine the image and its two-dimensional measure.
If \((x,y)\in A\), then \(x^2\le1\), so \(-1\le x\le1\) and \(T(x,y)=(x,0)\). Conversely, for every \(u\in[-1,1]\), the point \((u,0)\) lies in \(A\), since \(u^2+0^2\le1\). Hence $$ T(A)=\{(u,0):-1\le u\le1\}. $$
The image is exactly the diameter, not the whole square \([-1,1]\times[-1,1]\). The matrix \(\begin{pmatrix}1&0\\0&0\end{pmatrix}\) has determinant \(1\cdot0-0\cdot0=0\), so the image has two-dimensional measure zero. Directly, for every \(\varepsilon>0\), the segment is contained in \([-1,1]\times[-\varepsilon,\varepsilon]\), of area \(2\cdot2\varepsilon=4\varepsilon\). Letting \(\varepsilon\) tend to zero confirms the result.
Worked Example: A Box Projected onto a Plane
In \(\mathbb{R}^3\), let \(A=[-1,2]\times[0,3]\times[2,5]\), and let \(T(x,y,z)=(x,y,0)\). Find the image and its three-dimensional measure.
For each \((x,y,z)\in A\), the output has first coordinate in \([-1,2]\), second coordinate in \([0,3]\), and third coordinate zero. Every pair \((x,y)\in[-1,2]\times[0,3]\) occurs, since \(z=2\) is an allowed choice. Therefore $$ T(A)=[-1,2]\times[0,3]\times\{0\}. $$
The matrix is \(\begin{pmatrix}1&0&0\\0&1&0\\0&0&0\end{pmatrix}\), with determinant \(1\cdot1\cdot0=0\), and its range is the plane \(z=0\). The image has zero three-dimensional measure. For a direct check, it is contained, for every \(\varepsilon>0\), in \([-1,2]\times[0,3]\times[-\varepsilon,\varepsilon]\). This covering box has volume \(3\cdot3\cdot2\varepsilon=18\varepsilon\), tending to zero. Thus the image has zero ambient volume even though its two-dimensional footprint is a rectangle of side lengths \(3\) and \(3\).
A Common Error: Measuring the Bounding Box
A singular image can lie inside a large box without filling that box. In the square example, the image lies in \([0,2]\times[0,4]\), whose area is \((2-0)(4-0)=8\). It would be wrong to report \(8\) as the image's measure: that computes the area of the bounding box, not the diagonal segment. The segment is covered by rectangles with total area at most \(8/N\), as calculated above, so its area is zero.
Another tempting shortcut is to apply the determinant-scaling formula for invertible maps and substitute a zero determinant. That theorem assumes the map is invertible; its hypotheses fail for a singular map. The zero-measure conclusion is correct, but it requires a different justification: show that the image lies in a lower-dimensional subspace and use the nullity of that subspace. This reasoning also settles measurability: in the complete Lebesgue measure space, every subset of the null range is measurable, even if the image is not described as a Borel set.
| Question | Correct interpretation |
|---|---|
| Which measure is zero? | Ambient \(n\)-dimensional Lebesgue measure. |
| Must the image fill its range? | No; its extent depends on the original set. |
| Is boundedness needed for nullity? | No; every image is a subset of the null range. |
| Why is the image measurable? | Completeness makes every subset of a measurable null set measurable. |
Check Your Understanding
For each question, distinguish the ambient measure from the size of the image inside its range.
- For \(T(x,y)=(0,y)\) and \(A=[-2,2]\times[1,3]\), describe \(T(A)\) and determine its two-dimensional Lebesgue measure.
- Why does the theorem still imply that \(T(A)\) is measurable if \(A\) is not assumed measurable?
- For a singular map on \(\mathbb{R}^4\) whose range has dimension \(2\), what is the four-dimensional measure of the image of any set?
- For \(T(x,y,z)=(x,0,0)\), identify the range and explain why every image has zero three-dimensional Lebesgue measure.
- Explain why the volume of a box containing a lower-dimensional image does not, in general, equal the measure of the image.