Diagonal Scaling and Volume Factors

How Coordinate Scaling Changes Volume

A diagonal map stretches each coordinate axis independently. If the scale factors are $s_1,\dots,s_n$, a box with side lengths $\ell_1,\dots,\ell_n$ has transformed side lengths $|s_1|\ell_1,\dots,|s_n|\ell_n$, provided no factor is zero. This suggests that its volume is multiplied by $\prod_{j=1}^n|s_j|$. The same rule holds for every Lebesgue measurable set; the proof below derives it from box covers rather than assuming a general change-of-variables theorem.

Throughout, $n\ge1$, and $m$ denotes Lebesgue measure on $\mathbb{R}^n$, with outer measure $m^*$. A box means a product of half-open intervals with finite side lengths, and its volume is the product of those lengths. A set is Lebesgue measurable if it belongs to the Lebesgue sigma-algebra $\mathcal{L}$. We use the earlier results that coordinate reflections preserve Lebesgue measure and that Lebesgue measure is complete: every subset of a measurable null set is measurable. We also use the defining box-cover construction of $m^*$.

Definition: For real numbers $s_1,\dots,s_n$, the diagonal map $D:\mathbb{R}^n\to\mathbb{R}^n$ is $D(x_1,\dots,x_n)=(s_1x_1,\dots,s_nx_n)$. Its coordinate volume factor is $c=\prod_{j=1}^n|s_j|$. The factors may be positive, negative, or zero.

Positive Coordinate Factors: The Box-Cover Argument

Proposition: Suppose $s_j>0$ for every $j$, and let $D(x_1,\dots,x_n)=(s_1x_1,\dots,s_nx_n)$. For every set $A\subseteq\mathbb{R}^n$, including sets of infinite outer measure, $$ m^*(D(A))=\left(\prod_{j=1}^n s_j\right)m^*(A). $$

Proof. Write $c=\prod_{j=1}^n s_j>0$. First, cover $A$ by a countable family of half-open boxes $Q_k=\prod_{j=1}^n[a_{kj},b_{kj})$. Since every $s_j$ is positive, $D(Q_k)=\prod_{j=1}^n[s_ja_{kj},s_jb_{kj})$, which is again a half-open box. Its volume is $$ \prod_{j=1}^n s_j(b_{kj}-a_{kj})=c\,m(Q_k). $$ The boxes $D(Q_k)$ cover $D(A)$, so the definition of outer measure gives $m^*(D(A))\le c\sum_k m(Q_k)$. Taking the infimum over all countable box covers of $A$ yields $m^*(D(A))\le c\,m^*(A)$. This inequality also holds if $m^*(A)=\infty$, as an inequality in the extended nonnegative reals.

For the reverse inequality, $D^{-1}$ is a positive diagonal map with factors $1/s_j$, whose product is $1/c$. Apply the inequality just proved to $D(A)$ and $D^{-1}$: $m^*(A)=m^*(D^{-1}(D(A)))\le (1/c)m^*(D(A))$. Multiplication by $c>0$ gives $c\,m^*(A)\le m^*(D(A))$. Combining both inequalities proves the equality, including when either side is infinite. $\square$

For measurable sets, we also need to know that the image remains measurable. A positive diagonal map is a homeomorphism, so it sends Borel sets to Borel sets. It sends null sets to null sets by the proposition. Every Lebesgue measurable set $E$ can be written $E=B\triangle N$, where $B$ is Borel and $N$ is a subset of a Borel null set. Thus $D(E)$ differs from the Borel set $D(B)$ by a subset of a null set. Completeness makes $D(E)$ Lebesgue measurable. Applying the proposition to $E$ now gives the measure formula.

Worked Example: Stretching a Three-Dimensional Box

Let $E=[0,1)\times[-1,2)\times[2,4)\subset\mathbb{R}^3$, with Lebesgue measure $m$, and set $D(x_1,x_2,x_3)=(2x_1,\tfrac13x_2,-4x_3)$. Compute $m(D(E))$ in two ways.

Directly, the transformed coordinate intervals have lengths $2\cdot1=2$, $\tfrac13\cdot3=1$, and $4\cdot2=8$. The last interval is reversed by the negative factor, but its length is still $8$; changing inclusion of an endpoint does not change its Lebesgue measure. Thus the transformed volume is $2\cdot1\cdot8=16$. Check: the direct product $16$ agrees with the factor calculation $(2\cdot\tfrac13\cdot4)\cdot6=16$.

Negative Factors and General Invertible Diagonal Maps

If some $s_j<0$ but none is zero, separate each sign from its magnitude. Write $D=R\circ P$, where $P$ has positive factors $|s_j|$, and $R$ reflects exactly those coordinates for which $s_j<0$. Coordinate reflections preserve Lebesgue measure, so they preserve outer measure as well: the same reflection argument applies to box covers, or equivalently to the defining outer measure. Consequently, the positive-factor proposition gives the scaling rule for $D$.

Theorem: If every $s_j\ne0$, then for every Lebesgue measurable $E\subseteq\mathbb{R}^n$, $D(E)$ is Lebesgue measurable and $$ m(D(E))=\left(\prod_{j=1}^n|s_j|\right)m(E). $$ The equality allows $m(E)=\infty$.

Proof. The factorization $D=R\circ P$ shows that $D$ is a homeomorphism. The positive map $P$ sends Borel sets to Borel sets and null sets to null sets by the proposition. The reflection $R$ preserves Borel sets and Lebesgue measure, and hence preserves null sets. Using the Borel-set-plus-null-set description of Lebesgue measurable sets and completeness, $D(E)$ is measurable. Finally, the positive-factor proposition and reflection invariance give $m(D(E))=m(P(E))=(\prod_j|s_j|)m(E)$. This remains valid for infinite measure because the outer-measure identity was proved with extended values. $\square$

Worked Example: A Countable Union of Stretched Rectangles

In $\mathbb{R}^2$, let $E=\bigcup_{k=1}^{\infty}[k,k+1)\times[0,2^{-k})$, and let $D(x,y)=(3x,2y)$. Compute the measure of $E$ and of its image.

The rectangles in the union are pairwise disjoint: if $k\ne\ell$, their first-coordinate intervals $[k,k+1)$ and $[\ell,\ell+1)$ are disjoint. The $k$-th rectangle has area $1\cdot2^{-k}=2^{-k}$. By countable additivity, $m(E)=\sum_{k=1}^\infty2^{-k}=1$; indeed, the geometric series has first term $1/2$ and ratio $1/2$, so its sum is $(1/2)/(1-1/2)=1$.

The image of the $k$-th rectangle is $[3k,3k+3)\times[0,2^{1-k})$, with area $3\cdot2^{1-k}$. These image rectangles remain disjoint because their first-coordinate intervals are disjoint. Their total area is $\sum_{k=1}^\infty3\cdot2^{1-k}=3\cdot2\sum_{k=1}^\infty2^{-k}=6$. The theorem independently gives $(|3|\cdot|2|)m(E)=6\cdot1=6$. Check: both direct summation and the scaling factor give $m(D(E))=6$.

Zero Factors Collapse Sets onto Null Hyperplanes

A zero factor is not an invertible scaling, so the inverse-map proof does not apply. Instead, the image is contained in a coordinate hyperplane. We first verify that such a hyperplane is null using covers that respect the half-open box convention.

Lemma: For each $j\in\{1,\dots,n\}$, the hyperplane $H_j=\{y\in\mathbb{R}^n:y_j=0\}$ has Lebesgue measure zero.

Proof. Fix a positive integer $N$. The portion of $H_j$ on which $|y_i|\le N$ for every $i\ne j$ is covered, for any $\delta>0$, by the box whose $j$-th interval is $[-\delta,\delta)$ and whose other intervals are $[-N,N+1)$. These intervals do cover the specified coordinates, including the endpoint $N$. The box has volume $2\delta(2N+1)^{n-1}$; when $n=1$, the empty product is $1$, so this is $2\delta$. Letting $\delta\downarrow0$ shows that this bounded portion has outer measure zero. The full hyperplane is the countable union of these portions over positive integers $N$, so countable subadditivity gives $m^*(H_j)=0$. Since $H_j$ is closed, it is Borel and therefore measurable, with $m(H_j)=0$. $\square$

Corollary: For any diagonal map $D$ with at least one zero factor and any Lebesgue measurable $E\subseteq\mathbb{R}^n$, $D(E)$ is measurable and $m(D(E))=0$. Thus $m(D(E))=(\prod_j|s_j|)m(E)=0$, with the right side interpreted as zero even if $m(E)=\infty$.

Proof. If $s_j=0$, every point of $D(E)$ has $j$-th coordinate zero, so $D(E)\subseteq H_j$. The hyperplane is a measurable null set by the lemma. Completeness of Lebesgue measure implies that its subset $D(E)$ is measurable and null. The product of the absolute factors is zero, and zero times any value of $m(E)$, finite or infinite, is interpreted here as zero. $\square$

Worked Example: A Map That Collapses a Box

Let $E=[0,1)\times[0,3)\times[0,2)\subset\mathbb{R}^3$, and let $D(x_1,x_2,x_3)=(2x_1,0,-x_3)$. Find the image and its measure.

For $x_1\in[0,1)$, $2x_1\in[0,2)$. The second coordinate is always $0$. For $x_3\in[0,2)$, $-x_3\in(-2,0]$. The coordinate $x_2\in[0,3)$ has no effect on the image. Hence $D(E)=[0,2)\times\{0\}\times(-2,0]$, which is contained in the hyperplane $y_2=0$, so $m(D(E))=0$.

The original volume is $m(E)=1\cdot3\cdot2=6$, while the factor is $|2|\cdot|0|\cdot|-1|=2\cdot0\cdot1=0$. Thus the formula gives $0\cdot6=0$, agreeing with the hyperplane argument. Check: the image has zero thickness in its second coordinate, and the computed measure is zero.

A Common Mistake: Keeping the Sign

Volume cannot be negative. For example, the map $(x,y)\mapsto(-2y,3x)$ has coordinate scale magnitudes $2$ and $3$, so its volume factor is $2\cdot3=6$. Using the signed product $(-2)\cdot3=-6$ would incorrectly predict a negative measure for every set of positive finite measure. The correction is to take the absolute value of each coordinate factor. Negative factors reverse orientation along an axis but do not produce negative volume.

The rule is specific to diagonal maps, whose coordinate actions scale independently. It supplies the exact volume factor for anisotropic stretching—different amounts of scaling in different directions—and applies to arbitrary Lebesgue measurable sets, not only to boxes. The zero-factor case is qualitatively different from invertible stretching: it collapses the image into a lower-dimensional hyperplane, which has zero $n$-dimensional Lebesgue measure.

Check Your Understanding

For each question, state whether the map is invertible and use absolute coordinate factors when computing volume.

For $D(x,y)=(\tfrac12x,5y)$, by what factor does the measure of a measurable set change?
A rectangle has side lengths $4$ and $3$. Under $(x,y)\mapsto(-2x,\tfrac13y)$, what is its image area?
Why does a diagonal map with one zero factor send every set into a null set?
Explain why the signed product of the coordinate factors cannot be a volume factor when one factor is negative.
For a positive diagonal map, identify the box-cover step that gives the upper bound for outer measure, and explain how the inverse map gives the reverse bound.