Pushforwards of Lebesgue Measure Under Elementary Maps

Preimages Determine the Pushforward

A measurable map need not preserve Lebesgue measure: it may stretch intervals, fold different parts of its domain onto the same set, or send an entire domain to one point. The pushforward records the measure of a target set by measuring its preimage. For the elementary maps considered here, the preimage calculation gives an explicit density and makes the effect of stretching and folding visible.

Let $I=[u,v]\subset\mathbb{R}$, where $u\le v$, and let $m_I$ be Lebesgue measure restricted to $I$: $m_I(E)=m(E)$ for Lebesgue-measurable $E\subseteq I$. This is a finite measure with total mass $v-u$, possibly zero if $u=v$. If $T:I\to\mathbb{R}$ is Borel measurable, its pushforward is the measure on $\mathcal{B}_{\mathbb{R}}$ defined by $\nu(B)= (T_{\#}m_I)(B)=m(\{x\in I:T(x)\in B\})$. Thus one computes $\nu(B)$ by finding the preimage, not by measuring the image set $T(B)$.

In each formula below, a density is with respect to Lebesgue measure on the target. Integrating that density over a Borel set $B$ gives the pushforward mass of $B$. Endpoints of intervals may be included or excluded in a density formula without changing its measure. A density that is unbounded near zero can still be integrable there and need not indicate an atom at zero.

Affine Maps: Constant Stretching

Proposition: Let $I=[u,v]$, let $m_I$ be restricted Lebesgue measure, and set $T(x)=cx+d$. If $c\ne0$, then for every $B\in\mathcal{B}_{\mathbb{R}}$, $$ (T_{\#}m_I)(B)=\int_B \frac{1}{|c|}\chi_J(y)\,dm(y), \qquad J=[\min(cu+d,cv+d),\max(cu+d,cv+d)]. $$ If $c=0$, then $T_{\#}m_I=(v-u)\delta_d$, where $\delta_d$ is unit mass at $d$.

Proof. Suppose first that $c\ne0$. The preimage of $B$ in $I$ is $I\cap\{(y-d)/c:y\in B\}$. The affine scaling property of Lebesgue measure, $m(sA+t)=|s|m(A)$ for Borel $A$ and $s\ne0$, therefore gives $m(I\cap T^{-1}(B))=|c|^{-1}m(B\cap T(I))$. The set $T(I)$ is precisely $J$, so this equals the stated density integral. If $c=0$, then $T(x)=d$ for every $x\in I$. The preimage of $B$ is $I$ when $d\in B$, and is empty otherwise. Its measure is $(v-u)\chi_{\{d\}}(B)$, which is the measure $(v-u)\delta_d(B)$. This also covers $u=v$, when the pushforward is zero. $\square$

Worked Example: An Affine Stretch and Translation

Take $I=[0,2]$ and $T(x)=3x-2$. The image interval is $J=[-2,4]$, since $T(0)=-2$ and $T(2)=4$. For every Borel set $B$, the affine formula gives $(T_{\#}m_I)(B)=\frac13m(B\cap[-2,4])$. In particular, for $B=[-1,2]$, the inverse map is $x=(y+2)/3$, so the preimage in $I$ is $[1/3,4/3]$. Its length is $4/3-1/3=1$, and the density formula gives $\frac13(2-(-1))=1$ as well.

Check: the total pushforward mass is $\frac13(4-(-2))=2$, equal to $m([0,2])=2$. The factor $1/3$ compensates for the map tripling lengths.

Absolute Value: Counting the Branches

The map $T(x)=|x|$ folds the negative half-line onto the positive half-line. A positive target value $y$ has possible preimages $y$ and $-y$; each contributes only if it lies in the source interval. The two branches are distinct for $y>0$. The target value zero has only the preimage $x=0$, a singleton of Lebesgue measure zero, so this map creates no atom at zero.

Proposition: Let $I=[u,v]$ be bounded, $m_I$ be restricted Lebesgue measure, and $T(x)=|x|$. For every $B\in\mathcal{B}_{\mathbb{R}}$, $$ (T_{\#}m_I)(B)=\int_{B\cap(0,\infty)} \bigl(\chi_I(y)+\chi_I(-y)\bigr)\,dm(y). $$ In particular, the pushforward has no atom at zero.

Proof. The preimage of $B\cap(0,\infty)$ splits into its positive and negative parts. On the positive part $T(x)=x$, so its measure is $m(B\cap(0,\infty)\cap I)$. On the negative part, reflection $x\mapsto -x$ preserves Lebesgue measure, and its measure is $m(B\cap(0,\infty)\cap(-I))$, where $-I=\{-x:x\in I\}$. These two contributions are exactly the two terms in the displayed integral. The remaining possible preimage, for target value zero, is contained in $\{0\}$, whose Lebesgue measure is zero. The positive and negative parts are disjoint, so their measures add. $\square$

Worked Example: Folding the Interval [-2,1]

Let $I=[-2,1]$. For $0<y<1$, both $y$ and $-y$ belong to $I$, so the density is $2$. For $1<y<2$, only $-y$ belongs to $I$, so the density is $1$. For $y>2$, neither branch belongs to $I$. Consequently, up to endpoint values that do not affect the measure, the density is $2$ on $(0,1)$, $1$ on $(1,2)$, and $0$ elsewhere.

For $B=(1/4,1)$, both inverse branches contribute. Each branch gives an interval of length $1-1/4=3/4$: the preimage is $(-1,-1/4)\cup(1/4,1)$, up to endpoints. Its measure is $3/4+3/4=3/2$. The density calculation gives $2(1-1/4)=3/2$. The total mass is $2\cdot1+1\cdot1=3$, which matches $m([-2,1])=3$.

Check: there is no additional mass at zero: its preimage is the singleton $\{0\}$, of measure zero.

The Square Map: The Density Changes with the Inverse

For $T(x)=x^2$, the inverse branches away from zero are $x=\sqrt y$ and $x=-\sqrt y$. Their slopes are not constant: the inverse stretches more near $y=0$. This produces the factor $1/(2\sqrt y)$ for each branch that belongs to the source interval. We first justify that factor from preimage lengths, then apply it to an arbitrary bounded interval.

Proposition: Let $I=[u,v]$ be bounded, $m_I$ be restricted Lebesgue measure, and $T(x)=x^2$. For every Borel $B\subseteq\mathbb{R}$, $$ (T_{\#}m_I)(B) = \int_{B\cap(0,\infty)} \frac{\chi_I(\sqrt y)+\chi_I(-\sqrt y)}{2\sqrt y}\,dm(y). $$ The pushforward has no atom at zero. The displayed density is integrable on bounded target intervals despite its possible singularity at zero.

Proof. First consider the positive branch $x\in[0,r]$, where $r\ge0$. For a target interval $(s,t)\subseteq(0,r^2)$, the preimage is $(\sqrt{s},\sqrt{t})$, whose length is $\sqrt{t}-\sqrt{s}$. Direct integration gives $\int_s^t (2\sqrt y)^{-1}\,dm(y)=\sqrt{t}-\sqrt{s}$. The same equality holds for intervals that reach zero by taking $s=0$, since $\int_0^t(2\sqrt y)^{-1}\,dm(y)=\sqrt t$, and for intervals cut off at $r^2$ by restricting to the source range. Thus the two finite Borel measures on $[0,r^2]$, one defined by the preimage and the other by this density, agree on intervals. By the standard uniqueness theorem for finite measures agreeing on a generating family of intervals (the pi-lambda theorem), they agree on every Borel set. For a positive source interval $[a,b]\subseteq[0,\infty)$, subtract the corresponding identities for $[0,a)$ and $[0,b]$; the result is the same density restricted to $y\in[a^2,b^2]$. Reflection gives the identical contribution for a negative source interval. Split $I$ into its negative and positive parts. Their only possible overlap is $x=0$, which has measure zero, and adding their contributions gives the two indicators in the formula. Finally, $T^{-1}(\{0\})\cap I\subseteq\{0\}$, so zero has no atom. The integral near zero is finite because $\int_0^t(2\sqrt y)^{-1}\,dm(y)=\sqrt t$. $\square$

Worked Example: Squaring the Interval [-1,2]

For $I=[-1,2]$, both inverse branches lie in $I$ when $0<y<1$, while only $+\sqrt y$ lies in $I$ when $1<y<4$. Thus, except at endpoints, the density is $1/\sqrt y$ on $(0,1)$, $1/(2\sqrt y)$ on $(1,4)$, and zero elsewhere.

For $B=[1,9/4]$, only the positive branch contributes except at the single endpoint $y=1$, which does not change the integral. The pushforward mass is $\int_1^{9/4}(2\sqrt y)^{-1}\,dm(y)=\sqrt{9/4}-\sqrt1=3/2-1=1/2$. Directly, $x^2\in[1,9/4]$ with $x\in[-1,2]$ means $x\in[1,3/2]$, up to endpoints, and this interval has length $3/2-1=1/2$.

Check: the total mass computed from the density is $\int_0^1y^{-1/2}\,dm(y)+\int_1^4(2\sqrt y)^{-1}\,dm(y)=2+(2-1)=3$, equal to the length of $[-1,2]$.

Common Mistakes and What the Formulas Show

Ignoring multiplicity. For the absolute-value map on $[-2,1]$, using density $1$ on $(0,2)$ would give total mass $2$, although the source has measure $3$. The correction is to count each available preimage: the density is $2$ on $(0,1)$ and $1$ on $(1,2)$, giving total mass $3$.

Counting zero twice as a branch of positive and negative values. For $y>0$, the branches $+\sqrt y$ and $-\sqrt y$ are distinct. At $y=0$, they coincide at $x=0$. In the square-map formula the integral is over $y>0$, and $x=0$ contributes no mass because its preimage is a singleton. Equivalently, if branch sets are written explicitly, define the strictly negative branch by $R_-(B)=\{-\sqrt y:y\in B\cap(0,\infty)\}$; then it does not contain zero. The positive branch may include zero, but any such inclusion changes no Lebesgue measure. The two branches on $y>0$ are disjoint, and zero is not an atom.

These computations distinguish stretching from folding. A nonconstant affine map has one inverse branch and a constant density $1/|c|$. Absolute value has a density equal to the number of available branches. Squaring has the same branch count, but each branch is weighted by $1/(2\sqrt y)$, reflecting the changing inverse scale. The total mass in every case remains $v-u$, since pushforward measures assign the whole target space the measure of its entire preimage $I$.

Check Your Understanding

In each question, compute using the preimage first, then compare with the relevant density formula.

For $I=[1,4]$ and $T(x)=2x+1$, find the pushforward density and its total mass.
For $I=[-3,-1]$ and $T(x)=|x|$, determine the density on the positive half-line and compute the pushforward mass of $(2,3)$.
For $I=[0,4]$ and $T(x)=x^2$, compute the density and the pushforward mass of $(1,4)$.
Explain why the square map on any bounded interval has no atom at zero, even when zero belongs to the source.
For $I=[-2,1]$, identify the interval of positive target values for which the absolute-value map has two preimages in $I$, and state its density there.