Composition and Invariance of Push-Forward Measures

Push-Forwards Under Composition

A push-forward records how much measure a map assigns to sets in its target: the mass of a target set is the mass of its preimage. This preimage viewpoint makes composition particularly simple. It also gives a precise test for invariance: a map preserves a measure exactly when every measurable set has a preimage of the same measure.

Let \((X,\mathcal{M},\mu)\) be a measure space, with no assumption that \(\mu\) is finite or sigma-finite. Let \((Y,\mathcal{N})\) be a measurable space, and let \(T:X\to Y\) be measurable, meaning \(T^{-1}(B)\in\mathcal{M}\) for every \(B\in\mathcal{N}\). The push-forward \(T_\#\mu\) is the measure on \((Y,\mathcal{N})\) defined by \((T_\#\mu)(B)=\mu(T^{-1}(B))\). Its values may be infinite.

Theorem (Composition rule): Let \((X,\mathcal{M},\mu)\), \((Y,\mathcal{N})\), and \((Z,\mathcal{P})\) be a measure space and two measurable spaces. If \(T:X\to Y\) and \(S:Y\to Z\) are measurable, then \((S\circ T)_\#\mu=S_\#(T_\#\mu)\) as measures on \((Z,\mathcal{P})\).

Proof. Both sides are measures on \((Z,\mathcal{P})\), so it suffices to show that they give the same value to each \(C\in\mathcal{P}\). By the definition of push-forward and the preimage identity for composition,

((S\circ T)_\#\mu)(C) =\mu((S\circ T)^{-1}(C)) =\mu(T^{-1}(S^{-1}(C))).

On the other hand, \(S^{-1}(C)\in\mathcal{N}\), and therefore

(S_\#(T_\#\mu))(C) =(T_\#\mu)(S^{-1}(C)) =\mu(T^{-1}(S^{-1}(C))).

The two values agree for every \(C\in\mathcal{P}\), including when the common value is \(+\infty\). Hence the measures are equal. \(\square\)

The order in this formula matters: first \(T\) acts on points of \(X\), then \(S\) acts on points of \(Y\). On the measure side, \(T_\#\) is applied first and \(S_\#\) second.

Worked Example: Composing a Dilation and a Translation

Take Lebesgue measure \(m\) on \((\mathbb{R},\mathcal{L})\), where \(\mathcal{L}\) is the Lebesgue sigma-algebra. This measure is complete and sigma-finite, but not finite. Define \(T(x)=2x\) and \(S(y)=y+1\). Both maps are measurable, and \(S\circ T(x)=2x+1\). Compute the push-forward masses of \(E=[1,3)\) by both routes.

Directly, \((S\circ T)^{-1}(E)=[0,1)\), since \(1\le 2x+1<3\) is equivalent to \(0\le x<1\). Thus \(((S\circ T)_\#m)(E)=m([0,1))=1\).

For the successive push-forwards, \(S^{-1}(E)=[0,2)\). The affine push-forward formula for Lebesgue measure is \(m(aA+b)=|a|m(A)\) for Lebesgue measurable \(A\) and \(a\ne0\), equivalently \(m((x\mapsto ax+b)^{-1}(A))=|a|^{-1}m(A)\). This formula, established in the elementary affine push-forward calculation, gives \((T_\#m)([0,2))=\tfrac12m([0,2))=1\). Consequently \((S_\#(T_\#m))(E)=1\), in agreement with the composition rule.

Check: the direct preimage has length \(1-0=1\); the successive calculation gives \(\tfrac12(2-0)=1\). The endpoints are handled by the half-open intervals, and changing endpoints would not change these Lebesgue measures.

Measure-Preserving Maps

For a map from a space to itself, invariance asks whether the push-forward measure equals the original measure. This definition concerns measurable preimages, not the measure of the image of a set. No bijectivity is required for the definition, although translations, reflections, and permutations considered below are bijections.

Definition (Measure-preserving map): Let \((X,\mathcal{M},\mu)\) be a measure space, and let \(T:X\to X\) be measurable. The map \(T\) preserves \(\mu\), or \(\mu\) is \(T\)-invariant, if \(T_\#\mu=\mu\). Equivalently, \(\mu(T^{-1}(A))=\mu(A)\) for every \(A\in\mathcal{M}\). These equalities are exact equalities of extended nonnegative numbers, not merely almost-everywhere statements.

Invariance is stable under composition. If \(T\) and \(S\) are measurable maps from \(X\) to itself and each preserves \(\mu\), then the composition theorem gives \((S\circ T)_\#\mu=S_\#(T_\#\mu)=S_\#\mu=\mu\). The identity map preserves every measure, since its preimage of \(A\) is \(A\).

Proposition: Let \((X,\mathcal{M},\mu)\) be a measure space and \(T:X\to X\) a measurable bijection. If \(T^{-1}\) is also measurable and \(T\) preserves \(\mu\), then \(T^{-1}\) preserves \(\mu\).

Proof. Fix \(B\in\mathcal{M}\). Since \(T^{-1}\) is measurable, \((T^{-1})^{-1}(B)=T(B)\) belongs to \(\mathcal{M}\). Apply the invariance of \(T\) to the measurable set \(T(B)\). It gives \(\mu(T^{-1}(T(B)))=\mu(T(B))\). Because \(T\) is a bijection, \(T^{-1}(T(B))=B\), so \(\mu(T(B))=\mu(B)\). Finally, \((T^{-1})^{-1}(B)=T(B)\), and hence \(((T^{-1})_\#\mu)(B)=\mu(B)\). This holds for every measurable \(B\), proving that \(T^{-1}\) preserves \(\mu\). \(\square\)

The measurability of the inverse is needed here: it ensures that the image \(T(B)\), which appears as a preimage under \(T^{-1}\), is measurable. A measurable bijection need not have a measurable inverse on arbitrary measurable spaces.

Translations, Reflections, and Dilations

Worked Example: Translations and Reflections Preserve Lebesgue Measure

On \((\mathbb{R},\mathcal{L},m)\), consider the translation \(T_a(x)=x+a\) for a fixed real number \(a\), and the reflection \(R(x)=-x\). These are measurable bijections with measurable inverses. The affine formula stated above gives, for every \(E\in\mathcal{L}\),

m(T_a^{-1}(E))=m(E-a)=m(E), \qquad m(R^{-1}(E))=m(-E)=m(E).

Thus both maps preserve Lebesgue measure, including on sets of infinite measure. For an explicit check, take \(E=[2,5)\) and \(a=3\). Then \(T_3^{-1}(E)=[-1,2)\), whose measure is \(2-(-1)=3\); \(E\) itself has measure \(5-2=3\). For reflection, \(R^{-1}(E)=(-5,-2]\), which has measure \((-2)-(-5)=3\) as well.

Check: the preimages and the original interval each have length \(3\). The reflected interval has a different endpoint convention, but endpoints have Lebesgue measure zero, so its measure is still \(3\).

Not every measurable bijection preserves Lebesgue measure. For the dilation \(D_c(x)=cx\), where \(c\ne0\), the same affine formula gives \(m(D_c^{-1}(E))=|c|^{-1}m(E)\). In particular, \(D_c\) preserves Lebesgue measure precisely when \(|c|=1\). The cases \(c=1\) and \(c=-1\) are the identity and reflection; if \(|c|\ne1\), any interval of positive finite length has its measure changed.

Worked Example: A Dilation Does Not Preserve Measure

Let \(D_2(x)=2x\) on \((\mathbb{R},\mathcal{L},m)\), and take \(E=[0,2)\). Its measure is \(m(E)=2-0=2\). The preimage is \(D_2^{-1}(E)=[0,1)\), because \(0\le2x<2\) is equivalent to \(0\le x<1\). Therefore \((D_{2\#}m)(E)=m([0,1))=1\), which is not \(m(E)=2\). So \(D_2\) is not measure-preserving.

Check: the affine formula gives \((D_{2\#}m)(E)=\tfrac12m(E)=\tfrac12\cdot2=1\), matching the direct preimage calculation. For \(D_{-1}\), the same formula gives factor \(|-1|^{-1}=1\), consistent with reflection invariance.

Invariance on Finite and Countable Spaces

On a finite set, invariance under a permutation has a direct description in terms of the masses of individual points. Let \(X=\{1,\ldots,n\}\), let \(\mathcal{M}\) be all subsets of \(X\), and define \(\mu(A)=\sum_{i\in A}w_i\), where each weight \(w_i\) is nonnegative and finite. This is a finite measure. If \(\sigma:X\to X\) is a permutation, then for any \(A\subseteq X\),

(\sigma_\#\mu)(A) =\mu(\sigma^{-1}(A)) =\sum_{j\in A}w_{\sigma^{-1}(j)}.

Proposition: In this finite weighted space, a permutation \(\sigma\) preserves \(\mu\) if and only if \(w_{\sigma^{-1}(j)}=w_j\) for every \(j\in X\). Equivalently, the weights are constant along each cycle of the permutation.

Proof. If \(\sigma\) preserves \(\mu\), apply \(\mu(\sigma^{-1}(A))=\mu(A)\) to the singleton \(A=\{j\}\). Since \(\sigma^{-1}(\{j\})=\{\sigma^{-1}(j)\}\), this gives \(w_{\sigma^{-1}(j)}=w_j\). Conversely, suppose this equality holds for every \(j\). For every subset \(A\subseteq X\), the displayed sum then gives \(\mu(\sigma^{-1}(A))=\sum_{j\in A}w_j=\mu(A)\). Thus \(\sigma\) preserves \(\mu\). Iterating the equality around each finite cycle shows that all weights in that cycle agree; conversely, cyclewise constancy implies the equality at every point. \(\square\)

Worked Example: A Permutation Preserving a Weighted Measure

Take \(X=\{1,2,3,4\}\), with weights \(w_1=2,\ w_2=1,\ w_3=2,\ w_4=1\), and let \(\sigma\) exchange \(1\) with \(3\) and \(2\) with \(4\). The two cycles are \((1\ 3)\) and \((2\ 4)\). The weights agree on each cycle: \(w_1=w_3=2\) and \(w_2=w_4=1\). The criterion therefore proves that \(\sigma\) preserves \(\mu\).

For a direct check, choose \(A=\{1,2,4\}\). Its mass is \(\mu(A)=w_1+w_2+w_4=2+1+1=4\). Since \(\sigma^{-1}=\sigma\), its preimage is \(\{3,4,2\}\), with mass \(w_3+w_4+w_2=2+1+1=4\). The two calculations agree. If instead a permutation exchanged \(1\) and \(2\), the singleton \(\{1\}\) would have preimage \(\{2\}\), of mass \(1\) rather than \(2\); that permutation would not preserve \(\mu\).

Check: the selected set and its preimage both have mass \(4\), and the unequal singleton weights in the alternative exchange provide a specific failure of invariance.

A related fact explains why translations preserve counting measure on the integers. On \((\mathbb{Z},\mathcal{P}(\mathbb{Z}),\mu)\), let \(\mu\) be counting measure, so \(\mu(A)\) is the number of elements of \(A\) if \(A\) is finite and \(+\infty\) if \(A\) is infinite. Every bijection \(\sigma:\mathbb{Z}\to\mathbb{Z}\) preserves this measure: its restriction gives a bijection from \(\sigma^{-1}(A)\) to \(A\), so the two sets have the same finite cardinality or are both infinite. This applies to integer translations and reflection. The measure is sigma-finite but not finite, since \(\mathbb{Z}\) is a countable union of singletons, each of measure \(1\).

A Common Pitfall: Using Images Instead of Preimages

The definition of \(T_\#\mu\) uses \(T^{-1}(A)\), not \(T(A)\). Confusing the two can reverse the scaling factor. For example, under \(D_2(x)=2x\), the image of \(E=[0,2)\) is \(D_2(E)=[0,4)\), of measure \(4\). But the push-forward mass assigned to \(E\) is \(m(D_2^{-1}(E))=m([0,1))=1\), not \(4\). The image calculation answers a different question: it measures the size of the transformed set. The push-forward calculation asks how much original mass lands inside the target set.

For any measurable bijection \(T:X\to X\), the correct test is to take an arbitrary measurable \(A\) and compare \(\mu(T^{-1}(A))\) with \(\mu(A)\). Testing only a few convenient sets can reveal failure, as the dilation example does, but proving invariance requires the equality for every measurable set. This criterion is useful in probability, where an invariant probability measure describes a distribution unchanged by a transformation, and in analysis, where changes of variables often reduce to understanding exactly how a map alters measure.

Check Your Understanding

For each question, state the spaces and use preimages when computing push-forward masses.

For measurable maps \(T:X\to Y\) and \(S:Y\to Z\), show directly that \((S\circ T)^{-1}(C)=T^{-1}(S^{-1}(C))\) for a measurable \(C\subseteq Z\), and use it to verify the composition rule.
On \((\mathbb{R},\mathcal{L},m)\), does \(x\mapsto -x+4\) preserve Lebesgue measure? Give the affine factor and test it on one interval.
On a four-point space with weights \(3,2,3,2\), which transpositions preserve the weighted measure? Explain using the cycle criterion.
Suppose \(T\) is a measure-preserving measurable bijection and \(T^{-1}\) is measurable. Identify where measurability of \(T^{-1}\) enters the proof that \(T^{-1}\) also preserves the measure.
For \(D_3(x)=3x\) and \(E=[0,3)\), compute \(m(E)\) and \(m(D_3^{-1}(E))\). Does \(D_3\) preserve Lebesgue measure?