Measures Transferred by Measurable Maps
A measurable map lets us use a measure on one space to assign masses to measurable sets in another. The essential operation is to take the inverse image of a target set: this is guaranteed to be measurable in the domain, so the original measure can be applied to it. The resulting measure records how much domain mass is sent into each target set, including when the map is many-to-one or the domain and target are different spaces.
Let \((X,\mathcal{M},\mu)\) and \((Y,\mathcal{N})\) be measure and measurable spaces, respectively. The measure \(\mu\) may be finite or infinite; neither finiteness nor sigma-finiteness is needed for the definition. A map \(T:X\to Y\) is \((\mathcal{M},\mathcal{N})\)-measurable if \(T^{-1}(A)\in\mathcal{M}\) for every \(A\in\mathcal{N}\). Here \(T^{-1}(A)=\{x\in X:T(x)\in A\}\). The target need not be a subset of the domain.
The formula is meaningful because measurability of \(T\) ensures \(T^{-1}(A)\in\mathcal{M}\). It also assigns a value in \([0,\infty]\), since \(\mu\) is a measure. To confirm that the definition really produces a measure, we check the empty set and countable additivity.
Why the Push-Forward Is a Measure
Proof. First, \(T^{-1}(\varnothing)=\varnothing\), so \((T_\#\mu)(\varnothing)=\mu(\varnothing)=0\). Now let \((A_j)_{j\ge1}\) be pairwise disjoint members of \(\mathcal{N}\). Their inverse images are measurable. They are pairwise disjoint: if \(x\in T^{-1}(A_i)\cap T^{-1}(A_j)\), then \(T(x)\in A_i\cap A_j\), which is impossible for \(i\ne j\). Also, inverse images preserve unions, so \(T^{-1}(\bigcup_{j\ge1}A_j)=\bigcup_{j\ge1}T^{-1}(A_j)\). Countable additivity of \(\mu\) therefore gives $$ (T_\#\mu)\left(\bigcup_{j\ge1}A_j\right) =\mu\left(\bigcup_{j\ge1}T^{-1}(A_j)\right) =\sum_{j\ge1}\mu(T^{-1}(A_j)) =\sum_{j\ge1}(T_\#\mu)(A_j). $$ Thus \(T_\#\mu\) is a measure. Finally, \(T^{-1}(Y)=X\), so \((T_\#\mu)(Y)=\mu(X)\). \(\square\)
This identity determines the total mass exactly, including the possibility of infinite mass. In particular, \(T_\#\mu\) is finite if and only if \(\mu\) is finite, and their total masses agree. If \(\mu\) is a probability measure, then \(T_\#\mu\) is also a probability measure. The definition does not require \(T\) to be one-to-one or onto.
Worked Example: The Identity Map
Let \((X,\mathcal{M},\mu)\) be any measure space and let \(T:X\to X\) be the identity map \(T(x)=x\). This map is measurable because \(T^{-1}(A)=A\) for every \(A\in\mathcal{M}\). Hence $$ (T_\#\mu)(A)=\mu(T^{-1}(A))=\mu(A),\qquad A\in\mathcal{M}. $$ Thus \(T_\#\mu=\mu\) on every measurable set. In particular, its total mass is \((T_\#\mu)(X)=\mu(X)\), whether that value is finite or infinite. Check: the general total-mass formula gives \((T_\#\mu)(X)=\mu(T^{-1}(X))=\mu(X)\), the same result.
Computing Values from Inverse Images
For a particular target set \(A\), computation means determining the domain set \(T^{-1}(A)\) and then measuring it. The measurable image \(T(E)\) of a domain set \(E\) is not what appears in the definition; it may not even belong to \(\mathcal{N}\). Keep the domain and target roles distinct: \(\mu\) measures subsets of \(X\), while \(T_\#\mu\) measures subsets of \(Y\).
Worked Example: A Constant Map
Take \(X=[0,1]\), \(\mathcal{M}\) the Lebesgue sigma-algebra on \([0,1]\), and \(\mu\) Lebesgue measure restricted to \([0,1]\). Take \(Y=\mathbb{R}\) with its Borel sigma-algebra \(\mathcal{B}_{\mathbb{R}}\), and define \(T(x)=0\) for every \(x\in X\). For a Borel set \(A\), if \(0\in A\), then every \(x\in[0,1]\) maps into \(A\), so \(T^{-1}(A)=[0,1]\) and \((T_\#\mu)(A)=1\). If \(0\notin A\), then \(T^{-1}(A)=\varnothing\) and \((T_\#\mu)(A)=0\). Therefore $$ (T_\#\mu)(A)= \begin{cases} 1,&0\in A,\\ 0,&0\notin A. \end{cases} $$ This is the Dirac measure \(\delta_0\) on \(\mathcal{B}_{\mathbb{R}}\). Check: since \(0\in\mathbb{R}\), the formula gives \((T_\#\mu)(\mathbb{R})=1\); independently, \(\mu(X)=m([0,1])=1\).
Worked Example: Inclusion of a Measurable Subset
Let \(X=\mathbb{R}\) with Lebesgue sigma-algebra \(\mathcal{L}\) and Lebesgue measure \(m\). Set \(S=[-1,2]\), equip \(S\) with the restricted sigma-algebra \(\mathcal{L}|_S=\{E\cap S:E\in\mathcal{L}\}\), and let \(\mu\) be the restriction of \(m\) to \(S\). Use the inclusion map \(T:S\to\mathbb{R}\), \(T(x)=x\), with target sigma-algebra \(\mathcal{L}\). It is measurable: for \(A\in\mathcal{L}\), \(T^{-1}(A)=A\cap S\in\mathcal{L}|_S\). Thus $$ (T_\#\mu)(A)=\mu(A\cap S)=m(A\cap[-1,2]),\qquad A\in\mathcal{L}. $$ For example, with \(A=[0,3]\), the inverse image is \([0,2]\), so \((T_\#\mu)([0,3])=m([0,2])=2\). The total mass is \((T_\#\mu)(\mathbb{R})=\mu(S)=m([-1,2])=2-(-1)=3\). Check: setting \(A=\mathbb{R}\) in the displayed set formula gives \(m(\mathbb{R}\cap[-1,2])=3\), agreeing with the total-mass identity.
Finite Mass and Sigma-Finiteness
Finiteness transfers directly because the total masses are equal. Sigma-finiteness needs more care. A measure is sigma-finite if its space is a countable union of measurable sets, each of finite measure. A sigma-finite measure on the domain need not give a sigma-finite push-forward: many domain regions can be sent into a single target set of infinite mass.
Proof. By definition, each \(A_n\) is measurable for \(T_\#\mu\), and \((T_\#\mu)(A_n)=\mu(T^{-1}(A_n))<\infty\). Since these sets cover \(Y\), they are a countable cover of \(Y\) by sets of finite \(T_\#\mu\)-measure. This is precisely sigma-finiteness. \(\square\)
Worked Example: Sigma-Finite Domain, Non-Sigma-Finite Push-Forward
Let \(X=\mathbb{N}=\{1,2,\ldots\}\) with all subsets measurable and counting measure \(\mu\), and let \(Y=\{*\}\) with sigma-algebra \(\{\varnothing,Y\}\). Define \(T(n)=*\) for every \(n\). The domain measure is sigma-finite because \(X=\bigcup_{n\ge1}\{n\}\) and \(\mu(\{n\})=1\). The map is measurable, and $$ (T_\#\mu)(\varnothing)=0,\qquad (T_\#\mu)(Y)=\mu(T^{-1}(Y))=\mu(\mathbb{N})=\infty. $$ This target measure is not sigma-finite. Indeed, the only measurable sets are \(\varnothing\), of measure \(0\), and \(Y\), of measure \(\infty\); no countable collection of finite-measure measurable sets can cover \(Y\). Check: every cover of \(Y\) by sets in \(\{\varnothing,Y\}\) must include \(Y\), whose push-forward measure is infinite.
A Common Mistake: Measuring the Image
A tempting but generally incorrect calculation is to replace \(T^{-1}(A)\) by \(T(A)\). These sets have different roles: \(A\) is a target set, while \(T^{-1}(A)\) is a domain set that records exactly which points are sent into \(A\). The map may also identify many domain points, so the image of a set does not count how much mass is sent into a target set.
For a concrete incorrect calculation, use the constant-map example and take \(A=\{0\}\). The wrong rule would give \(\mu(T(A))=\mu(\{0\})=0\), since \(\{0\}\) has Lebesgue measure zero in the domain \([0,1]\). But the definition gives $$ (T_\#\mu)(\{0\})=\mu(T^{-1}(\{0\}))=\mu([0,1])=1. $$ The correction is to find the full inverse image first, not the image of \(A\). This example also shows why even when a set can be viewed as belonging to both spaces, applying the domain measure to \(T(A)\) does not compute the push-forward.
The inverse-image formula is useful precisely because it applies uniformly to all measurable target sets and automatically preserves countable additivity. It also makes the mass statement immediate: all of \(X\) maps into \(Y\), so the target’s total mass is exactly the domain’s total mass. The calculation always depends on the target sigma-algebra: only sets \(A\in\mathcal{N}\) are assigned push-forward values.
Check Your Understanding
For each question, start with the target set and determine its inverse image under the map.
- State the two conditions on \(T\) and on the measurable spaces that make \((T_\#\mu)(A)=\mu(T^{-1}(A))\) well-defined.
- Prove directly from the definition that \((T_\#\mu)(Y)=\mu(X)\).
- If \(\mu(X)=5\), what is the total mass of \(T_\#\mu\)? Does the answer depend on whether \(T\) is one-to-one?
- For the constant map from \([0,1]\) to \(\mathbb{R}\) with value \(0\), compute \((T_\#\mu)([-1,0])\) and \((T_\#\mu)((0,1])\).
- Why does sigma-finiteness of \(\mu\) alone not guarantee sigma-finiteness of \(T_\#\mu\)?