Measurable Maps into Product Spaces

Measurability Coordinate by Coordinate

A map into a product space has several component functions, one for each coordinate. The product sigma-algebra is designed so that measurability of the whole map can be tested using those components. This criterion does not require a measure on the source, nor any finiteness or sigma-finiteness assumption; it is a statement about measurable spaces.

Let $I$ be an index set, and for each $i\in I$ let $(Y_i,\mathcal{N}_i)$ be a measurable space. The product set is $Y=\prod_{i\in I}Y_i$. For each $i$, the coordinate projection is the map $\pi_i:Y\to Y_i$ given by $\pi_i(y)=y_i$. A coordinate cylinder associated with $i$ and $B\in\mathcal{N}_i$ is the subset $\pi_i^{-1}(B)$ of $Y$.

Definition: The product sigma-algebra $\bigotimes_{i\in I}\mathcal{N}_i$ is the sigma-algebra on $Y$ generated by all coordinate cylinders: $$ \bigotimes_{i\in I}\mathcal{N}_i = \sigma\!\left(\{\pi_i^{-1}(B): i\in I,\ B\in\mathcal{N}_i\}\right). $$ A map $F:(X,\mathcal{M})\to(Y,\mathcal{A})$ is measurable if $F^{-1}(A)\in\mathcal{M}$ for every $A\in\mathcal{A}$.

When $I$ is finite, measurable rectangles $B_1\times\cdots\times B_k$, with $B_j\in\mathcal{N}_j$, generate this same product sigma-algebra. For any index set, including an infinite one, coordinate cylinders are the defining generators. The theorem below follows directly from that definition.

The Coordinate Criterion

Theorem: Let $(X,\mathcal{M})$ be a measurable space, let $(Y_i,\mathcal{N}_i)$, $i\in I$, be measurable spaces, and equip $Y=\prod_{i\in I}Y_i$ with $\bigotimes_{i\in I}\mathcal{N}_i$. A map $F:X\to Y$ is measurable if and only if every coordinate map $F_i=\pi_i\circ F:X\to Y_i$ is measurable. No measure on $X$, and no finiteness assumption, is required.

Proof. First suppose $F$ is measurable. Fix $i\in I$ and $B\in\mathcal{N}_i$. The cylinder $\pi_i^{-1}(B)$ belongs to the product sigma-algebra by its definition. Therefore $$ F_i^{-1}(B) =(\pi_i\circ F)^{-1}(B) =F^{-1}\!\left(\pi_i^{-1}(B)\right)\in\mathcal{M}. $$ This holds for every $B\in\mathcal{N}_i$, so $F_i$ is measurable.

Conversely, suppose every $F_i$ is measurable. Define $$ \mathcal{D}=\{A\subseteq Y:F^{-1}(A)\in\mathcal{M}\}. $$ This is a sigma-algebra: $F^{-1}(Y)=X$, $F^{-1}(A^c)=X\setminus F^{-1}(A)$, and $F^{-1}(\bigcup_{n=1}^{\infty}A_n)=\bigcup_{n=1}^{\infty}F^{-1}(A_n)$. Thus it contains $Y$, is closed under complements, and is closed under countable unions. For every generator $\pi_i^{-1}(B)$, where $B\in\mathcal{N}_i$, we have $$ F^{-1}\!\left(\pi_i^{-1}(B)\right)=F_i^{-1}(B)\in\mathcal{M}. $$ Hence each generator belongs to $\mathcal{D}$. Since $\mathcal{D}$ is a sigma-algebra containing all the generators, it contains the sigma-algebra they generate. Thus $F^{-1}(A)\in\mathcal{M}$ for every $A\in\bigotimes_{i\in I}\mathcal{N}_i$, which is the measurability of $F$. This argument also covers $I=\varnothing$: the product is a singleton, its sigma-algebra is $\{\varnothing,Y\}$, and the coordinate condition is vacuous. $\square$

The proof explains why it is enough to check the coordinate maps: the preimages of the product generators are exactly coordinate preimages. It does not assert that an arbitrary subset of the product is itself a cylinder, or that checking only one selected rectangle suffices.

Worked Applications

Worked Example: An Affine Map into a Product of Real Lines

Take $(X,\mathcal{M})=(\mathbb{R},\mathcal{B}_{\mathbb{R}})$ and $Y=\mathbb{R}\times\mathbb{R}$ with $\mathcal{B}_{\mathbb{R}}\otimes\mathcal{B}_{\mathbb{R}}$. Define $F(t)=(t,1-t)$. Both coordinate functions are continuous, hence Borel measurable: the preimage of every open set under a continuous function is open, and the Borel sigma-algebra is generated by the open sets. The coordinate criterion therefore makes $F$ measurable.

For a direct inverse-image calculation, take the measurable rectangle $E=(0,2]\times[0,1)$. Its preimage is found by imposing both coordinate conditions: $$ \begin{aligned} F^{-1}(E) &=\{t:0<t\leq 2\text{ and }0\leq 1-t<1\}\\ &=\{t:0<t\leq 2\text{ and }t\leq 1\text{ and }t>0\}\\ &=(0,1]. \end{aligned} $$ The endpoint $t=1$ is included because $1-t=0$ belongs to $[0,1)$; $t=0$ is excluded because the first interval is open at zero. The answer is Borel measurable, as required. Substitution checks the boundary: $F(1)=(1,0)\in E$, while $F(0)=(0,1)\notin E$.

Worked Example: A Measurable Map with a Discontinuous Coordinate

Let $X=[0,1]$ carry the Lebesgue sigma-algebra $\mathcal{L}$, and let the target be $\{0,1\}\times\mathbb{R}$ with $\mathcal{P}(\{0,1\})\otimes\mathcal{B}_{\mathbb{R}}$. Define $F(x)=(\chi_{\mathbb{Q}}(x),x)$, where $\mathbb{Q}$ is the set of rational numbers. The first coordinate is measurable because its level sets are $\mathbb{Q}\cap[0,1]$ and its complement in $[0,1]$; both are Borel and therefore Lebesgue measurable. The second coordinate is the identity map on $[0,1]$, which is measurable. The criterion gives measurability of $F$, even though its first coordinate is discontinuous.

For the rectangle $\{1\}\times[0,\tfrac12]$, the strictness and inclusion of endpoints give $$ F^{-1}\!\left(\{1\}\times[0,\tfrac12]\right) =\mathbb{Q}\cap[0,\tfrac12]. $$ This set is countable, hence has Lebesgue measure zero. Indeed, if its points are listed as $q_1,q_2,\ldots$, then for any $\varepsilon>0$ they can be covered by intervals whose lengths sum to at most $\varepsilon$, for example intervals of lengths $\varepsilon 2^{-j}$ around $q_j$; consequently its outer measure is at most $\varepsilon$, for every $\varepsilon>0$, and its measure is zero. The calculation also confirms that the endpoints $0$ and $\tfrac12$ are included whenever they are rational.

Worked Example: A Map from a Counting-Measure Space

Let $X=\mathbb{N}=\{1,2,\ldots\}$ with sigma-algebra $\mathcal{P}(\mathbb{N})$ and counting measure, and use the target $\{0,1\}\times\mathbb{R}$ with $\mathcal{P}(\{0,1\})\otimes\mathcal{B}_{\mathbb{R}}$. Define $$ F(n)=(n\bmod 2,1/n), $$ where $n\bmod 2$ is $0$ for even $n$ and $1$ for odd $n$. Every subset of the source is measurable, so each coordinate map is measurable. Thus $F$ is measurable.

For the rectangle $\{0\}\times(\tfrac13,1]$, the first condition requires $n$ to be even. The second requires $\tfrac13<1/n\leq1$. Since $n\geq1$, the upper inequality always holds; multiplying the strict lower inequality by the positive number $3n$ gives $n<3$. The only positive even integer below $3$ is $2$. Therefore $$ F^{-1}\!\left(\{0\}\times(\tfrac13,1]\right)=\{2\}. $$ Its counting measure is $1$. Checking the value, $F(2)=(0,\tfrac12)$ belongs to the rectangle, whereas $F(1)=(1,1)$ fails the first condition and $F(4)=(0,\tfrac14)$ fails the strict lower bound. The strict inequality excludes $n=3$, for which $1/n=1/3$.

A Common Pitfall: Losing an Endpoint

In an inverse-image calculation, the coordinate conditions must be combined with their original strict and non-strict inequalities. For the first example, a tempting but incorrect computation is to report $F^{-1}((0,2]\times[0,1))=(0,1)$, on the grounds that the second coordinate must be positive. That changes the stated target interval: $[0,1)$ includes zero. The correct conditions are $0<t\leq2$ and $0\leq1-t<1$; together they give $0<t\leq1$, hence $(0,1]$. Substitution of $t=1$ verifies the included endpoint.

A related conceptual error is to replace the product sigma-algebra by some other sigma-algebra on the same underlying set without checking the target. The theorem concerns exactly $\bigotimes_{i\in I}\mathcal{N}_i$, generated by coordinate cylinders. The coordinate test follows from those generators; it should not be applied to an unspecified target sigma-algebra.

Why the Criterion Is Useful

For a map into a product, direct testing against every measurable subset of the target can be unwieldy. The theorem reduces the task to one coordinate at a time, and its proof identifies the precise reason this reduction is valid: inverse images preserve complements and countable unions, so measurability on the generators extends to their generated sigma-algebra. In applications, coordinates may have different forms—continuous functions, indicator functions, or functions on a discrete source—yet the same criterion applies.

There is no measure-theoretic hypothesis hidden in the statement. If $(X,\mathcal{M},\mu)$ is a measure space, $\mu$ may be finite, infinite, or absent from the question entirely; the criterion is unchanged. It asserts measurability everywhere in the inverse-image sense, not an almost-everywhere property. When checking an example, first state the sigma-algebras on the source and every coordinate target, then verify each coordinate map against all measurable sets in its coordinate space.

Check Your Understanding

Use the definition of the product sigma-algebra and the coordinate criterion in each question.

For $F:\mathbb{R}\to\mathbb{R}^2$, $F(t)=(t^2,t+1)$, with Borel sigma-algebras, explain why $F$ is measurable and compute $F^{-1}([0,1]\times[1,2])$.
Suppose $F:X\to Y_1\times Y_2$ is measurable. For $B\in\mathcal{N}_2$, write the inverse-image identity that proves the second coordinate map is measurable.
On $\mathbb{N}$ with counting measure, compute the preimage of $\{1\}\times[\tfrac14,\tfrac12]$ under $n\mapsto(n\bmod2,1/n)$.
Why does the coordinate criterion not require $X$ to have a measure, or its measure to be finite?
In the affine-map example, test $t=1$ directly against $(0,2]\times[0,1)$. Which endpoint error would result from omitting it?