Coordinate Marginals and Coordinate Strips
A measure on a product space assigns mass to sets in both coordinates at once. Its marginals record the mass seen in each coordinate separately: to find the first marginal on a measurable set \(A\), take the measure of the strip \(A\times Y\); to find the second on \(B\), take the measure of \(X\times B\). For finite product measures these strip masses follow directly from the defining rectangle formula. The key detail is that a marginal need not equal its factor measure: it includes the total mass of the other factor.
Throughout, let \((X,\mathcal{M},\mu)\) and \((Y,\mathcal{N},\nu)\) be finite measure spaces, meaning \(\mu(X)<\infty\) and \(\nu(Y)<\infty\). Write \(\mathcal{M}\otimes\mathcal{N}\) for the product sigma-algebra on \(X\times Y\), and let \(\lambda=\mu\times\nu\) be the finite product measure on that sigma-algebra. We use the rectangle identity \(\lambda(A\times B)=\mu(A)\nu(B)\) for \(A\in\mathcal{M}\) and \(B\in\mathcal{N}\). This identity is the product-measure property established in the earlier material on product measures.
These definitions are meaningful on the stated sigma-algebras: \(A\times Y\) belongs to \(\mathcal{M}\otimes\mathcal{N}\) whenever \(A\in\mathcal{M}\), and \(X\times B\) belongs to it whenever \(B\in\mathcal{N}\). In particular, the projections are measurable. For example, \(\pi_X^{-1}(A)=A\times Y\), so the preimage of every measurable set in \(X\) is measurable in the product space. The same reasoning applies to \(\pi_Y\).
The Marginal Formula
Proof. Since \(X\times Y\) is a measurable rectangle, the rectangle identity gives \(\lambda(X\times Y)=\mu(X)\nu(Y)<\infty\), so \(\lambda\) is finite. For each \(A\in\mathcal{M}\), \(A\times Y\) is a measurable rectangle. Applying that same identity with \(B=Y\) gives \(\lambda_X(A)=\lambda(A\times Y)=\mu(A)\nu(Y)\). This holds for every \(A\in\mathcal{M}\), which is precisely the equality of measures \(\lambda_X=\nu(Y)\mu\). For each \(B\in\mathcal{N}\), applying the rectangle identity with \(A=X\) gives \(\lambda_Y(B)=\lambda(X\times B)=\mu(X)\nu(B)\). This holds for every \(B\in\mathcal{N}\), so \(\lambda_Y=\mu(X)\nu\). The argument also covers zero total masses: if, for example, \(\nu(Y)=0\), the first marginal formula says \(\lambda_X(A)=0\) for every \(A\), while the second formula remains \(\lambda_Y(B)=\mu(X)\nu(B)\). \(\square\)
The scalars in these formulas matter. If both factor measures are probability measures, then \(\mu(X)=\nu(Y)=1\), so the marginals equal the factors. In the general finite case the product measure need not be a probability measure, and its marginals are scaled. The total mass is \(\lambda(X\times Y)=\mu(X)\nu(Y)\), consistent with either marginal formula applied to the whole factor space.
Worked Example: Probability Factors
Let \(X=Y=[0,1]\), each with its Lebesgue sigma-algebra, and let \(\mu\) and \(\nu\) have densities \(2x\) and \(3y^2\), respectively, with respect to Lebesgue measure. Compute the marginals and the product measure of the strip \(A\times Y\), where \(A=[0,\tfrac12]\), and the rectangle \(A\times B\), where \(B=(\tfrac13,1]\).
First compute the total masses directly: \(\mu(X)=\int_0^1 2x\,dx=[x^2]_0^1=1\), and \(\nu(Y)=\int_0^1 3y^2\,dy=[y^3]_0^1=1\). Thus both factors are probability measures. By the theorem, \(\lambda_X=\mu\) and \(\lambda_Y=\nu\). In particular, the first marginal gives \(\lambda_X(A)=\mu(A)=\int_0^{1/2}2x\,dx=[x^2]_0^{1/2}=1/4\). The second marginal gives \(\lambda_Y(B)=\nu(B)=\int_{1/3}^{1}3y^2\,dy=[y^3]_{1/3}^{1}=1-1/27=26/27\). The interval \(B\) excludes \(1/3\), which does not change its measure because a singleton has Lebesgue measure zero.
The strip has measure \(\lambda(A\times Y)=\mu(A)\nu(Y)=(1/4)(1)=1/4\). The rectangle has measure \(\lambda(A\times B)=\mu(A)\nu(B)=(1/4)(26/27)=13/54\). Check: the rectangle calculation factors into the two integrals, giving \((1/4)(26/27)=13/54\); enlarging \(B\) to all of \(Y\) replaces \(26/27\) by \(1\) and gives the strip value \(1/4\), as required.
Worked Example: Nonprobability Factors
Let \(X=[0,2]\) and \(Y=[0,3]\), equipped with their Lebesgue sigma-algebras. Let \(\mu\) have density \(1+x\) on \(X\), and let \(\nu\) have constant density \(2\) on \(Y\). Find both marginals and compute the masses of \([0,1]\times Y\) and \(X\times[1,2]\).
The total masses are \(\mu(X)=\int_0^2(1+x)\,dx=[x+x^2/2]_0^2=2+2=4\) and \(\nu(Y)=\int_0^3 2\,dy=[2y]_0^3=6\). Therefore the first marginal is \(\lambda_X=6\mu\), which has density \(6(1+x)\) on \(X\), and the second is \(\lambda_Y=4\nu\), which has density \(8\) on \(Y\).
For \(A=[0,1]\), \(\mu(A)=\int_0^1(1+x)\,dx=[x+x^2/2]_0^1=1+1/2=3/2\). Hence \(\lambda(A\times Y)=\mu(A)\nu(Y)=(3/2)(6)=9\). For \(B=[1,2]\), \(\nu(B)=\int_1^2 2\,dy=[2y]_1^2=4-2=2\). Hence \(\lambda(X\times B)=\mu(X)\nu(B)=(4)(2)=8\). Check: the total product mass is \(\lambda(X\times Y)=(4)(6)=24\). The first marginal assigns \(6\mu(X)=6(4)=24\) to \(X\), and the second assigns \(4\nu(Y)=4(6)=24\) to \(Y\), agreeing with that total.
Worked Example: Piecewise Densities and a Rectangle
On \(X=Y=[0,1]\), define \(\mu\) to have density \(2\) on \([0,\tfrac12]\) and density \(1\) on \((\tfrac12,1]\). Define \(\nu\) to have density \(3\) on \([0,\tfrac13]\) and density \(1\) on \((\tfrac13,1]\). Compute both marginal measures and the product measure of \(A\times B\), where \(A=(\tfrac14,\tfrac34]\) and \(B=(0,\tfrac23]\).
The factor totals are \(\mu(X)=2(1/2)+1(1/2)=1+1/2=3/2\) and \(\nu(Y)=3(1/3)+1(2/3)=1+2/3=5/3\). Thus \(\lambda_X=(5/3)\mu\) and \(\lambda_Y=(3/2)\nu\).
To compute \(\mu(A)\), split \(A\) at \(1/2\). The part \((1/4,1/2]\) has length \(1/2-1/4=1/4\) and density \(2\), so its mass is \(2(1/4)=1/2\). The part \((1/2,3/4]\) has length \(3/4-1/2=1/4\) and density \(1\), so its mass is \(1/4\). Therefore \(\mu(A)=1/2+1/4=3/4\). For \(B\), the part \((0,1/3]\) has mass \(3(1/3)=1\), and the part \((1/3,2/3]\) has mass \(1(2/3-1/3)=1/3\). Therefore \(\nu(B)=1+1/3=4/3\). It follows that \(\lambda(A\times B)=\mu(A)\nu(B)=(3/4)(4/3)=1\). Check: the same rectangle value computed in two stages is \(\lambda_X(A)\nu(B)=(5/3)(3/4)(4/3)=5/3\), which is not the rectangle measure: it incorrectly uses a first-marginal strip factor and then multiplies by \(\nu(B)\). The correct direct rectangle check is \(\mu(A)\nu(B)=(3/4)(4/3)=1\). The strip masses are instead \(\lambda(A\times Y)=(5/3)(3/4)=5/4\) and \(\lambda(X\times B)=(3/2)(4/3)=2\).
Zero Masses and a Common Scaling Error
The theorem does not require the factor measures to have positive total mass. If \(\nu(Y)=0\), then \(\lambda_X=0\), the zero measure on \((X,\mathcal{M})\). Also \(\lambda(X\times Y)=\mu(X)\nu(Y)=0\); since every measurable subset of \(X\times Y\) has measure at most this total mass, \(\lambda\) itself is the zero measure, and consequently \(\lambda_Y=0\). The formula \(\lambda_Y=\mu(X)\nu\) indeed gives zero because \(\nu\) is a nonnegative measure of total mass zero. The corresponding conclusions hold if \(\mu(X)=0\). Thus no division by a total mass is needed, and the formulas remain valid in degenerate cases.
A frequent error is to assume that the first marginal always equals \(\mu\). In the second example, taking \(A=[0,1]\) gives \(\mu(A)=3/2\), but the strip mass is \(\lambda(A\times Y)=9\), not \(3/2\). The missing multiplier is \(\nu(Y)=6\): \(\lambda(A\times Y)=\mu(A)\nu(Y)=(3/2)(6)=9\). Likewise, the second marginal must be multiplied by \(\mu(X)\). Only when the other factor has total mass \(1\), as in a probability measure, does that multiplier disappear.
Marginals are useful because they let us recover the mass seen in one coordinate without needing to analyze arbitrary subsets of the product space. The strip identities are also a direct check on calculations: taking \(A=X\) or \(B=Y\) must return the product's total mass. No interchange of integrals is involved here; the identities follow from the rectangle formula for the product measure.
Check Your Understanding
Use the definitions and the marginal formula to answer the following.
- If \(\mu(X)=5\) and \(\nu(Y)=2\), what are \(\lambda_X\), \(\lambda_Y\), and \(\lambda(X\times Y)\)?
- Suppose \(\mu(A)=3/4\) and \(\nu(Y)=8\). Find \(\lambda(A\times Y)\), and state the additional condition under which it equals \(\mu(A)\).
- If \(\mu(X)=0\), what are the product measure and both marginals? Explain why the formulas apply.
- For the piecewise-density example, compute \(\lambda(A\times Y)\) from the first marginal and compare it with the value in the worked example.
- Why does the first marginal equal \(\mu\) when both factors are probability measures, but not in general?