Sections of Sets and Functions

Fixing One Coordinate at a Time

In a product space, a section is obtained by fixing one coordinate and allowing the other to vary. This simple operation turns a subset of $X\times Y$ into a subset of one factor, and a function on $X\times Y$ into a function on one factor. The coordinate that is fixed matters: the two kinds of sections generally have different domains and different values.

Let $(X,\mathcal{M})$ and $(Y,\mathcal{N})$ be measurable spaces, with product sigma-algebra $\mathcal{M}\otimes\mathcal{N}$. A set $E\subseteq X\times Y$ need not be measurable for its sections to be defined. The definitions below are set-theoretic and require no measures, finiteness, or sigma-finiteness. If measures $\mu$ on $(X,\mathcal{M})$ and $\nu$ on $(Y,\mathcal{N})$ are also given, they do not change the definitions.

Definition: For $E\subseteq X\times Y$, the vertical section at $x\in X$ and the horizontal section at $y\in Y$ are $$ E_x=\{y'\in Y:(x,y')\in E\}\subseteq Y,\qquad E^y=\{x'\in X:(x',y)\in E\}\subseteq X. $$ The subscript indicates that the $X$-coordinate is fixed; the superscript indicates that the $Y$-coordinate is fixed. For a function $f:X\times Y\to S$, where $S$ may be $\mathbb{R}$, $[-\infty,\infty]$, or $\mathbb{C}$, its corresponding sections are $f_x:Y\to S$ and $f^y:X\to S$, defined by $f_x(y')=f(x,y')$ and $f^y(x')=f(x',y)$.

Thus $E_x$ is obtained by intersecting $E$ with the vertical line whose first coordinate is $x$, then recording the second coordinates of the intersection. The set $E^y$ is obtained by fixing the second coordinate instead. These descriptions remain valid when $X$ and $Y$ are abstract sets rather than subsets of the real line.

Sections of Rectangles and Set Operations

For a rectangle $A\times B$, membership separates into two conditions: $(x,y)\in A\times B$ exactly when $x\in A$ and $y\in B$. Fixing either coordinate therefore leaves either all of the opposite side of the rectangle or nothing.

Proposition: For $A\subseteq X$, $B\subseteq Y$, and $E,F\subseteq X\times Y$, $$ (A\times B)_x= \begin{cases} B,&x\in A,\\ \varnothing,&x\notin A, \end{cases} \qquad (A\times B)^y= \begin{cases} A,&y\in B,\\ \varnothing,&y\notin B. \end{cases} $$ Sections also preserve unions, intersections, and relative complements: $$ (E\cup F)_x=E_x\cup F_x,\quad (E\cap F)_x=E_x\cap F_x,\quad (E^c)_x=(E_x)^c, $$ where $E^c$ is the complement in $X\times Y$ and $(E_x)^c$ is the complement in $Y$. The corresponding identities hold for $E^y$, with complements taken in $X$.

Proof. A point $y\in Y$ belongs to $(A\times B)_x$ exactly when $(x,y)\in A\times B$, which holds exactly when $x\in A$ and $y\in B$. If $x\in A$, the allowed $y$'s are all of $B$; if $x\notin A$, there are none. This proves the first rectangle formula. Interchanging the roles of the coordinates proves the second. For the set identities, fix $x\in X$ and $y\in Y$. For example, $z\in(E\cup F)_x$ exactly when $(x,z)\in E\cup F$, which holds exactly when $(x,z)\in E$ or $(x,z)\in F$; this is equivalent to $z\in E_x\cup F_x$. Replacing “or” by “and” proves the intersection identity. Finally, $(x,z)\in E^c$ exactly when $(x,z)\notin E$, which is equivalent to $z\notin E_x$, proving the complement identity. The proofs for horizontal sections use the same membership equivalences with the second coordinate fixed. $\square$

Worked Example: Sections of a Rectangle at Its Boundary

Take $X=Y=\mathbb{R}$, $A=[-1,2]$, $B=(0,3]$, and $E=A\times B$. Compute the vertical sections at $x=-1,2,2.1$, and the horizontal sections at $y=0,3$.

Since $-1\in[-1,2]$, the rectangle formula gives $E_{-1}=B=(0,3]$. Since $2\in[-1,2]$, it also gives $E_2=(0,3]$. The endpoint $2$ is included in $A$. But $2.1\notin[-1,2]$, so $E_{2.1}=\varnothing$. For horizontal sections, $0\notin(0,3]$, hence $E^0=\varnothing$; $3\in(0,3]$, hence $E^3=[-1,2]$.

Check: The point $(2,3)$ belongs to $E$, since both coordinates meet their respective membership conditions, so it appears in both $E_2$ and $E^3$. The point $(2,0)$ does not belong to $E$, because $0\notin B$, which agrees with $E^0=\varnothing$. This verifies the endpoint decisions directly.

Sections of Graphs

Graphs give a useful contrast between the two directions of section. Let $T:D\to Y$ be any map with $D\subseteq X$, and let its graph be $\Gamma_T=\{(x,T(x)):x\in D\}$. Fixing $x$ leaves at most one point of the graph, while fixing $y$ asks for all inputs mapped to $y$.

Proposition: For every $x\in X$ and $y\in Y$, $$ (\Gamma_T)_x= \begin{cases} \{T(x)\},&x\in D,\\ \varnothing,&x\notin D, \end{cases} \qquad (\Gamma_T)^y=\{x\in D:T(x)=y\}. $$ In particular, a horizontal section of a graph is the preimage in $D$ of the singleton $\{y\}$.

Proof. A point $z\in Y$ belongs to $(\Gamma_T)_x$ exactly when $(x,z)\in\Gamma_T$. By the definition of the graph, this requires $x\in D$ and $z=T(x)$. Thus the section is $\{T(x)\}$ when $x\in D$, and is empty otherwise. A point $u\in X$ belongs to $(\Gamma_T)^y$ exactly when $(u,y)\in\Gamma_T$, which holds exactly when $u\in D$ and $T(u)=y$. This proves the second formula. $\square$

Worked Example: Sections of a Parabola

Let $D=[-2,2]$, $Y=\mathbb{R}$, $T(x)=x^2$, and $\Gamma_T=\{(x,x^2):x\in[-2,2]\}$. Compute the vertical section at $x=-1$, and horizontal sections at $y=-1,0,1,4,5$.

The vertical section at $-1$ is the singleton $(\Gamma_T)_{-1}=\{T(-1)\}=\{(-1)^2\}=\{1\}$. For horizontal sections, solve $x^2=y$ subject to $-2\le x\le2$. If $y=-1$, there is no real solution, so $(\Gamma_T)^{-1}=\varnothing$. If $y=0$, the equation has the sole solution $x=0$, so $(\Gamma_T)^0=\{0\}$. If $y=1$, its solutions are $x=-1$ and $x=1$, both in $D$, so $(\Gamma_T)^1=\{-1,1\}$. If $y=4$, the solutions are $x=-2$ and $x=2$, both included endpoints of $D$, so $(\Gamma_T)^4=\{-2,2\}$. If $y=5$, there is no solution in $D$, so $(\Gamma_T)^5=\varnothing$.

Check: The point $(-2,4)$ is on the graph because $(-2)^2=4$, and it is accounted for in $(\Gamma_T)^4$. No point of the graph has second coordinate $5$, since $x\in[-2,2]$ implies $0\le x^2\le4$. This confirms the empty section at $y=5$.

Sections of Functions and Indicators

For a function $f:X\times Y\to S$, the section $f_x$ is the function of the remaining variable $y$, with $x$ held fixed. Likewise, $f^y$ varies $x$ while holding $y$ fixed. These are functions on $Y$ and $X$, respectively; they are not values of $f$ at a single point.

Proposition: If $E\subseteq X\times Y$ and $\chi_E:X\times Y\to\{0,1\}$ is its indicator, then for every $x\in X$, $y\in Y$, and $z\in Y$, $u\in X$, $$ (\chi_E)_x(z)=\chi_{E_x}(z),\qquad (\chi_E)^y(u)=\chi_{E^y}(u). $$ These equalities hold pointwise, whether or not $E$ or its sections are measurable.

Proof. By definition, $(\chi_E)_x(z)=\chi_E(x,z)$, which is $1$ exactly when $(x,z)\in E$. By the definition of $E_x$, this is equivalent to $z\in E_x$, which is exactly when $\chi_{E_x}(z)=1$. If $(x,z)\notin E$, both indicator values are $0$. The horizontal identity follows by the same argument: $(\chi_E)^y(u)=1$ exactly when $(u,y)\in E$, which is equivalent to $u\in E^y$. $\square$

Worked Example: Indicator Sections of a Triangle

Let $X=Y=[0,2]$ and $E=\{(x,y)\in[0,2]^2:y\le x\}$. Compute $E_{3/2}$, $E^1$, and the corresponding sections of $\chi_E$. Test the points $(3/2,1)$ and $(1,3/2)$.

To find $E_{3/2}$, fix $x=3/2$ and impose the defining conditions $y\in[0,2]$ and $y\le3/2$. Thus $E_{3/2}=[0,3/2]$. To find $E^1$, fix $y=1$ and require $x\in[0,2]$ with $1\le x$. Thus $E^1=[1,2]$. The indicator proposition now gives $(\chi_E)_{3/2}=\chi_{[0,3/2]}$ as a function on $[0,2]$, and $(\chi_E)^1=\chi_{[1,2]}$ as a function on $[0,2]$.

At $(3/2,1)$, the condition $1\le3/2$ holds, so $\chi_E(3/2,1)=1$; equivalently $1\in E_{3/2}$ and $3/2\in E^1$. At $(1,3/2)$, the condition $3/2\le1$ fails, so $\chi_E(1,3/2)=0$; correspondingly $3/2\notin E_1$ and $1\notin E^{3/2}$.

Check: The boundary point $(1,1)$ belongs to $E$, because the inequality is non-strict and $1\le1$. It lies in both $E_1$ and $E^1$, which confirms that the endpoints in the computed sections are included.

Worked Example: Sections of a Two-Variable Function

On $X=Y=\mathbb{R}$, define the real-valued function $f(x,y)=x+2y$. Find the two sections through the point $(1,3)$, and verify that they return the original function value there.

Fixing $x=1$ gives the function $f_1:\mathbb{R}\to\mathbb{R}$ with $f_1(y)=f(1,y)=1+2y$. Fixing $y=3$ gives $f^3:\mathbb{R}\to\mathbb{R}$ with $f^3(x)=f(x,3)=x+2(3)=x+6$. Evaluating the first section at $y=3$ gives $f_1(3)=1+2(3)=1+6=7$. Evaluating the second at $x=1$ gives $f^3(1)=1+6=7$. Direct substitution also gives $f(1,3)=1+2(3)=7$.

Check: Both section evaluations agree with the direct value $7$. The domains also match the definition: $f_1$ varies its $Y$-argument, while $f^3$ varies its $X$-argument.

A Coordinate-Swapping Pitfall

A common error is to use the correct equation but assign its solutions to the wrong kind of section. For the graph $y=x^2$, the calculation “set $y=1$, so $x=-1$ or $x=1$” computes the horizontal section $\Gamma^1$, not the vertical section $\Gamma_1$. The correct vertical section fixes $x=1$, giving $\Gamma_1=\{1^2\}=\{1\}$. The correct horizontal section fixes $y=1$ and solves $x^2=1$, giving $\Gamma^1=\{-1,1\}$. In particular, one section is a set of second coordinates and the other is a set of first coordinates.

For measurable sets in a product sigma-algebra, the question of whether their sections are measurable is a separate issue from the set-theoretic definitions and computations here. Likewise, computing a section does not itself justify integrating over that section. Those steps require their own hypotheses and results. Keeping the fixed coordinate, the resulting domain, and the measurability question distinct prevents many errors in later uses of product measures.

Check Your Understanding

For each question, identify which coordinate is fixed before computing the section.

For $E=[0,1]\times(2,4]\subseteq\mathbb{R}^2$, compute $E_0$, $E_1^0$ (that is, $E^0$), and $E^3$, stating why each endpoint is included or excluded.
Let $T:[-1,1]\to\mathbb{R}$ be $T(x)=x^3$. Compute $(\Gamma_T)_0$, $(\Gamma_T)^0$, and $(\Gamma_T)^1$.
For $E=\{(x,y)\in[0,1]^2:x+y\le1\}$, compute $E_{1/4}$ and $E^{1/2}$.
Prove directly from the definitions that $(E\setminus F)_x=E_x\setminus F_x$ for arbitrary $E,F\subseteq X\times Y$.
If $f(x,y)=3x-y$, write $f_2$ and $f^{-1}$ explicitly, specifying the variable in each function’s domain.

What You'll Learn