Boundary Atoms and Open or Closed Regions

Why Boundary Inclusion Matters

The regions \(X\leq a\) and \(X<a\) differ only when \(X=a\). If \(X\) has positive probability at \(a\), the two probabilities differ; if it has no atom there, they agree. The same principle applies to joint regions, but their boundaries can contain entire faces, isolated corner points, or even curves carrying probability.

The key question is not simply whether a boundary looks small in the plane. It is whether the joint distribution assigns probability to the part of the boundary whose inclusion changes. Marginal atoms identify possible mass on coordinate-aligned faces, while joint point masses identify mass at corners. For other boundaries, the joint law itself matters.

Definition: A random variable \(X\) has an atom at \(a\) if \(P(X=a)>0\). For a pair \((X,Y)\), a joint point mass at \((a,c)\) is \(P(X=a,Y=c)>0\). Changing a region from open to closed changes its probability only through probability assigned to the points whose inclusion changes.

Marginal Atoms and CDF Jumps

Let \(F_X(t)=P(X\leq t)\) be the marginal CDF of \(X\). Its left limit at \(a\) is \(F_X(a-)=\lim_{t\uparrow a}F_X(t)\). This left limit equals \(P(X<a)\): for any sequence \(t_n\) increasing to \(a\), the events \(\{X\leq t_n\}\) increase to \(\{X<a\}\), so continuity of probability for increasing events gives the result. Consequently,

$$ P(X=a)=F_X(a)-F_X(a-). $$

Thus the jump size of the marginal CDF at \(a\) is exactly the probability on the vertical line \(X=a\), with no restriction on \(Y\). It provides an upper bound for the probability on any part of that line. More precisely, for any measurable set \(B\),

0\leq P(X=a,\;Y\in B)\leq P(X=a).

The bound can be strict: an atom in \(X\) does not say how its probability is distributed across values of \(Y\). Likewise, an atom of \(Y\) bounds the probability on a horizontal line but does not determine how much of that line lies within a particular region.

Theorem: For real numbers \(a<b\) and \(c<d\), the difference between the probability of the closed rectangle \([a,b]\times[c,d]\) and the probability of the open rectangle \((a,b)\times(c,d)\) is

\begin{aligned} &P\bigl((X,Y)\in[a,b]\times[c,d]\bigr) -P\bigl((X,Y)\in(a,b)\times(c,d)\bigr)\\ &=P(X=a,\ c\leq Y\leq d) +P(X=b,\ c\leq Y\leq d)\\ &\quad +P(a<X<b,\ Y=c) +P(a<X<b,\ Y=d). \end{aligned}

In particular, the two rectangle probabilities are equal if and only if each of the four terms on the right is zero. The formula includes corner points: each corner belongs to one of the two vertical-face terms, while the horizontal-face terms use strict inequalities in \(X\).

Proof. The open rectangle is contained in the closed rectangle. Their difference consists of points in the closed rectangle that are not in the open rectangle. Partition that difference into four disjoint sets: the left face \(X=a,\ c\leq Y\leq d\); the right face \(X=b,\ c\leq Y\leq d\); the bottom face \(a<X<b,\ Y=c\); and the top face \(a<X<b,\ Y=d\). These sets are disjoint because the horizontal faces exclude the vertical endpoints, while the vertical faces include their corner points. Additivity of probability over this disjoint partition gives the formula. Each term is nonnegative, so the difference is zero exactly when all four terms are zero. \(\square\)

This decomposition also explains why adding the probabilities of all four complete faces would be wrong: each corner would be counted twice. Assigning the corners to the vertical faces yields a disjoint partition and a correct sum.

Worked Examples With Rectangle Boundaries

Worked Example: An Atom on One Vertical Face

Suppose that with probability \(0.3\), \(X=0\) and \(Y\) is uniform on \([0,1]\). With probability \(0.7\), \((X,Y)\) is uniform on \([1,2]\times[0,1]\). Compare the probabilities of \([0,1]\times[0,0.5]\) and \((0,1)\times(0,0.5)\).

In the first mixture component, \(X=0\). The closed rectangle includes these points when \(0\leq Y\leq0.5\), which has conditional probability \(0.5\). Their unconditional probability is \(0.3(0.5)=0.15\). The open rectangle excludes them because it requires \(X>0\).

In the second component, \(X\) is uniform on \([1,2]\). It has probability zero of equalling the endpoint \(1\), and it cannot be less than \(1\). Thus neither rectangle receives probability from that component. Therefore the closed-rectangle probability is \(0.15\), the open-rectangle probability is \(0\), and their difference is \(0.15\).

Here \(P(X=0)=0.3\), but the relevant part of that vertical face has probability only \(0.15\), because the rectangle restricts \(Y\) to half of its uniform range. The marginal atom bounds the face probability; it does not supply the face probability by itself.

Worked Example: Corner Masses in a Finite Joint Distribution

Let \((X,Y)\) have the following joint point masses: \(P(X=0,Y=0)=0.10\), \(P(X=0,Y=1)=0.20\), \(P(X=1,Y=0)=0.15\), \(P(X=1,Y=1)=0.25\), and \(P(X=0.5,Y=0.5)=0.30\). The masses sum to \(0.10+0.20+0.15+0.25+0.30=1\). Compare the probabilities of \([0,1]^2\) and \((0,1)^2\).

All five points lie in \([0,1]^2\), so its probability is \(1\). The open square \((0,1)^2\) contains only \((0.5,0.5)\), so its probability is \(0.30\). The difference is \(1-0.30=0.70\).

The disjoint boundary decomposition gives the same answer. The left face \(X=0,\ 0\leq Y\leq1\) has mass \(0.10+0.20=0.30\). The right face \(X=1,\ 0\leq Y\leq1\) has mass \(0.15+0.25=0.40\). The bottom face with \(0<X<1,\ Y=0\) and the top face with \(0<X<1,\ Y=1\) each have mass zero. The total boundary mass is \(0.30+0.40=0.70\).

The corner points are included in the vertical-face probabilities. Counting complete vertical and horizontal faces without removing their endpoints would count those joint point masses twice.

When Boundary Probability Must Be Zero

For an absolutely continuous joint distribution with density \(f_{X,Y}\), any measurable region \(A\) has probability \(P((X,Y)\in A)=\int_A f_{X,Y}(x,y)\,dx\,dy\). Therefore, if the boundary being added or removed has two-dimensional Lebesgue measure zero, its probability is zero. In particular, this applies to the boundary of an ordinary rectangle under an absolutely continuous joint law. This conclusion depends on absolute continuity of the joint law; it does not follow merely from continuous marginal CDFs.

For coordinate-aligned faces, marginal atoms provide a direct test. For example, if \(P(X=a)=0\), then \(P(X=a,\ c\leq Y\leq d)=0\) for every interval \([c,d]\). But a boundary need not be a coordinate face, and the absence of marginal atoms does not rule out probability concentrated along a curve.

Worked Example: Continuous Marginals, Probability on a Boundary

Let \(U\) be uniform on \([0,1]\), and set \(X=U\) and \(Y=U\). Both marginal distributions are continuous: for every real number \(a\), \(P(X=a)=P(Y=a)=0\). Compare the regions \(R_{\leq}=\{(x,y):0\leq x\leq y\leq1\}\) and \(R_{<}=\{(x,y):0\leq x<y\leq1\}\).

Every possible outcome has \(X=Y=U\), so it belongs to \(R_{\leq}\). Hence \(P((X,Y)\in R_{\leq})=1\). No outcome satisfies \(X<Y\), so \(P((X,Y)\in R_{<})=0\). The difference is \(1\), entirely due to probability concentrated on the diagonal boundary \(x=y\).

This example shows why continuous marginal distributions are not a general guarantee that open and closed regions have equal probabilities. The law is concentrated on a diagonal, despite each coordinate having no atoms. For an arbitrary boundary, assess the probability assigned by the joint law, rather than inferring it from marginal continuity alone.

A Reliable Boundary Check

When comparing two regions that differ only in whether their boundaries are included, first identify the points whose membership changes. If one region is contained in the other, the probability difference is the probability of their set difference. For a rectangle, the disjoint-face formula reduces that calculation to probabilities on its four faces. Marginal atoms bound the vertical and horizontal face probabilities, while joint point masses can determine the contributions at corners.

For regions with other boundaries, use the relevant joint representation: a density gives zero probability to a boundary of area zero, while a singular component may put positive mass there. This is why a geometric observation such as “the boundary has zero area” is useful only when the law is absolutely continuous with respect to area.

Common Mistake

Common mistake: “The marginal distributions are continuous, so every boundary has probability zero.” Continuous marginals do rule out atoms on coordinate lines such as \(X=a\) or \(Y=c\). They do not rule out mass on a non-coordinate boundary: if \(X=Y=U\), then both marginals are continuous but \(P(X=Y)=1\). Check the joint probability on the particular boundary being included or excluded.

Key takeaway: Open and closed versions of a region have the same probability exactly when the part of the boundary on which membership changes has probability zero. For rectangles, use marginal atoms to assess faces and joint point masses to track corner contributions without double counting.

Check Your Understanding

Answer each question by identifying the relevant boundary probability.

Why is \(P(X\leq a)-P(X<a)\) equal to \(P(X=a)\), and how is this quantity read from the marginal CDF?
For a closed rectangle compared with its open interior, why should the horizontal-face terms exclude the vertical endpoints?
If \(P(X=a)=0\), what can be concluded about \(P(X=a,\ Y\in B)\) for a measurable set \(B\)?
Can two continuous marginal distributions have positive probability on a diagonal? Explain using an explicit joint construction.
What additional condition lets an area density imply that adding a zero-area boundary does not change probability?