Dirac Measures And Point-Mass Calculations

A Measure Concentrated at One Point

A measure can assign all of its mass to a single point, regardless of the size or shape of the surrounding space. For such a measure, calculating the measure of a set requires only one question: does the set contain that point? This simple rule makes the Dirac measure a useful model for point masses and a basic building block for more complicated measures.

Let $(X,\mathcal{M})$ be any measurable space, where $X$ is a set and $\mathcal{M}$ is a sigma-algebra on $X$. Fix $x\in X$. No assumption that the singleton $\{x\}$ belongs to $\mathcal{M}$ is needed to define the Dirac measure. The membership test $x\in A$ makes sense for every $A\in\mathcal{M}$, whether or not $\{x\}$ itself is measurable.

Definition: The Dirac measure at $x$, denoted by $\delta_x$, is the set function on $\mathcal{M}$ given by $$ \delta_x(A)= \begin{cases} 1,&x\in A,\\ 0,&x\notin A. \end{cases} $$ The measure space $(X,\mathcal{M},\delta_x)$ is finite, with $\delta_x(X)=1$, and therefore is sigma-finite. It need not be complete for an arbitrary choice of $\mathcal{M}$.

The definition depends on the chosen point and on set membership, not on a distance, topology, or coordinate system. In particular, it applies on a finite set, on $\mathbb{R}$ with its Borel sigma-algebra $\mathcal{B}_{\mathbb{R}}$, or on $\mathbb{R}$ with its Lebesgue sigma-algebra $\mathcal{L}$. When the measurable space is understood, $\delta_x$ assigns mass one to every measurable set containing $x$, and mass zero to every measurable set excluding $x$.

Why the Point-Mass Rule Defines a Measure

Proposition: For any measurable space $(X,\mathcal{M})$ and $x\in X$, the set function $\delta_x$ is a measure on $\mathcal{M}$.

Proof. By definition, $\delta_x(A)$ is either $0$ or $1$, so it is nonnegative for every $A\in\mathcal{M}$. Since $x\notin\varnothing$, we have $\delta_x(\varnothing)=0$. To prove countable additivity, let $(A_n)_{n\ge1}$ be a sequence of pairwise disjoint sets in $\mathcal{M}$. If $x\notin\bigcup_{n\ge1}A_n$, then $x\notin A_n$ for every $n$, so $\delta_x(\bigcup_n A_n)=0=\sum_n\delta_x(A_n)$. If $x\in\bigcup_{n\ge1}A_n$, then $x\in A_k$ for some $k$. Pairwise disjointness means there cannot be another index $j\ne k$ with $x\in A_j$. Thus $\delta_x(A_k)=1$, all other $\delta_x(A_n)=0$, and both $\delta_x(\bigcup_n A_n)$ and $\sum_n\delta_x(A_n)$ equal $1$. These cases establish countable additivity, so $\delta_x$ is a measure. $\square$

The proof isolates the key feature of disjoint unions: the chosen point can belong to at most one set in a pairwise disjoint family. For a finite disjoint family $A_1,\ldots,A_N$, the same reasoning gives $\delta_x(\bigcup_{j=1}^N A_j)=\sum_{j=1}^N\delta_x(A_j)$. For a countable disjoint family, the sum also has at most one nonzero term, so it is always either $0$ or $1$.

Proposition: Let $(X,\mathcal{M},\delta_x)$ be a Dirac measure space. For $A\in\mathcal{M}$, $\delta_x(A^c)=1-\delta_x(A)$. For a sequence $(A_n)_{n\ge1}$ in $\mathcal{M}$ that is pairwise disjoint, $$ \delta_x\left(\bigcup_{n\ge1} A_n\right)=\sum_{n\ge1}\delta_x(A_n). $$

Proof. Exactly one of $x\in A$ and $x\in A^c$ holds. If $x\in A$, then $\delta_x(A)=1$ and $\delta_x(A^c)=0=1-\delta_x(A)$. If $x\notin A$, then $x\in A^c$, so $\delta_x(A)=0$ and $\delta_x(A^c)=1=1-\delta_x(A)$. The disjoint-union identity is the countable-additivity property proved above: if the union excludes $x$, both sides are zero; if it contains $x$, exactly one set in the disjoint family contains $x$, and both sides are one. $\square$

These formulas are membership tests, not approximations. They require no calculation of lengths, areas, or limiting sizes of sets. In particular, for any measurable $A$, the answer is determined before any other property of $A$ is considered.

Worked Example: Intervals Containing a Fixed Point

On $(\mathbb{R},\mathcal{B}_{\mathbb{R}},\delta_3)$, calculate the measures of $(-\infty,3)$, $[3,5]$, $(3,5]$, and $(-\infty,3)\cup[3,5]$. All four sets are Borel measurable. Their endpoints are tested using the stated open or closed brackets.

The point defining the measure is $3$. It is excluded from $(-\infty,3)$ because the endpoint $3$ is open, so $\delta_3((-\infty,3))=0$. It is included in $[3,5]$ because the left endpoint is closed, so $\delta_3([3,5])=1$. It is excluded from $(3,5]$ because the left endpoint is open, so $\delta_3((3,5])=0$. The union $(-\infty,3)\cup[3,5]$ contains $3$ through its second set, giving measure $1$.

Check: Substituting the membership answers into the defining rule gives $0,1,0,1$, respectively. The complement rule also checks the first value: the complement of $(-\infty,3)$ is $[3,\infty)$, which contains $3$, so the two complementary measures are $0$ and $1$.

Worked Example: A Countable Partition of the Real Line

On $(\mathbb{R},\mathcal{B}_{\mathbb{R}},\delta_{5/2})$, let $I_k=[k,k+1)$ for each integer $k$. These intervals are pairwise disjoint and their union is $\mathbb{R}$. Calculate the measure of the union of the intervals with even indices and the measure of the union of those with odd indices.

The point $5/2=2.5$ satisfies $2\le 2.5<3$, so it belongs to $I_2=[2,3)$. The index $2$ is even. Thus the union over even integers contains $5/2$ and has $\delta_{5/2}$-measure $1$. The union over odd integers does not contain $5/2$ and has measure $0$. Because the intervals form a disjoint partition, countable additivity gives the same totals by summing the interval measures: in the even-index sum, $\delta_{5/2}(I_2)=1$ and every other term is $0$, so the sum is $1$; in the odd-index sum, every term is $0$, so the sum is $0$.

Check: The two unions are disjoint and together equal $\mathbb{R}$. Their measures add to $1+0=1$, which agrees with $\delta_{5/2}(\mathbb{R})=1$.

Worked Example: A Countable Disjoint Union with One Nonzero Term

On $(\mathbb{R},\mathcal{B}_{\mathbb{R}},\delta_0)$, define $A_1=(-\infty,0)$, $A_2=[0,1)$, and $A_n=[n-2,n-1)$ for integers $n\ge3$. Calculate the measure of $\bigcup_{n\ge1}A_n$ by evaluating each term in the countable-additivity sum.

The sets are pairwise disjoint: $A_1$ contains only negative real numbers, $A_2$ contains numbers from $0$ up to but not including $1$, and for $n\ge3$, $A_n$ begins at $n-2$ and ends just before $n-1$. They cover $\mathbb{R}$: $A_1$ covers the negative half-line, $A_2$ covers $[0,1)$, and the later intervals cover $[1,\infty)$. The point $0$ is excluded from $A_1$, included in $A_2$, and excluded from every $A_n$ with $n\ge3$. Therefore $$ \delta_0(A_1)=0,\qquad \delta_0(A_2)=1,\qquad \delta_0(A_n)=0\quad(n\ge3), $$ and countable additivity gives $\delta_0(\bigcup_{n\ge1}A_n)=0+1+\sum_{n\ge3}0=1$.

Check: The union is $\mathbb{R}$, whose measure is $\delta_0(\mathbb{R})=1$. This agrees with the sum, which has exactly one nonzero term.

Overlapping Unions and a Common Error

For any sequence of measurable sets $(A_n)$, not necessarily disjoint, the defining rule still gives $$ \delta_x\left(\bigcup_{n\ge1}A_n\right)= \begin{cases} 1,&x\in A_n\text{ for at least one }n,\\ 0,&x\notin A_n\text{ for every }n. \end{cases} $$ This is a statement about whether the union contains $x$. It does not say that the measure of a union equals the sum of the measures when the sets overlap. Countable additivity applies to disjoint sets; the sum for overlapping sets may count the same point repeatedly.

Common mistake. On $(\mathbb{R},\mathcal{B}_{\mathbb{R}},\delta_0)$, consider $A=[-1,1]$ and $B=[0,2]$. A wrong calculation is $\delta_0(A\cup B)=\delta_0(A)+\delta_0(B)=1+1=2$. The sets are not disjoint: both contain $0$. Their union is $[-1,2]$, which contains $0$, so the correct calculation is $\delta_0(A\cup B)=1$. The sum $1+1$ is not a countable-additivity calculation because its sets fail the disjointness condition.

A second point to keep in view is the distinction between a point mass and Lebesgue measure. For example, on $(\mathbb{R},\mathcal{L},m)$, a singleton has Lebesgue measure zero, whereas on $(\mathbb{R},\mathcal{L},\delta_x)$, a measurable set containing $x$, including $\{x\}$, has measure one. These are different measures on the same measurable space. The Dirac measure is finite and sigma-finite, but completeness is not automatic: it depends on the sigma-algebra. No completeness assumption is needed for the membership calculations or countable-additivity proof above.

Check Your Understanding

For each question, identify the point defining the Dirac measure and test its membership in the sets, paying attention to strict and non-strict endpoints.

On $(\mathbb{R},\mathcal{B}_{\mathbb{R}},\delta_2)$, calculate the measures of $(-\infty,2]$ and $(2,\infty)$.
Suppose $(A_n)_{n\ge1}$ is a pairwise disjoint sequence in $\mathcal{M}$ and $x\in A_7$. What are $\delta_x(\bigcup_n A_n)$ and $\sum_n\delta_x(A_n)$?
On $(\mathbb{R},\mathcal{B}_{\mathbb{R}},\delta_0)$, calculate the measure of $\bigcup_{n\ge1}[n,n+1)$.
Give an example of two non-disjoint measurable sets $A,B$ for which $\delta_x(A)+\delta_x(B)\ne\delta_x(A\cup B)$, and calculate all three values.
Why does defining $\delta_x(A)$ by membership not require $\{x\}\in\mathcal{M}$?

Dirac Measures And Point-Mass Calculations

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