Finite Sums Of Dirac Measures

From Point Masses to a Measure

A Dirac measure concentrates all its mass at one point. A finite weighted sum of Dirac measures allows several points to carry mass, with each weight specifying how much is placed there. The measure of a set is obtained by adding the weights of the points the set contains. This construction is useful for describing discrete distributions and for checking measure calculations in which repeated points or interval endpoints matter.

Let $(X,\mathcal{M})$ be a measurable space: $X$ is a set and $\mathcal{M}$ is a sigma-algebra of subsets of $X$. For $x\in X$, define the Dirac measure $\delta_x$ on $\mathcal{M}$ by $\delta_x(E)=1$ if $x\in E$ and $\delta_x(E)=0$ if $x\notin E$, for each $E\in\mathcal{M}$. This definition makes sense even when the singleton $\{x\}$ is not in $\mathcal{M}$; in that case it does not assert that $\{x\}$ is measurable.

Definition: Fix a finite list of points $x_1,\ldots,x_N\in X$ and finite weights $a_1,\ldots,a_N\ge 0$. The associated finite weighted sum of Dirac measures is the set function $\mu:\mathcal{M}\to[0,\infty)$ defined by $$ \mu(E)=\sum_{i=1}^{N}a_i\delta_{x_i}(E) =\sum_{i=1}^{N}a_i\chi_E(x_i),\qquad E\in\mathcal{M}. $$ Here $\chi_E(x_i)$ is $1$ when $x_i\in E$ and $0$ otherwise. If $N=0$, the empty sum defines the zero measure.

The list of points need not be distinct. If a point occurs more than once, each occurrence contributes its own weight whenever that point belongs to $E$. Weights equal to zero are allowed and contribute nothing. The underlying set $X$ need not be finite, and the sigma-algebra need not be the power set of $X$.

Why the Construction Satisfies the Measure Axioms

Theorem: For any measurable space $(X,\mathcal{M})$, any finite list $x_1,\ldots,x_N\in X$, and any finite weights $a_i\ge0$, the set function $\mu(E)=\sum_{i=1}^N a_i\delta_{x_i}(E)$ is a measure on $\mathcal{M}$. Its total mass is $\mu(X)=\sum_{i=1}^N a_i$, so it is a finite measure (and therefore sigma-finite). No completeness assumption is required, and the construction does not guarantee completeness.

Proof. If $N=0$, then $\mu(E)=0$ for every $E\in\mathcal{M}$, which satisfies both measure axioms. Suppose now that $N\ge1$. Since no point belongs to the empty set, $\delta_{x_i}(\varnothing)=0$ for every $i$, and therefore $\mu(\varnothing)=\sum_{i=1}^N a_i\cdot0=0$.

Let $(E_k)_{k\ge1}$ be a sequence of pairwise disjoint sets in $\mathcal{M}$, and write $E=\bigcup_{k=1}^\infty E_k$, which belongs to $\mathcal{M}$. For a fixed $i$, if $x_i\notin E$, then $\delta_{x_i}(E)=0$, and $x_i$ belongs to none of the $E_k$. If $x_i\in E$, disjointness means it belongs to exactly one $E_k$. In either case, $$ \delta_{x_i}(E)=\sum_{k=1}^\infty\delta_{x_i}(E_k). $$ Consequently, $$ \mu(E)=\sum_{i=1}^N a_i\sum_{k=1}^\infty\delta_{x_i}(E_k). $$ For each positive integer $K$, the finite sum over $k=1,\ldots,K$ can be reordered with the finite sum over $i=1,\ldots,N$: $$ \sum_{i=1}^N a_i\sum_{k=1}^K\delta_{x_i}(E_k) =\sum_{k=1}^K\sum_{i=1}^N a_i\delta_{x_i}(E_k) =\sum_{k=1}^K\mu(E_k). $$ Let $K$ increase to infinity. Each partial sum is nondecreasing, and its limit is the corresponding nonnegative series; because the sum over $i$ has only finitely many terms, its limit is the sum of the individual limits, with $+\infty$ permitted for a series. Thus $\mu(E)=\sum_{k=1}^\infty\mu(E_k)$. This proves countable additivity. Finally, every $x_i$ belongs to $X$, so $\delta_{x_i}(X)=1$ and $\mu(X)=\sum_{i=1}^N a_i<\infty$. $\square$

Proposition: If several listed points are equal, their weights can be combined without changing the measure. In particular, for $x\in X$, if $\{x\}\in\mathcal{M}$, then $\mu(\{x\})$ is the sum of the weights attached to occurrences of $x$.

Proof. Group the indices $1,\ldots,N$ according to their point values. For any $E\in\mathcal{M}$, every index in the group for a point $y$ has $\delta_y(E)=\chi_E(y)$. The contribution of that group is therefore the sum of its weights multiplied by the single value $\chi_E(y)$. Replacing the repeated occurrences of $y$ by one occurrence whose weight is their sum leaves $\mu(E)$ unchanged for every $E$. If $\{x\}$ is measurable, then $\delta_{x_i}(\{x\})=1$ exactly when $x_i=x$; all other terms are zero. The stated formula for $\mu(\{x\})$ follows. $\square$

Worked Set Calculations

Worked Example: Repeated Locations on the Real Line

On $(\mathbb{R},\mathcal{B}_{\mathbb{R}})$, let $\mu=2\delta_0+3\delta_2+\delta_2$. Compute $\mu([0,2))$, $\mu([0,2])$, and $\mu(\mathbb{R})$.

The list has one point at $0$ with weight $2$, and two occurrences of $2$ with weights $3$ and $1$. The interval $[0,2)$ contains $0$ but not $2$, so $$ \mu([0,2))=2\cdot1+3\cdot0+1\cdot0=2. $$ The interval $[0,2]$ contains both locations, including its right endpoint $2$, so $$ \mu([0,2])=2\cdot1+3\cdot1+1\cdot1=6. $$ The whole space contains all three listed occurrences, giving $$ \mu(\mathbb{R})=2\cdot1+3\cdot1+1\cdot1=6. $$ Check: combining the two weights at $2$ gives $3+1=4$. Thus the measure has mass $2$ at $0$ and $4$ at $2$, which again gives $\mu([0,2))=2$, $\mu([0,2])=2+4=6$, and total mass $2+4=6$.

Worked Example: An Open Endpoint Excludes Its Atom

On $(\mathbb{R},\mathcal{B}_{\mathbb{R}})$, take $$ \mu=2\delta_{-1}+\frac12\delta_1+3\delta_{5/2}. $$ Find the measures of $(-1,5/2)$ and $(-1,5/2]$.

The open interval $(-1,5/2)$ excludes both endpoints: it contains neither $-1$ nor $5/2$, but it does contain $1$. Hence $$ \mu((-1,5/2))=2\cdot0+\frac12\cdot1+3\cdot0=\frac12. $$ The interval $(-1,5/2]$ still excludes $-1$ and contains $1$, but now includes $5/2$. Therefore $$ \mu((-1,5/2])=2\cdot0+\frac12\cdot1+3\cdot1=\frac72. $$ Check: changing the right endpoint from open to closed adds exactly the weight $3$ at $5/2$, and $\frac12+3=\frac72$. A common incorrect calculation assigns $\frac72$ to the open interval by counting the atom at $5/2$; that calculation uses the membership rule for $(-1,5/2]$, not for $(-1,5/2)$. The endpoint notation determines whether that atom is included.

Worked Example: A Finite Space with a Zero Weight

Let $X=\{a,b,c,d\}$, $\mathcal{M}=\mathcal{P}(X)$, and $$ \mu=2\delta_a+0\delta_b+\frac13\delta_c+\frac23\delta_c+4\delta_d. $$ Compute $\mu(\{a,c\})$, $\mu(\{b,d\})$, and the total mass.

The set $\{a,c\}$ includes $a$ and both occurrences of $c$, but excludes $b$ and $d$. Thus $$ \mu(\{a,c\})=2\cdot1+0\cdot0+\frac13\cdot1+\frac23\cdot1+4\cdot0 =2+\frac13+\frac23=3. $$ The set $\{b,d\}$ includes $b$ and $d$, but excludes $a$ and $c$. Its measure is $$ \mu(\{b,d\})=2\cdot0+0\cdot1+\frac13\cdot0+\frac23\cdot0+4\cdot1=4. $$ The whole space contains every listed location, so $$ \mu(X)=2+0+\frac13+\frac23+4=2+0+1+4=7. $$ Check: combining the two weights at $c$ gives $\frac13+\frac23=1$, and the zero weight at $b$ contributes nothing. The resulting masses at $a,c,d$ are $2,1,4$, respectively; their sum is $2+1+4=7$. The set $\{a,c\}$ has mass $2+1=3$, while $\{b,d\}$ has mass $0+4=4$.

Reading the Formula Reliably

For a finite weighted sum, there is no need to determine whether a set contains “most” of the space or to approximate its measure. Check membership point by point. Each occurrence contributes its weight if and only if its point belongs to the set; otherwise it contributes zero. This rule also explains why repeated locations must either be counted occurrence by occurrence or first combined into one weight, but not both.

On $(\mathbb{R},\mathcal{B}_{\mathbb{R}})$, intervals are Borel sets, so the construction applies directly to them. When computing their measure, endpoint brackets are part of the membership test: a square bracket includes that endpoint and a parenthesis excludes it. For a general measurable space, the measure of a singleton can be discussed as $\mu(\{x\})$ only when $\{x\}\in\mathcal{M}$. The defining formula for $\mu(E)$, however, applies to every $E\in\mathcal{M}$ without requiring that each listed singleton be measurable.

The construction also gives a direct check on whether proposed weights could define a finite measure: the weights must be nonnegative, and their finite sum is the total mass. A negative weight can destroy nonnegativity, which is required of a measure; an infinite weight is excluded here because the stated construction is a finite measure. These conditions are about the weights, not about the number of points in $X$.

Check Your Understanding

For each question, identify exactly which listed points belong to the set before adding their weights.

On $(\mathbb{R},\mathcal{B}_{\mathbb{R}})$, let $\mu=4\delta_{-2}+2\delta_0+\delta_3$. Compute $\mu([-2,3))$ and $\mu([-2,3])$.
Suppose $\mu=5\delta_x+2\delta_x+\delta_y$, where $x\ne y$. What are the combined masses at $x$ and $y$, and what is $\mu(X)$?
For a finite weighted sum of Dirac measures, prove directly from the definition that $\mu(\varnothing)=0$.
Why does a zero weight attached to a point make no contribution to the measure of any measurable set?
If $\{x\}$ is measurable, express $\mu(\{x\})$ using the weights attached to $x$. Why is the measurability condition relevant to this expression?

What You'll Learn