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For a finite measure on a measurable space whose singletons are measurable sets, the set of point atoms of a measure is at most countable. For each positive integer $m$, let $D_m$ consist of the points with singleton measure at least $1/m$. If $D_m$ contained more than $m\mu (\Omega )$ distinct points, finite additivity on those disjoint singletons would exceed the total measure. Thus each $D_m$ is finite.

Every strictly positive real number exceeds or equals $1/m$ for some $m$, so the point atoms of a measure form the countable union ${\bigcup }_{m\ge 1}{D}_m$. This proves that the set is countable. It also shows that a cumulative distribution function has only countably many jumps. The result does not say the measure is concentrated on those points: a non-atomic measure may assign them total measure zero.

Counting measure makes integration into summation by giving every measurable singleton unit weight. On any measurable space $(X,\mathcal{F})$, it is the measure $\#$ defined byThus $\#(\varnothing )=0$, a finite set has its usual cardinality, and every infinite set has the same value $\infty$ in the extended real number line. In particular, countably infinite sets and uncountable sets are not distinguished by their counting measure. Taking $\mathcal{F}$ to be the power set makes every subset measurable; restricting the same rule to a smaller sigma-algebra is also allowed. ^[ref]

Why is this a countably additive measure? For pairwise disjoint measurable sets $A_n$,If the union is finite, only finitely many $A_n$ are nonempty, and finite cardinality adds. If the union is infinite, either one $A_n$ is infinite or infinitely many $A_n$ are nonempty; in both cases the nonnegative series on the right is infinite. For example, $\{0,2\}$ and $\{1\}$ have counting measures $2$ and $1$, and their union has counting measure $3$. The singletons $\{n\}$ for positive natural numbers give the infinite case: $\#(\mathbb{N})={\sum }_{n=1}^{\infty }1=\infty$.

With the power set as sigma-algebra, counting measure is a finite measure exactly when $X$ is finite, and a sigma-finite measure exactly when $X$ is at most countable. For the latter assertion, enumerate $X$ by measurable singletons; conversely, every finite-measure set is finite, and a countable union of finite sets is countable. Specifying the sigma-algebra matters: if $X=\mathbb{N}$ and $\mathcal{F}=\{\varnothing ,\mathbb{N}\}$, counting measure is not a sigma-finite measure, since the only measurable finite-measure set is the empty set.

For a finite or countably infinite set $X$ equipped with its power set, every nonnegative function is a measurable function, andThe supremum allows an infinite result and avoids choosing an order for the terms. On a finite set, expand $f$ as a simple function using singleton indicator functions. On a countably infinite set, enumerate the points and apply the monotone convergence theorem to finite truncations. Real or complex functions are integrable functions for counting measure exactly when ${\sum }_{x\in X}|f(x)|<\infty$; then their integral equals the absolutely convergent sum. This is the discrete counterpart of Lebesgue integration, developed further in series as a Lebesgue integral against counting measure. ^[ref]

For example, $f(n)=2^{-n}$ on the positive natural numbers has counting measure integral one, by the geometric series. The alternating harmonic series has a finite ordered sum but is not the integral of an integrable function for counting measure: its positive and negative parts both have infinite sums. Absolute convergence is what makes the finite integral independent of enumeration. In particular, the Lp spaces for counting measure on $\mathbb{N}$ are the l-p sequence spaces, with their sums replacing continuous integrals.

Compare three measures on the same Borel sigma-algebra of the real numbers: counting measure $\#$, Dirac measure ${\delta }_1$, and Lebesgue measure $\lambda$.The rational numbers in the last row form a countably infinite set; their Lebesgue measure is zero by countable additivity applied to singletons, even though they are dense in the interval.

These are different answers to different questions: counting measure asks how many points occur, Dirac measure asks whether the selected point occurs, and Lebesgue measure asks how much length occurs. A null set therefore depends on its measure. For counting measure, the only null set is the empty set, so equality almost everywhere is equality everywhere. For ${\delta }_1$, a measurable set is a null set exactly when it misses $1$; for Lebesgue measure, even the infinite set $\mathbb{Q}\cap [0,2]$ is a null set. Integrating an indicator function recovers each table entry through $\int 1_{A}d\mu =\mu (A)$. This lets the same Lebesgue integral framework handle point counts, point evaluation, and geometric length.

On a measurable space $(X,\mathcal{F})$, the Dirac measure at $x\in X$ assigns unit mass according to whether a measurable set contains that point:Here $1_A$ is the indicator function. This is a probability measure, since ${\delta }_x(X)=1$. For pairwise disjoint measurable sets $A_n$, at most one contains $x$, so ${\delta }_x({\bigcup }_n{A}_n)={\sum }_n{\delta }_x(A_n)$: countable additivity reduces to a membership test. The definition works even when the singleton $\{x\}$ is not a measurable set. ^[ref]

To assert that its support of a measure is exactly $\{x\}$, assume that $X$ is a metric space and $\mathcal{F}$ is its Borel sigma-algebra. The singleton $\{x\}$ is then a closed set and a Borel set, with ${\delta }_x(\{x\})=1$. Every open neighbourhood of $x$ has mass one; every other point $y$ has an open ball of radius less than its distance from $x$, with mass zero. Thus $\mathrm{supp}{\delta }_x=\{x\}$. An abstract measurable space has no specified topology, so this topological conclusion is not part of the bare definition. Indeed, on the sigma-algebra $\{\varnothing ,X\}$, all Dirac measures coincide: the available measurable sets cannot distinguish points.

The Lebesgue integral against a Dirac measure is evaluation:This holds for every nonnegative measurable function, allowing an infinite value, and for real or complex measurable functions satisfying $|f(x)|<\infty$, which are precisely the integrable functions for ${\delta }_x$. For a nonnegative simple function, only the cell containing $x$ contributes to its integral; increasing simple functions and the monotone convergence theorem extend the identity. Splitting into positive and negative parts, and then real and imaginary parts, gives the remaining cases. This is why a Dirac measure turns integration into a point query.

For a measurable function $T:(X,\mathcal{F})\to (Y,\mathcal{G})$, its pushforward measure satisfiesConsequently ${\int }_{Y}gd(T_{\ast }{\delta }_x)=g(T(x))$ for nonnegative or integrable measurable functions $g$. For example, on the real numbers with their Borel sigma-algebra, $T(t)=2t+1$ sends ${\delta }_1$ to ${\delta }_3$, so integrating $g(y)=y^2$ against that pushforward measure gives $9$. ^[ref]

Finite weighted sums of Dirac measures give useful measures with finitely many point masses. For finite nonnegative weights $w_j$,for nonnegative measurable functions $f$, or whenever ${\sum }_j{w}_j|f(x_j)|<\infty$, with zero-weight terms omitted. Repeated locations have their weights added. On the real numbers, this is the finite version of the measure underlying the left-continuous cumulative function of an atomic measure; a positive-weight singleton is an atom of a measure. Unit weights at the distinct points of a finite set recover counting measure on that set. ^[ref]

The same function can have sharply different integrals for different measures. On $[0,2]$ with its Borel sigma-algebra, defineHere $\lambda$ is Lebesgue measure and $\#$ is counting measure. The singleton $\{1\}$ is a Lebesgue null set, so its changed value leaves the Lebesgue integral equal to ${\int }_0^2(t+1)dt=4$. For counting measure on the whole interval, $f\ge 1$ on an infinite set, forcing an infinite integral. On the smaller measure space $D=\{0,1,2\}$ with its power set and counting measure, the restricted function instead has integral $1+5+3=9$.

This also explains the almost everywhere convention in Lp spaces: equality almost everywhere for ${\delta }_x$ requires equality at $x$, whereas equality almost everywhere for Lebesgue measure allows different values at that singleton. A point value therefore cannot be read unambiguously from an equivalence class in an Lp space formed using Lebesgue measure.

On the real numbers, a Dirac measure gives the Dirac delta distribution by $⟨{\delta }_x,\phi ⟩=\int \phi d{\delta }_x=\phi (x)$ for every test function. The name Dirac delta function does not mean an ordinary function that is zero away from $x$ and infinitely high at $x$. Such a nonnegative measurable function still has Lebesgue integral zero, because its nonzero values occur on a Lebesgue null set. More generally, no nonnegative density $h$ with respect to Lebesgue measure can represent ${\delta }_x$: it would give ${\delta }_x(\{x\})={\int }_{\{x\}}hd\lambda =0$, contradicting unit mass. The measure and its associated distribution act on different kinds of inputs, namely measurable sets and test functions, but express the same point evaluation. ^[ref]

On a measurable space $(X,\mathcal{F})$, a measure is a function $\mu :\mathcal{F}\to [0,\infty ]$ with $\mu (\varnothing )=0$ and countable additivity: for pairwise disjoint measurable sets $A_n$,Infinite size is permitted. Nonnegativity implies that if measurable sets satisfy $A\subseteq B$, then $\mu (A)\le \mu (B)$: decompose $B$ into $A$ and $B\setminus A$. For overlapping measurable sets, use countable subadditivity rather than adding their full sizes twice.

For example, counting measure gives $\mu (\{a,b,c\})=3$, while a Dirac measure ${\delta }_a$ gives that same set size $1$ if it contains $a$. Under Lebesgue measure, $[0,1/4]\cup [3/4,1]$ has size $1/4+1/4=1/2$. These are all measures, but they answer different questions: how many objects, whether a distinguished object is present, or how much length is occupied.

A subset $A$ of a measurable space $(X,\Sigma )$ is non-measurable when $A\notin \Sigma$. On the real line, unless another sigma-algebra is specified, this commonly means that $A$ is not a Lebesgue measurable set. By Completeness of Lebesgue measure, such a set cannot be contained in a Lebesgue null set.

For an example, use the axiom of choice to select one representative from each equivalence class of $[0,1]$ under $x\sim y$ if $x-y\in \mathbb{Q}$, and call the resulting set $V$. The translates $V+q$, for $q\in \mathbb{Q}\cap [-1,1]$, are pairwise disjoint, their union contains $[0,1]$, and their union is contained in $[-1,2]$. If $V$ were a Lebesgue measurable set, translation invariance and countable additivity would make that union have measure zero when $m(V)=0$ and infinite measure when $m(V)>0$. Both conclusions contradict these inclusions. Thus $V$ is non-measurable.

A Poisson random measure with intensity the sigma-finite measure $\mu$ is a random nonnegative integer-valued measure $M$ such that, for each outcome outside a single null set, $A\mapsto M(A)$ is countably additive, and for each measurable $A$, $M(A)$ is a measurable random variable. For every measurable $A$ with $\mu (A)<\infty$,and for every finite collection of pairwise disjoint finite-intensity measurable sets, their counts are independent random variables. These are integer-valued counts, with possible value infinity on infinite-intensity sets. In particular $\mu (A)=0$ implies $M(A)=0$ almost surely. The intensity identity is $\mathbb{E}M(A)=\mu (A)$, also in the extended sense.

For completeness, choose an increasing finite-intensity exhaustion $E_m\uparrow E$. Each $M(E_m)$ is finite almost surely, simultaneously for all $m$, so $M$ is a sigma-finite measure almost surely. If $\mu (A)=\infty$, then $\mu (A\cap E_m)\to \infty$, and for any fixed integer $K$,Hence $M(A)=\infty$ almost surely. Independence for counts on arbitrary disjoint measurable sets follows by this exhaustion; an infinite count is a constant extended value. This convention completes the definition on a general measurable space, not just on bounded subsets of Euclidean space.

On a measurable space $(E,\mathcal{E})$ with a sigma-finite measure $\mu$, a Poisson random measure $N$ is a random counting measure such that $N(A)$ has the Poisson distribution with parameter $\mu (A)$ for every measurable $A$ of finite intensity, and the counts on finitely many pairwise disjoint measurable sets are independent. The sample measures are countably additive; repeated atoms are allowed if the intensity has atoms.

When $\mu (A)=\infty$, exhaustion by finite-intensity subsets gives $N(A)=\infty$ almost surely. Indeed their counts increase, and the Poisson distribution probability of a count at most any fixed integer tends to zero as the parameter tends to infinity. Independent Poisson counts on disjoint sets, with means prescribed by $\mu$, characterize the Poisson random measure.

The Radon-Nikodym theorem for positive measures says: if $\mu$ and $\nu$ are sigma-finite measures on the same measurable space, and $\nu$ is absolutely continuous with respect to $\mu$, then there is a nonnegative measurable function $g$, unique $\mu$-almost everywhere, such thatHere sigma-finiteness means that the space is a countable union of measurable sets of finite measure; absolute continuity of measures, written $\nu \ll \mu$, means that $\mu (E)=0$ implies $\nu (E)=0$. The function $g=d\nu /d\mu$ is the Radon-Nikodym derivative. For a finite signed or complex measure $\nu$ of finite total variation norm of a measure, absolutely continuous with respect to a sigma-finite $\mu$, the corresponding density belongs to $L^1(\mu )$. This follows by applying the positive theorem to the positive and negative parts of the real and imaginary parts of $\nu$.

For every measure space and $1<p<\infty$, $(L^p{)}^{\ast }$ is isometrically $L^q$, where $1/p+1/q=1$. We use the complex-linear pairingWith the convention $\int h\overline{g}$, the same identification is conjugate-linear in $g$. By Hölder's inequality, ${\Phi }_g$ is a bounded linear functional with $\Vert {\Phi }_g\Vert \le \Vert g{\Vert }_q$. If $g\ne 0$, setinterpreting the numerator as zero where $g=0$. The identity $(q-1)p=q$ gives $\Vert h{\Vert }_p=1$ and ${\Phi }_g(h)=\Vert g{\Vert }_q$. ThusIt remains to represent an arbitrary $\Phi \in (L^p{)}^{\ast }$, rather than merely produce functionals from $L^q$.

We prove Lp duality on an arbitrary measure space without imposing sigma-finiteness on $\mu$. Write $C=\Vert \Phi \Vert$. For every measurable set $E$ with $\mu (E)<\infty$, define a complex measure on $E$ byIt is countably additive: for disjoint $F_j\subseteq E$, the partial sums of their indicator functions tend in $L^p$ to $1_{\bigcup F_j}$, because the measure of the omitted tail tends to zero. It is absolutely continuous with respect to $\mu {|}_E$, since indicator functions of null sets represent zero in $L^p$.

Its total variation norm of a measure is finite. For any finite measurable partition $E={\bigcup }_j{E}_j$, choose scalars $c_j$ of modulus $1$ with $c_j{\nu }_E(E_j)=|{\nu }_E(E_j)|$. ThenTaking the supremum over partitions gives the variation bound. Since $\mu {|}_E$ is finite, the Radon-Nikodym theorem supplies $g_E\in L^1(E)$ with ${\nu }_E(F)={\int }_F{g}_{E}d\mu$. By uniform approximation with simple functions,for every bounded measurable $h$ supported in $E$.

To improve $g_E$ from $L^1$ to $L^q$, test with the bounded functionIf $I_N={\int }_{E\cap \{|g_E|\le N\}}|g_E{|}^{q}d\mu$, thenThus $I_N^{1/q}\le C$ when $I_N>0$, and the same bound is trivial when it is zero. The monotone convergence theorem givesIf $E,F$ both have finite measure, the densities $g_E,g_F$ agree almost everywhere on $E\cap F$: their integrals over every measurable subset of the intersection equal the same functional value. This is uniqueness in the Radon-Nikodym theorem.

We now perform support localization of an Lp functional. SetChoose finite-measure sets $E_n$ whose displayed integrals tend to $M$, and let $D_n=E_1\cup \cdots \cup E_n$, $S={\bigcup }_n{D}_n$. If $M=0$, take $E_n=\varnothing$. Compatibility allows us to define a measurable function $g$ on $S$ by taking $g_{D_n}$ on the disjoint measurable sets $D_n\setminus D_{n-1}$, and put $g=0$ off $S$. It agrees almost everywhere with $g_{D_n}$ on every $D_n$. Moreover,Indeed each integral is at most $M$, and it is at least the integral over $E_n$, which tends to $M$.

For any finite-measure set $F\subseteq \Omega \setminus S$, compatibility on the disjoint union $D_n\cup F$ givesLetting $n\to \infty$ forces $g_F=0$ almost everywhere. For an arbitrary finite-measure set $E$, compatibility on $E\cap D_n$ and the preceding conclusion on $E\setminus S$ show that $g_E=g$ almost everywhere on $E$. Consequently $\Phi (h)=\int hgd\mu$ for every simple function supported on a finite-measure set.

Those simple functions are dense in $L^p$ even for this arbitrary measure space. To see the needed finite-support property, for $h\in L^p$ the sets $\{|h|>1/n\}$ have finite measure, bounded by $n^p\Vert h{\Vert }_p^p$. First truncate $h$ to such sets and to bounded values, then approximate by simple functions; the discarded $L^p$ integral tends to zero. Continuity of $\Phi$ and Hölder's inequality therefore extend the representation to every $h\in L^p$.

We have constructed $g\in L^q$ with $\Phi ={\Phi }_g$, and the previously proved norm identity gives $\Vert g{\Vert }_q=\Vert \Phi \Vert$. It also proves uniqueness: if ${\Phi }_g={\Phi }_{\tilde{g}}$, then $\Vert g-\tilde{g}{\Vert }_q=0$. This completes the isometric duality of Lp spaces.

The dominated sequence actually converges to zero in norm, and hence weakly. The assumptions give $|f_n{|}^p\le |f{|}^p\in L^1$ and $|f_n{|}^p\to 0$ almost everywhere. The dominated convergence theorem yieldsFor every $\Phi \in (L^p{)}^{\ast }$, $|\Phi (f_n)|\le \Vert \Phi \Vert \Vert f_n{\Vert }_p\to 0$. Therefore

On the same measurable space, absolute continuity of measures meansThe Radon-Nikodym theorem then supplies a nonnegative measurable density $Z$ with ${\mathbb{E}}_{\mathbb{P}}Z=1$ and $\tilde{\mathbb{P}}(A)={\mathbb{E}}_{\mathbb{P}}[Z{1}_A]$. The converse follows directly from this formula. Strict positivity of the Radon-Nikodym derivative gives equivalence of the two probability measures; absolute continuity of measures alone need not give equivalence.

A point process records random locations by a random measure that counts them. Examples include arrival times of a queue, positions of stars, and the finite eigenvalue configuration of a random matrix. The count in an observation window is a random variable; its expected value is supplied by the intensity measure of a point process, while its fluctuations and relationships with other windows require more information.

For a precise spatial convention, take $E={\mathbb{R}}^d$ with its Borel sigma-algebra and a probability space $(\Omega ,\mathcal{F},\mathbb{P})$. A locally finite point process $\Pi$ is a measurable function from $\Omega$ to the space ${\mathcal{N}}_{\mathrm{lf}}(E)$ of locally finite point measures, equipped with the sigma-algebra generated by evaluations $\nu \mapsto \nu (A)$ for Borel sets $A$. Each realization has the formwhere $J(\omega )$ is finite or countable and repeated locations are allowed; ${\delta }_x$ is a Dirac measure. Here a point measure is a counting measure with possible multiplicities, rather than the counting measure of a set of distinct locations.^[ref]

Local finiteness means ${\Pi }_{\omega }(B)<\infty$ for every bounded Borel set $B$, on a single almost sure event. In ${\mathbb{R}}^d$ it suffices to check the countable collection of balls of integer radius centered at the origin, and this is equivalent to finiteness on every compact set. On a general measurable space there is no notion of boundedness without additional structure: state a metric or an explicit family of observation windows. Another convention on locally compact spaces requires finiteness on compact sets; this need not coincide with finiteness on all bounded sets outside ${\mathbb{R}}^d$. Local finiteness does not require finitely many points in all of $E$.^[ref]

A random element is a measurable function from a probability space $(\Omega ,\mathcal{F},\mathbb{P})$ to a measurable space $(E,\mathcal{E})$. Its law is the pushforward measure ${\mathbb{P}}_X(A)=\mathbb{P}(X^{-1}(A))$ for $A\in \mathcal{E}$. Real-valued random variables and vector-valued random variables are examples; a stochastic sample path can also be a random element when its path-space sigma-algebra is specified. For a Polish space $E$, the usual choice is its Borel sigma-algebra. A large deviation principle concerns the laws of a family of such random elements.

A standard Borel space is a measurable space isomorphic to a Borel set of a Polish space, with its Borel sigma-algebra. This measurable regularity permits the usual constructions of random samples and random counting measures, without requiring a particular compatible topology.

Total variation distance asks how much two probability distributions can disagree about the same event. For probability measures $P,Q$ on one measurable space $(S,\mathcal{F})$, its definition isThe supremum ranges over all measurable sets, so $d_{\mathrm{TV}}(P,Q)\le \epsilon$ guarantees that every yes-or-no prediction has probability error at most $\epsilon$. This metric takes values between zero and one; it is zero exactly when the two probability measures agree. This makes it useful for comparing a sampling algorithm's output with its target probability distribution, and for defining the mixing time of a Markov chain.

On a finite set, add the excess probability mass wherever $P$ exceeds $Q$. That excess is also half the sum of all absolute differences:The last expression reveals the common probability mass that the two probability distributions can share.^[ref] The coupling characterization of total variation distance turns this overlap into an actual joint construction; minimum sum of errors in a simple hypothesis test gives it a statistical interpretation. One quantity connects probability bookkeeping, comparison of sampling methods, and the limits of a statistical test.