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For a probability measure on the real line with characteristic function $\phi$, averaging $1-\cos (tx)$ over $[-h,h]$ gives $1-\sin (hx)/(hx)$. This is at least $1/2$ when $|x|\ge 2/h$. Integrating gives the displayed bound. Consequently pointwise convergence of characteristic functions to a function continuous at zero implies tightness of probability measures, by the dominated convergence theorem on $[-h,h]$ and a separate tail bound for the finitely many initial measures.

Maximize the integral of a continuous strictly convex function among probability measures with a fixed barycenter. The supporting measure lemma for affine upper envelopes implies that the maximizing measure has zero integral of the nonnegative envelope gap. Strict convexity makes this gap positive at every nonextreme point, so the measure is concentrated on the extreme boundary. Metrizability makes that boundary Borel and supplies the continuous strictly convex function.

A weakly closed uniformly tight family of Borel probability measures on a Polish space is weakly compact. Embed the underlying space into a compact metric closure, take subsequential pushforward limits, and use the closed-set Portmanteau theorem on the common tightness compact sets. The limit retains full mass in the original space; inverse continuity of the probability pushforward embedding theorem then returns the weak limit.

A compact group is a topological group whose underlying topological space is a compact Hausdorff space. The circle group is a compact group: the unit circle is a closed bounded subset of the complex plane, so the Heine-Borel theorem gives compactness. The additive group $(\mathbb{R},+)$ is a locally compact group and a Hausdorff space, but is not a compact group: the open cover $\{(-n,n):n\ge 1\}$ has no finite subcover. A finite group with the discrete topology is another compact group; compactness does not require a smooth manifold structure.

The great advantage of a compact group is that one can average over every element without choosing a preferred origin. Precisely, every locally compact group that is a Hausdorff space admits a nonzero left Haar measure $\mu$, a regular Borel measure finite on compact sets, unique up to a positive scalar. Left invariance means $\mu (gA)=\mu (A)$ for every Borel set $A$ and every $g\in G$. There is also a right Haar measure; in general left and right Haar measures can differ. On a compact group, $0<\mu (G)<\infty$, and normalizing to $\mu (G)=1$ gives a unique regular Borel probability measure invariant under both left and right translation. See Tornier, Haar Measures, Theorem 2.2 for existence and uniqueness, and Ram's notes on left and right Haar measure for the distinction.

For a finite group $G$, normalized Haar measure assigns each element mass $1/|G|$, so integration is simply $|G{|}^{-1}{\sum }_{g\in G}f(g)$. For the circle group, the corresponding integral is uniform angle averaging, computed below. On $(\mathbb{R},+)$, Lebesgue measure is a Haar measure, but its total mass is infinite: no positive scalar multiple makes it a probability measure. Thus the existence of Haar measure does not make every topological group a compact group.

Historically, Alfréd Haar introduced the general invariant-measure construction in Der Massbegriff in der Theorie der kontinuierlichen Gruppen, Annals of Mathematics 34 (1933), pp. 147-169. This bibliographic identification is also recorded in André Weil's archived seminar notes, Mesure de Haar, reference Haa33. The conceptual step is to let group symmetry itself determine the weights used in integration, rather than relying on coordinates in an ambient Euclidean space.

The logarithm is strictly concave on positive real numbers, as its second derivative is negative. Thus the Jensen inequality gives $\int \log fd\mu \le \log \int fd\mu$ for a probability measure and an integrable positive $f$, whenever the left side is defined. It also controls the negative logarithmic part of convex combinations of strictly positive functions.

Equip the Boolean hypercube with uniform probability measure and graph distance divided by its graph diameter. These spaces form a Lévy family of graphs. The vertex-isoperimetric inequality in the discrete cube reduces the complement of a positive-density set's radius-$\epsilon n$ graph neighbourhood to a binomial distribution tail. For fixed density $\delta >0$, Chebyshev's inequality puts a lower density quantile at least $n/2-C_{\delta }\sqrt{n}-O(1)$. After enlarging by $\epsilon n$, the omitted mass is at most $n/[4(\epsilon n-C_{\delta }\sqrt{n}-O(1){)}^2]$, which tends to zero.

Equip the symmetric group $S_n$ with its uniform probability measure and unnormalized Hamming distance. If $f$ has Lipschitz constant $K$, it obeys the displayed two-sided bound, and each one-sided tail has the same exponential without the factor two. Reveal a uniform permutation one image at a time. Two possible next images have completion sets paired by swapping those two values, changing exactly two coordinates. Thus the corresponding conditional means differ by at most $2K$. Each increment of the Doob exposure martingale has conditional range length at most $2K$. Applying the conditional Hoeffding lemma gives $\mathbb{E}{e}^{\lambda (f-\mathbb{E}f)}\le e^{n{\lambda }^2{K}^2/2}$; optimizing the exponential Markov bound proves the bound. A mere absolute-increment bound of $2K$ gives a weaker constant, so using the range length matters.

For a metric space with a probability measure $\mu$, the concentration function is $\alpha (\epsilon )={\sup }_{\mu (A)\ge 1/2}(1-\mu (A_{\epsilon }))$, where $A_{\epsilon }$ is the metric neighbourhood of radius $\epsilon$. A sequence exhibits concentration of measure if these functions tend to zero for every fixed positive radius. The metric scale is part of this definition.

For events $A,B$ with $\mathbb{P}(B)>0$, conditional probability isRestrict attention to the outcomes in $B$, then renormalize their probability measure so its total mass becomes one. The resulting rule $A\mapsto \mathbb{P}(A\mid B)$ is itself a probability measure. The definition and its connection to independent events are developed in MIT 18.05, Reading 3: Conditional Probability, Independence and Bayes' Theorem.

For the two-dice model, let $A$ mean the sum is at least nine and $B$ mean the first result is six. Of the six outcomes in $B$, four belong to $A$: the second die can be 3, 4, 5, or 6. Thus $\mathbb{P}(A\mid B)=2/3$. Reversing the information gives a different answer: four of the ten outcomes in $A$ have first result six, so $\mathbb{P}(B\mid A)=2/5$. Both ratios use the same overlap; their denominators describe different information.

The definition requires positive conditioning mass. For a continuously distributed random variable $X$, it cannot directly define $\mathbb{P}(A\mid X=x)$ when $\mathbb{P}(X=x)=0$. A regular conditional distribution or conditional density supplies the appropriate extension, with the usual almost everywhere qualification. Nor does conditioning by itself justify causal inference: changing which outcomes are considered is different from physically changing how the experiment is generated.

Continuity of probability says that nested events can be approximated by their finite stages without losing probability in the limit of a sequence. In a probability space $(\Omega ,\mathcal{F},\mathbb{P})$, let $A_n,B_n\in \mathcal{F}$. ThenThe sigma-algebra ensures that both limiting events belong to $\mathcal{F}$. These are continuity from below of a measure and continuity from above of a measure. The first holds even for an infinite measure; the second requires finite measure of the first set, which is automatic for a probability measure because $\mathbb{P}(B_1)\le 1$.^[ref]

For a concrete model, toss a fair coin through a sequence of independent Bernoulli trials, and let $A_n$ be the event of at least one head in the first $n$ tosses. Its complement $B_n=A_n^c$ is the event of $n$ initial tails. On the sample space of infinite head-or-tail sequences,Every finite waiting horizon has positive tail probability, yet the event of waiting forever has probability zero. That intersection is not empty: it contains the all-tails sequence. Thus a zero-probability event need not be impossible as an outcome, and eventual success almost surely does not supply a deterministic deadline. The waiting time for the first head has a geometric distribution with success probability $1/2$. After ten tosses, the remaining tail probability is exactly $1/1024<1/1000$; nine tosses leave $1/512>1/1000$.

A proof outline exposes what makes countable additivity stronger than finite additivity of a set function. For increasing events, put $D_1=A_1$ and $D_n=A_n\setminus A_{n-1}$ for $n\ge 2$. These set differences are pairwise disjoint events, their union is $A$, and the first $n$ pieces have union $A_n$. ConsequentlyFor decreasing events, their complements increase to $B^c$; applying the increasing result and the complement rule of probability gives $\mathbb{P}(B_n)=1-\mathbb{P}(B_n^c)\to 1-\mathbb{P}(B^c)=\mathbb{P}(B)$. The finite-total-mass condition is essential for this subtraction. For comparison, the counting measure of each set $\{n,n+1,\dots \}$ of natural numbers is infinite, although their intersection is empty and has measure zero.

Conversely, on a sigma-algebra, a nonnegative normalized set function satisfying finite additivity of a set function and $R_n\downarrow \varnothing \Rightarrow \mathbb{P}(R_n)\to 0$ satisfies countable additivity. For pairwise disjoint events $D_j$ with $E={\bigcup }_j{D}_j$, the remainders $R_n=E\setminus {\bigcup }_{j=1}^n{D}_j$ decrease to the empty set, andTaking the limit of a sequence recovers the infinite sum. This equivalence explains why continuity of probability is part of the foundations rather than a separate assumption about smooth curves.

One useful application controls infinitely many possible errors. If events $E_j$ satisfy ${\sum }_j\mathbb{P}(E_j)<\infty$, then the events $C_n={\bigcup }_{j\ge n}{E}_j$ decrease, and the union bound gives $\mathbb{P}(C_n)\le {\sum }_{j\ge n}\mathbb{P}(E_j)\to 0$. Continuity of probability therefore makes $\mathbb{P}({\bigcap }_n{C}_n)=0$: errors occur infinitely often with probability zero. This is the Borel-Cantelli first lemma, and it requires no independence. For the illustrative bounds $\mathbb{P}(E_j)\le 2^{-j}$, the probability of any error at index $n$ or later is at most $2^{1-n}$, an explicit truncation guarantee.

The word continuity here concerns nested sets, not the absence of atoms in a probability distribution. For any real-valued random variable $X$, the events $\{X\le x+1/n\}$ decrease to $\{X\le x\}$. Hence its cumulative distribution function is right-continuous even when it has jumps, as for a geometric distribution. The nesting hypothesis matters: an arbitrary sequence of events need not admit either displayed interchange of limit and probability.

A countable algebra of sets generating the intended sigma-algebra. Standard Borel spaces admit such algebras by taking finite Boolean combinations from a countable topological base. Finite linear combinations of its indicators are dense in $L^2$ for every Borel probability measure, by the Monotone class theorem. Countability permits a common full-measure convergence set for all test indicators.

A density ratio compares two probability density functions relative to the same underlying measure. More generally it is a Radon-Nikodym derivative when one probability measure has absolute continuity of measures with respect to the other.

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]

For a probability measure on a standard Borel space and a measurable map $\pi$ to another standard Borel space, there is an almost-everywhere unique measurable probability kernel $y\mapsto {\mu }_y$ supported on ${\pi }^{-1}\{y\}$. It integrates to $\mu$, and $\int fd{\mu }_{\pi (x)}$ realizes conditional expectation given $\pi$. These measures are the conditional measures of a factor.

In Euclidean space, push an optimal transport plan $\pi$ forward by $(x,y)\mapsto (1-t)x+ty$. The resulting probability measures form a constant speed curve for the p-Wasserstein distance:If the plan is induced by a transport map $T$, this becomes the displayed title formula.

For discounted one-period asset gains $Y$, a dominated martingale measure is a probability measure $Q$ absolutely continuous with respect to the physical probability measure $P$, under which the gains have zero expectation. Its Radon-Nikodym derivative $Z$ satisfies $Z\ge 0$, ${\mathbb{E}}_{P}Z=1$ and ${\mathbb{E}}_P[ZY]=0$. An equivalent martingale measure additionally has $Z>0$ almost surely. Absolute continuity permits some positive-probability physical states to receive zero pricing probability; equivalence does not.

For an $N$-by-$N$ Hermitian matrix, the empirical spectral measure puts mass $1/N$ at each eigenvalue, counted with multiplicity. It is a probability measure. Integrating a test function against it gives the average of that function evaluated at the eigenvalues.

Two probability measures are equivalent when each is absolutely continuous with respect to the other, or equivalently when they have the same null events.