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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.

Suppose one observation comes from either $P$ or $Q$, each with prior probability $1/2$. A decision rule that chooses $P$ on an event $E$ and $Q$ otherwise has success probabilityOn a finite set, choosing $E=\{x:P(x)>Q(x)\}$ therefore gives $\frac{1}{2}(1+d_{\mathrm{TV}}(P,Q))$. This is the equal-prior version of minimum sum of errors in a simple hypothesis test; allowing a statistical test to use an auxiliary random variable cannot improve it, because for each outcome the better of the two decisions already maximizes its contribution.

For the four-outcome example, choose $P$ after $A$ and $Q$ after $B$ or $C$, with either decision allowed after $D$. Choosing $Q$ after $D$ yields $\frac{1}{2}(0.4+0.9)=0.65$. The best success probability is 65%, with a 35% error probability. A smaller total variation distance means the two sources are harder to distinguish from one observation. Unequal prior probabilities, unequal costs, or multiple observations change this particular decision problem; the $65\%$ figure is not a universal bound for every experiment.

Consider the probability mass functions $P=(0.4,0.1,0.2,0.3)$ and $Q=(0.1,0.3,0.3,0.3)$ on the finite set $S=\{A,B,C,D\}$. Their difference is $(0.3,-0.2,-0.1,0)$. ThusThe event $E=\{A\}$ attains this difference: $P(E)=0.4$ and $Q(E)=0.1$. Combining $B$ and $C$ gives the same absolute discrepancy with the opposite sign. Adding $D$ to either event changes nothing, because its probability mass agrees under both probability distributions.

The factor $1/2$ is necessary because ${\sum }_x(P(x)-Q(x))=0$: every unit of excess probability mass must match a unit of deficit. For any event, including a negative term cannot increase its signed discrepancy, while omitting a positive term cannot increase it either. Therefore $\{x:P(x)>Q(x)\}$ attains the positive maximum. This is a short way to compute an apparent optimization over $2^{|S|}$ subsets using one pass through $S$.

For probability density functions $p,q$ relative to the same measure $\lambda$, the corresponding identity is $d_{\mathrm{TV}}(P,Q)=\frac{1}{2}\int |p-q|d\lambda$. Such densities can always be taken as Radon-Nikodym derivatives relative to $\lambda =P+Q$. Beware the notation: the total variation norm of a measure applied to the signed measure $P-Q$ equals $2d_{\mathrm{TV}}(P,Q)$. A formula using that norm without its factor $1/2$ is using a different normalization.

Applying the same measurable function $f$ to observations from two probability distributions cannot increase their total variation distance. Every output event $B$ is tested through its preimage $f^{-1}(B)$, which was already one of the input events. ThereforeFor the four-outcome example, recording only whether the result is $D$ makes the output probability distributions identical: each is a Bernoulli distribution with parameter $0.3$. The original total variation distance was $0.3$, but this coarse measurement leaves zero discrepancy. Discarding information can make two sources less distinguishable.

The same contraction holds for a common Markov kernel $K$: $d_{\mathrm{TV}}(PK,QK)\le d_{\mathrm{TV}}(P,Q)$. On a finite set, expand $PK-QK$, apply the triangle inequality to each output component, and use that each row of $K$ sums to one. A common randomized postprocessing step cannot increase distinguishability either.

For a finite graph vertex set $W$ in a locally finite graph, let ${\partial }^{+}W$ be its exterior graph neighbours. The number of trifurcation vertices in percolation lying in $W$ is at most $|{\partial }^{+}W|$.

For each cluster meeting $W$, contract its connected components of a graph outside $W$ to terminal graph vertices, keeping only those adjacent to $W$. The resulting incidence graph is finite and connected. Each trifurcation graph vertex in $W$ separates at least three groups of terminals, because every infinite branch must leave the finite set $W$. Take a minimal subtree connecting all terminals. Each such trifurcation is unavoidable in that subtree and has degree of a vertex at least three. All leaves are terminals. The tree identity ${\sum }_v(\mathrm{deg}(v)-2)=-2$ bounds the number of these branching graph vertices by the number of leaves minus two, hence by the number of terminals. Different terminals, even across different clusters, can be assigned different exterior graph neighbours. Summing proves the bound. For square boxes in ${\mathbb{Z}}^2$, the boundary has order $n$ and the volume has order $n^2$.

For a finite set of primes choose $0<g(p)<1$, put $k(p)=g(p)/(1-g(p))$, and let $W={\prod }_p(1+k(p))$. Give each squarefree integer $d$ on these primes probability $k(d)/W$. Inclusion of each prime is represented by a Bernoulli random variable with parameter $g(p)$; these are independent random variables, so $\mathbb{E}[\log d]={\sum }_{p}g(p)\log p$. If this is at most $\theta \log L$ with $0<\theta <1$, the Markov inequality givesThis turns a full Euler product into a lower bound for the truncated normalizing sum of a Selberg sieve.

If $B$ is a countably infinite set and $A$ is a finite set with at least two elements, the set of surjective functions $B\to A$ is uncountable. Fix finitely many values to cover $A$, then encode an arbitrary infinite binary sequence using two values of $A$ on the remaining inputs.

Let $X_1,X_2,\dots$ be independent and identically distributed with probability law $P$, and let $\mathcal{G}$ be a class of measurable $P$-integrable functions. Write $P_{n}g=n^{-1}{\sum }_{i=1}^{n}g(X_i)$ and $Pg=\mathbb{E}g(X_1)$. A uniform law of large numbers concernswith measurability of the supremum assumed. A uniform weak law uses convergence in probability; a uniform strong law uses almost sure convergence. The mode and sampling assumptions are part of the statement.

For a finite class, the strong law of large numbers for each function holds simultaneously and gives the uniform strong law. For infinite classes, pointwise convergence alone is insufficient. For example, let $P$ be the uniform distribution on $[0,1]$ and take indicators of all finite subsets. Each fixed function has expectation zero and empirical mean tending to zero almost surely. Yet for every sample the finite set $\{X_1,\dots ,X_n\}$ belongs to the class and its empirical mean is one. The displayed supremum is therefore one for every $n$.

Conditions such as finite bracketing of a function class at every positive width control this complexity. The Glivenko-Cantelli theorem gives the uniform strong law for indicators of half-lines, the class determining the empirical distribution function.

A uniformly continuous function has one input tolerance valid throughout its domain of a function. For $f:D\to \mathbb{R}$, with $D\subseteq \mathbb{R}$, uniform continuity meansChoose the output tolerance $\epsilon$ first, then a positive $\delta$ valid for every pair of function inputs. It may depend on $\epsilon$, the function and its domain of a function, but not on the pair’s location. The absolute values measure input and output distance.^[ref]

Compare the quantifier order with the epsilon-delta definition of continuity:Both lines end with $|x-a|<\delta \Rightarrow |f(x)-f(a)|<\epsilon$. In the first, $\delta$ can depend on the point $a$; in the second, it is chosen before $a$. Thus uniform continuity implies continuity at every point. Each requested $\epsilon$ gets its own shared tolerance.

For an exact example, take the polynomial function $f(x)=x^2$ on the closed interval $D=[0,2]$. For every pair of function inputs in that interval, polynomial factorization givesConsequently $\delta =\epsilon /4$ works everywhere on this domain of a function. With $\epsilon =0.4$, the common tolerance is $\delta =0.1$. Near the right endpoint, for example, $x=1.92$ and $y=2$ have input distance $0.08$ and output distance $4-1.92^2=0.3136<0.4$. The bound is sufficient; it does not claim the largest admissible input window at each point.

On the whole real line, the same polynomial function is a continuous function but fails uniform continuity. Take the sequences $x_n=n$ and $y_n=n+1/n$, for positive integers $n$. ThenFor any proposed $\delta >0$, choose an integer $n>1/\delta$. The inputs are closer than $\delta$, yet their outputs violate the fixed tolerance $\epsilon =1$. The sequences need not themselves have a finite limit: their mutual distance is what tends to zero. This is the obstruction captured by the sequential criterion for uniform continuity. Restricting the domain of a function to $[0,2]$ removes these escaping pairs.

A uniformly continuous function need not be a bounded function. The identity function $f(x)=x$ on the real line has $|f(x)-f(y)|=|x-y|$, so $\delta =\epsilon$ always works, although its image of a function is the unbounded real line. Uniform continuity controls changes between nearby inputs, not the overall size of outputs. Likewise, an unbounded domain of a function does not by itself prevent uniform continuity.

The square root $g(x)=\sqrt{x}$ supplies a different common tolerance on the interval $[0,\infty )$. Assuming $x\ge y\ge 0$, put $u=\sqrt{x}$ and $v=\sqrt{y}$. Since $0\le u-v\le u+v$,Interchanging the inputs when necessary provesThus $\delta ={\epsilon }^2$ guarantees uniform continuity on the entire domain of a function, including zero. For $\epsilon =0.4$, any input distance below $0.16$ suffices. The pairs $(0,0.09)$ and $(4,4.09)$ both have input distance $0.09$; their output distances are respectively $0.3$ and $0.09/(\sqrt{4.09}+2)<0.0225$. The common guarantee accommodates quite different local sensitivities.

This Hölder continuity bound allows a curved tolerance rule. It is weaker than a Lipschitz bound: at zero the output change divided by the input change isNo finite Lipschitz constant works near zero, even though uniform continuity does. A Lipschitz function does have uniform continuity, choosing $\delta =\epsilon /L$ when its Lipschitz bound $L$ is positive; when $L=0$ it is a constant function.^[ref]

For general metric spaces, replace the absolute values by the respective distance functions. The Heine-Cantor theorem gives an existence guarantee: every continuous function on a compact metric space is uniformly continuous. In particular, continuity on a finite closed interval provides a common tolerance, even when no simple formula for it is known.^[ref]

For a short proof outline, fix $\epsilon >0$. Continuity gives each point $a$ an open ball of radius $r_a>0$ within which outputs differ from $f(a)$ by less than $\epsilon /2$. The open balls of half these radii still cover the domain of a function. Compactness supplies a finite subcover. Take $\delta$ to be the smallest half-radius in this finite set. Two inputs closer than $\delta$ lie together in one of the original open balls, so the triangle inequality bounds their output distance by $\epsilon$. The beautiful step is that the compact set turns local choices into finitely many choices, whose minimum stays positive.

A shared tolerance also makes uniformly continuous functions preserve Cauchy sequences: after choosing $\delta$ for $\epsilon$, all sufficiently late input pairs are closer than $\delta$, so all corresponding output pairs are closer than $\epsilon$. Unlike a convergent sequence argument, this does not require an input limit inside the domain of a function. It is why uniform continuity is useful for extending a function toward missing endpoints and for the decay argument below.

A pointwise periodic self-map has a uniform positive period exactly when its finite cycle lengths are bounded. A uniform period is divisible by every length; conversely, bounded lengths divide the least common multiple of $1,\dots ,M$ for some bound $M$. Arbitrarily many bounded-length cycles are allowed. Every such map on a finite set has a uniform period. On any infinite set containing a countably infinite subset, cycles of unbounded finite lengths and fixed points elsewhere give a counterexample. Thus, with the usual axiom of choice assumption, the sets on which every pointwise periodic self-map has a uniform period are exactly finite sets.

For a number field $K$, a finite set $S$ and $m\ge 2$, let $K(S,m)$ consist of classes $[a]\in K^{\ast }/K^{\ast m}$ whose valuations outside $S$ are divisible by $m$. It fits into $0\to {\mathcal{O}}_{K,S}^{\times }/({\mathcal{O}}_{K,S}^{\times }{)}^m\to K(S,m)\to \mathrm{Cl}({\mathcal{O}}_{K,S})[m]\to 0$. The S-unit group is finitely generated and the ideal class group is finite, so $K(S,m)$ is finite. When $S$ contains primes dividing $m$, this controls unramified multiplicative Kummer classes outside $S$.

Given finitely many distinct places of a number field $v_1,\dots ,v_r$, the diagonal image of $K$ is dense in ${\prod }_i{K}_{v_i}$. One can therefore simultaneously prescribe finite congruences and real signs. This is useful in the ray class exact sequence and in finding representatives of an ideal class prime to a prescribed finite set of prime ideals.

Let a finite group $G$ act on a finite set $X$, with an orbit-invariant weight $w:X\to R$ in a commutative ring. Double counting gives the division-free identityIndeed, each $x$ is counted $|G_x|$ times, and each orbit contributes $|O||G_x|=|G|$ copies of its common weight. If $|G|$ is invertible in $R$, dividing gives the weighted average formula of Burnside's lemma. For formal monomial weights one can first count over the integers and then specialize into any commutative ring; this avoids dividing by zero in modular characteristic.

Start with the highest weight and process weights by increasing height below it. For each simple root, compute the upper endpoint from the already known higher weights, use the weight string endpoint relation, and append all downward steps. Repeat until the finite set stabilizes. This recovers the weights of a finite-dimensional Irreducible Lie algebra representation without determining weight multiplicities.

Consider the following finite set of subsets, ordered by subset containment:Every nonempty chain in a partial order has a greatest element within that chain in a partial order, hence an upper bound in $P$. For the empty chain in a partial order, choose $\varnothing \in P$. Thus the hypothesis of Zorn's lemma holds. The two maximal elements are $\{a,b\}$ and $\{b,c\}$: neither is a proper subset of another member of $P$. There is no greatest element, because it would have to contain both, and no member of $P$ contains $\{a,b,c\}$.

For the chain in a partial order $C=\{\varnothing ,\{b\}\}$, the member $\{b\}$ is already an upper bound, although it is not a maximal element of $P$. Both orange members are also upper bounds for $C$. A chain in a partial order need not have a unique upper bound, and the upper bound used to check the hypothesis need not be the eventual maximal element.

The restriction to chains in a partial order matters. The subset $\{\{a\},\{c\}\}\subseteq P$ has no upper bound in $P$, but its two members are incomparable, so it is not a chain in a partial order. Requiring every subset of $P$ to have an upper bound would be much stronger: taking that subset to be $P$ itself would force a greatest element.