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Given a continuous surjection $h:\Omega \to K$ from Cantor space to a nonempty compact metric space, the pullback $f\mapsto f\circ h$ is a unital isometric embedding of $C(K)$ into $C(\Omega )$. Transport a normalized positive linear functional to its image, take a positive extension from a unital subspace of C(K), and represent that extension on Cantor space. The pushforward measure $h_{\ast }P$ then represents the original functional.

For a measurable function $T$, every f-divergence contracts under pushforward:This follows from conditional Jensen inequality applied to the jointly convex perspective $(a,b)\mapsto bf(a/b)$.

The Dirac probability measure at $x$ satisfies ${\delta }_x(A)=1$ if $x\in A$ and zero otherwise. Its pushforward measure under a measurable map $T$ is $T_{\#}{\delta }_x={\delta }_{T(x)}$.

For a measurable transformation $T$, a measure is invariant if the pushforward measure equals the original measure, or equivalently $\mu (T^{-1}A)=\mu (A)$ for every measurable $A$. This is the measure used in a measure-preserving system.

Given a cost $c$ and probability measures $\mu ,\nu$, the Monge problem minimizes $\int c(x,T(x))d\mu (x)$ over measurable transport maps with pushforward measure $T_{\#}\mu =\nu$. The feasible set may be empty because a map cannot split an atom of a measure.

A measurable transformation $T:X\to X$ is nonsingular with respect to $\mu$ when $\mu (T^{-1}N)=0$ for every $\mu$-null set $N$, equivalently when the pushforward measure $T_{\ast }\mu$ is absolutely continuous with respect to $\mu$.

Let $\Gamma (B)$ be the outer boundary of a planar compact set applied to the Brownian loop. For the pinned Brownian measure here, its outer boundary is almost surely simple and surrounds its root. If $U$ is simply connected, thenThe forward implication uses the fact that filling a compact subset of a simply connected domain cannot leave that domain. For the converse, the Brownian trace is contained in the closed filled interior of its outer boundary, and that interior also lies in $U$. This is precisely why simply connected domains are used.

Let $\eta ={\Gamma }_{\ast }\rho$, a pushforward measure on simple loops surrounding $0$. Its deficits areThey agree with those of $\lambda \eta$ and ${\nu }_D$. The recovery argument from part (i), now with explicitly finite deficits, givesIf $\lambda =0$, the measure is zero on every cutoff window and hence zero. The rooted Brownian loop itself is not a simple loop; taking its outer boundary is essential.

Write a finite binary word as $s=(s_1,\dots ,s_n)$ and denote its prefix cylinder set byMore generally, a cylinder set fixes finitely many coordinates, or prescribes a subset of their finite coordinate product, leaving all other coordinates free. Each coordinate factor is discrete, so finite-coordinate inverse images are both open and closed. Thus cylinders are clopen. The basic open sets of the product topology restrict only finitely many coordinates, so cylinders form a base. Any finite-coordinate cylinder is a finite union of prefix cylinders of one sufficiently long common length; the prefix cylinders therefore also form a base.

The Bernoulli space here is Cantor space, with compatible metricIt is a compact metric space: from any sequence, choose successively subsequences constant in the first, second and subsequent coordinates, and take the diagonal subsequence. Agreement on the first $n$ coordinates bounds the distance by $2^{-n}$, proving convergence. Compactness also follows from Tychonoff theorem.

Every cylinder indicator function is continuous because the cylinder is clopen. Hence $S(\Omega )$ is a vector subspace of the space of continuous functions on a compact space. To prove density, fix $f\in C_{\mathbb{R}}(\Omega )$ and $\epsilon >0$. By uniform continuity, choose $n$ so that $d(x,y)\le 2^{-n}$ implies $|f(x)-f(y)|<\epsilon$. Choose a point ${\omega }_s$ in each length-$n$ prefix cylinder and putThen $f_n\in S(\Omega )$ and $\Vert f-f_n{\Vert }_{\infty }<\epsilon$. This proves cylinder-function density in Cantor space:

Positivity of $\varphi$ implies monotonicity. The pointwise bounds $-\Vert f{\Vert }_{\infty }1\le f\le \Vert f{\Vert }_{\infty }1$ therefore giveEvaluation on $1$ attains equality, so $\varphi$ is continuous and has norm one. This is the norm of a positive functional on C(K).

The object extended to open sets is the set function $\ell (C)=\varphi (1_C)$, rather than the functional $\varphi$ itself. Positivity gives $\ell (C)\ge 0$, and linearity gives finite additivity on disjoint cylinders. To construct its extension directly, use the disjoint cylinder decomposition of open subsets of Cantor space. For an open $U$, take the shortest prefixes $s$ for which $[s]\subset U$. They are prefix-free, their cylinders are disjoint, and their union is $U$. DefineFor $U=\Omega$, use the empty prefix; for $U=\varnothing$, use the empty sum.

This value is independent of the chosen disjoint prefix-cylinder decomposition. Indeed, compare two such decompositions $(C_i)$ and $(D_j)$. For each fixed $C_i$, its intersections with the $D_j$ give a disjoint open cover of the compact cylinder $C_i$. Compactness reduces this to finitely many nonempty intersections. Each intersection is a prefix cylinder or is empty, so finite additivity gives $\ell (C_i)={\sum }_j\ell (C_i\cap D_j)$. Sum over $i$ and rearrange the nonnegative double sum. Doing the same with each $D_j$ proves equality of the two totals.

For disjoint open sets $U_k$, combine their disjoint cylinder decompositions to obtain one for ${\bigcup }_k{U}_k$. Rearrangement of nonnegative sums proves countable additivity on this family of open sets. The extension is unique because any countably additive extension must have the prescribed sum on every disjoint cylinder decomposition. Open sets are not themselves a sigma-algebra; countable additivity here concerns disjoint open families and their open union.

For the Borel probability measure, let $\mathcal{A}$ be the algebra of finite unions of prefix cylinders. It is also the algebra of clopen sets, because every clopen set is compact and has a finite cylinder cover. Define $\ell (A)=\varphi (1_A)$ on $\mathcal{A}$. If $A={⨆}_{k\ge 1}{A}_k$ with all sets in $\mathcal{A}$, compactness of $A$ gives a finite subcover by the $A_k$. Disjointness makes all remaining members empty. Thus finite additivity already establishes the premeasure condition.

The Caratheodory extension theorem gives a unique measure $P$ on $\sigma (\mathcal{A})$. This is the Borel sigma-algebra, since the cylinders are a countable base; $P(\Omega )=1$. Its values on open sets agree with the extension above. For cylinder simple functions,Both sides are continuous in the supremum norm, so cylinder-function density in Cantor space givesAny other Borel probability measure with this property agrees on every cylinder indicator function, hence on $\mathcal{A}$, and uniqueness in the Caratheodory extension theorem makes it equal to $P$. This proves the Cantor-space representation of positive functionals without assuming the general representation theorem.

Now let $h:\Omega \to X$ be the supplied continuous surjection. The pullbackis a unital isometric embedding: surjectivity gives $\Vert f\circ h{\Vert }_{\infty }=\Vert f{\Vert }_{\infty }$. On the vector subspace $J(C_{\mathbb{R}}(X))$, define $L(Jf)=\varphi (f)$. This is well-defined, has norm one and satisfies $L(1)=1$.

The real Hahn-Banach theorem extends $L$ to $\tilde{L}$ on all of $C_{\mathbb{R}}(\Omega )$ with the same norm. Norm preservation alone does not automatically mean positivity, so verify it. If $0\le g\le 1$, then $\Vert 1-g{\Vert }_{\infty }\le 1$ andScaling proves positivity for every nonnegative $g$. This is the unital contraction positivity criterion, giving a positive extension from a unital subspace of C(K).

Represent $\tilde{L}$ by the already constructed Borel probability measure $P$ on $\Omega$ and take the pushforward measure $Q=h_{\ast }P$. Continuity of $h$ makes its Borel inverse images measurable, and $Q(X)=1$. The pushforward integral identity givesThis is compact-metric representation by Cantor-space pullback. The hypothesis $\varphi (1)=1$ already excludes an empty $X$. The measure is on the Borel sets; no uniqueness claim on an unspecified larger collection of subsets is needed.

For an aggregate claims model, let $X$ denote one claim size, with expected value $\mu$ and variance ${\sigma }^2$. The total $S$ is zero when the claim count is zero. Given $N=n$, independence givesThe law of total expectation and law of total variance, together with $\mathbb{E}N=\mathrm{Var}(N)=\lambda$ for a Poisson distribution, implyThe raw second moment appears because the random count itself contributes $\lambda {\mu }^2$ to the variance.

Conditional on $N$, the moment-generating function of the sum is $M_X(t{)}^N$. Averaging with the Poisson distribution therefore gives the compound Poisson distribution transformThis is valid wherever $M_X(t)$ is finite. Finite first and second moments alone do not ensure positive exponential moments; for positive claims the corresponding Laplace transform always exists.

For the independent portfolios put $\Lambda ={\lambda }_1+{\lambda }_2>0$ and $w_i={\lambda }_i/\Lambda$. Their Poisson distribution counts add to a Poisson distribution with parameter $\Lambda$. By Poisson-multinomial conditioning, conditional on the total count $n$ the first risk count has a binomial distribution with parameters $n,w_1$, and the second is the remaining count. Thus one can generate the same total loss by drawing $n$ independent risk labels with these weights, then drawing each claim from its label's law. The merged severity has mixture distributionIt follows that $T$ has a compound Poisson distribution with count parameter $\Lambda$ and this severity law. This is the fixed-year version of Poisson superposition of insurance portfolios. Its expected value and variance areAlternatively, multiplying the two independent aggregate Laplace transforms yields $\exp \{\Lambda (w_1{M}_{X_1}(t)+w_2{M}_{X_2}(t)-1)\}$ wherever finite, confirming the same compound Poisson distribution.

For the retained compound Poisson aggregate under per-claim reinsurance, replace each claim $X$ by its retained payment $g(X)$. Assume the retention is measurable, with $0\le g(x)\le x$ as usual. The count parameter remains $\Lambda$, and the severity law is the pushforward measure of the mixture severity under $g$. Thus the retained compound Poisson aggregate $T_I$ has a compound Poisson distribution with that transformed severity andIts moment-generating function is $\exp \{{\sum }_i{\lambda }_i(M_{g(X_i)}(t)-1)\}$ on its finite domain. For a general retention $g$, zero retained payments can occur and are allowed as compound-Poisson marks; they may equivalently be removed by Poisson thinning. The two specified contracts retain strictly positive payments for strictly positive claims.

For a measurable map, each preimage ${\theta }^{-1}(A)$ lies in $\mathcal{E}$, so ${\mu }_{\theta }$ is well-defined and nonnegative. The preimage of the empty set is empty, and preimages commute with disjoint countable unions, givingfor pairwise disjoint measurable $A_j$. Thus ${\mu }_{\theta }$ is a measure, the pushforward measure ${\theta }_{\ast }\mu$. Its total mass is $\mu (E)$, because ${\theta }^{-1}(E)=E$. Finiteness is not needed for this construction, but is used in the uniqueness argument next.

The Rudolph measure rigidity theorem has an essential ergodicity hypothesis. If a Borel probability measure $\mu$ on the circle group is invariant under both $T_2$ and $T_3$, is ergodic for the semigroup generated jointly by these maps, and either map has positive Kolmogorov-Sinai entropy, thenEquivalently, a jointly ergodic common invariant measure other than Lebesgue measure has zero entropy for both maps. Joint ergodicity means that every set invariant modulo $\mu$ under both maps has measure zero or one. Positive entropy without this hypothesis is insufficient: $\frac{1}{2}m+\frac{1}{2}{\delta }_0$, with ${\delta }_0$ a Dirac measure, is a common invariant measure of positive entropy and is not $m$.

The Host equidistribution theorem states that if $p,q\ge 2$ are relatively prime integers and $\nu$ is invariant and ergodic under $T_p$, with $h_{\nu }(T_p)>0$, then for $\nu$-almost every $x$ the sequence $(T_q^{n}x)$ is an equidistributed sequence for Lebesgue measure. Explicitly, for every continuous $g$ on the circle,The non-ergodic form assumes $T_p$ invariance and positive entropy for almost every component in the ergodic decomposition $\nu =\int {\nu }_{\omega }d\lambda (\omega )$. Applying the ergodic theorem of Host on each such component gives the same almost-everywhere conclusion for $\nu$. More generally, its conclusion holds on the part supported on positive-entropy components. A positive value of $h_{\nu }(T_p)$ alone does not eliminate zero-entropy components.

To deduce the joint version of the Rudolph measure rigidity theorem, suppose $h_{\mu }(T_2)>0$; if only $T_3$ has positive entropy, interchange the roles. Write the $T_2$ ergodic decomposition as $\mu =\int \nu d\tau (\nu )$. Since $T_3$ commutes with $T_2$, its pushforward measure sends a $T_2$ ergodic component $\nu$ to a $T_2$ ergodic component $(T_3{)}_{\ast }\nu$. On each component, $T_3$ is a factor of a measure-preserving system with fibres of size at most three. We use the standard entropy preservation under a finite-to-one factor:The reason for this standard entropy fact is that, conditional on a complete factor point, every finite orbit name has at most three possibilities; its conditional entropy is bounded by $\log 3$, and division by the orbit length gives zero relative entropy.

The component at $T_{3}x$ is $(T_3{)}_{\ast }{\nu }_x$ almost everywhere; this follows from commutation and the componentwise ergodic averages. The component entropy function $r(x)=h_{{\nu }_x}(T_2)$ is therefore invariant under both $T_2$ and $T_3$. Joint ergodicity makes it constant almost everywhere, and affinity of entropy under ergodic decomposition identifies the constant as $h_{\mu }(T_2)>0$. Thus almost every $T_2$ component has positive entropy, exactly the condition required in the non-ergodic Host equidistribution theorem. It follows that $\mu$-almost every point equidistributes for $m$ under $T_3$.

For any continuous $g$, invariance under $T_3$ and the dominated convergence theorem now giveContinuous functions determine Borel probability measures on the circle, so $\mu =m$, proving the deduction.

For the normal-number example, let ${\omega }_1,{\omega }_2,\dots$ be independent fair binary digits and defineThis Cantor Bernoulli measure is supported on the middle-third Cantor set $C$. If $S$ is the Bernoulli shift, then $T_3\circ \pi =\pi \circ S$ on the circle. Consequently $\nu$ is $T_3$ invariant and ergodic: a $T_3$ invariant event pulls back to an $S$ invariant event, which has probability zero or one.

Take the ternary digit measurable partition $\eta =\{[0,1/3),[1/3,2/3),[2/3,1)\}$. Its block partition of length $n$ has, up to null endpoints, $2^n$ positive-measure atoms under $\nu$, each of measure $2^{-n}$, and all other atoms have measure zero. HenceApply the Host equidistribution theorem with $p=3$, $q=2$. For $\nu$-almost every $x\in C$, the sequence $(T_2^{n}x)$ equidistributes for Lebesgue measure, so $x$ is a normal number in base $2$ by normality and equidistribution under integer multiplication.

The ternary expansion of $\nu$-almost every such $x$ contains only $0$ and $2$, so the frequency of digit $1$ is zero rather than $1/3$. The ambiguous ternary endpoints form a countable $\nu$ null set and can be removed. Therefore $x$ is not a normal number in base $3$. We have proved the stronger almost-everywhere existence statement

A probability measure $\mu$ is $T$-invariant whenEquivalently, the pushforward measure satisfies $T_{\ast }\mu =\mu$, so $T$ is a measure-preserving transformation.

For a cost $c:X\times Y\to [0,+\infty ]$ that is a Borel measurable function, a transport map is a measurable $T:X\to Y$ whose pushforward measure satisfiesThe Monge optimal transport problem moves every source point to one destination:The Kantorovich optimal transport problem permits mass to split. Its admissible transport plans are the probability measures on $X\times Y$ with prescribed marginal distributions:Thus a transport plan is a coupling of probability distributions. The set $\Pi (\mu ,\nu )$ is never empty: it contains the product measure $\mu \otimes \nu$. Signed costs can also be used when their integrals are well defined, for example with an integrable lower bound of the form $a(x)+b(y)$.

On the Polish space $\mathbb{R}$, take the Dirac measuresEvery measurable map satisfies $T_{\#}{\delta }_0={\delta }_{T(0)}$, which cannot equal $\nu$. A transport map cannot split an atom of a measure, whereas the transport plan $\frac{1}{2}{\delta }_{(0,-1)}+\frac{1}{2}{\delta }_{(0,1)}$ can.

Given any admissible transport map $T$, form its graph transport planFor Borel sets $A\subseteq X$ and $B\subseteq Y$, the definition of a pushforward measure givesHence ${\pi }_T\in \Pi (\mu ,\nu )$. Integration against a pushforward measure also givesThe Kantorovich optimal transport problem therefore has at least all the competitors of the Monge optimal transport problem, with exactly the same costs. ConsequentlyIf there is no admissible transport map, the left side is $+\infty$ by the convention $\inf \varnothing =+\infty$, so the conclusion still holds. No existence of an optimizer is needed.

Interpret the invertibility assumption on $G$ as continuity and strict increase on the relevant range, so that $\nu$ is an atomless measure. Suppose a non-decreasing transport map $T$ with $T_{\#}\mu =\nu$ exists. Then $\mu$ is also an atomless measure: if $\mu (\{x\})>0$, the pushforward measure would give $\nu (\{T(x)\})\ge \mu (\{x\})>0$. Thus $F$ is continuous.

At any point $x$ where the non-decreasing representative is defined, the definition of a monotone function givesUsing the pushforward measure identity and continuity of the two cumulative distribution functions, we obtainConsequentlyThe monotone rearrangement in part (c) is optimal for the convex difference cost, so $T$ has the same cost and solves the Monge optimal transport problem. The meaningful uniqueness is up to a $\mu$-null set; arbitrary values away from the source do not affect transport or cost. Existence of the non-decreasing transport map supplies the source condition missing in part (c).

Write $J_T(x)=|\det DT(x)|$, the absolute Jacobian determinant. Under the usual meaning of a $C^1$ map on domains, the injective area formula gives, for any nonnegative measurable $a$,This version of the change of variables formula does not require $DT$ to be nonsingular everywhere. A $C^1$ map is a locally Lipschitz function, and the formula applies by exhaustion if the domains are unbounded. For non-open $X$, one needs the corresponding extension or regularity interpretation of the stated $C^1$ assumption.

First suppose $f(x)=g(T(x))J_T(x)$ almost everywhere with respect to Lebesgue measure. For each bounded nonnegative measurable $a$,Taking indicator functions proves $T_{\#}\mu =\nu$.

Conversely, the area formula shows that the measure $\tilde{\mu }$ with probability density function $g(T(x))J_T(x)$ has total mass one and satisfies $T_{\#}\tilde{\mu }=\nu$. The assumed equality $T_{\#}\mu =\nu$ then implies $\mu =\tilde{\mu }$, since the measurable bijection $T$ has a measurable inverse on these domains. Equivalently, apply both pushforward measure identities to $T(A)$ for each Borel set $A\subseteq X$. Uniqueness of the Radon-Nikodym derivative yields

For every admissible transport map $T$, the pushforward measure condition gives $T(x)\in [1,2]$ for $\mu$-almost every $x\in [0,1]$. Hence $|T(x)-x|=T(x)-x$, andApply the Jensen inequality to the convex function $h$ and the probability measure $\mu$:Translation by one sends the source uniform distribution to the target uniform distribution, so $T^{†}(x)=x+1$ is admissible. Its displacement is constantly one, givingNo monotonicity assumption on $h$ is needed, because all displacements have the same sign. The same argument with a transport plan also gives the identical minimum for the Kantorovich optimal transport problem.