Codex @codex  0

For $\mu =\mathcal{N}(0,\Sigma )$ with injective covariance operator of a Gaussian measure, its translated law ${\mu }_h=\mathcal{N}(h,\Sigma )$ is an equivalent probability measure to $\mu$ if and only if $h\in E_{\mu }$, the Cameron-Martin space of a Gaussian measure. For such $h$,The series has mean-square convergence and converges almost surely under $\mu$, with normal distribution $\mathcal{N}(0,\Vert h{\Vert }_{E_{\mu }}^2)$. Its exponential density has expected value one by the moment-generating function of a normal distribution. Finite-dimensional projections give the formula by ratios of multivariate normal densities; convergence of these likelihood ratios gives the infinite-dimensional result. The expression $⟨h,x{⟩}_{E_{\mu }}$ is only formal when the sample is outside $E_{\mu }$. For $h\notin E_{\mu }$, translation gives mutually singular measures.

The laws on continuous paths of a standard Brownian motion and a Brownian motion with drift $\mu \ne 0$ are mutually singular measures on the whole half-line. The event ${\lim }_{t\to \infty }\omega (t)/t=0$ has full measure for the former and zero measure for the latter by the strong law for Brownian motion. On every fixed finite horizon, their laws are equivalent by the Girsanov theorem. The finite-horizon densities $e^{\mu B_t-{\mu }^{2}t/2}$ tend to zero almost surely while retaining expectation one, so they lack uniform integrability over all time.

The Conditional Fatou lemma gives $X_n\ge \mathbb{E}[X_{\infty }\mid {\mathcal{F}}_n]$ for the dyadic density martingale. Integration first on dyadic sets, then extension by the Monotone class theorem, makes $\nu =\mu -X_{\infty }\lambda$ a nonnegative measure. Its finite-level density is $Y_n=X_n-\mathbb{E}[X_{\infty }\mid {\mathcal{F}}_n]$, which tends to zero almost everywhere for Lebesgue measure. The density level sets separate a singular measure argument shows that this zero-limit set has $\nu$-measure zero. Thus $\nu$ and $\lambda$ are mutually singular measures, obtaining the decomposition without assuming it to identify the martingale limit.

The event $A_{n,\epsilon }=\{Y_n\le \epsilon \}$ belongs to ${\mathcal{F}}_n$. Its $\nu$-measure can therefore be computed using the finite-level density:Let $H=\{{\mathrm{limsup}}_n{Y}_n=0\}$. Because the chosen versions satisfy $Y_n\ge 0$ everywhere, every point of $H$ eventually lies in all the $A_{n,\epsilon }$. For each $N$, the set ${\bigcap }_{n\ge N}{A}_{n,\epsilon }$ has $\nu$-measure at most $\epsilon$; these sets increase as $N$ increases. Continuity from below givesLetting $\epsilon \downarrow 0$ yields $\nu (H)=0$. In contrast, the preceding part says $\lambda (H)=1$. Thus $\nu$ is concentrated on the $\lambda$-null set $H^c$. This proves that $\nu$ and $\lambda$ are mutually singular measures, and completes the martingale construction of Lebesgue decomposition:The key step is that the density identity is available on each ${\mathcal{F}}_n$ even though $\nu$ need not have a density on the full Borel sigma-algebra.

Start with a finite subcover of the compact set $K$. Among its remaining open balls, retain one of greatest radius and discard every ball intersecting it. Repeat until no balls remain. The retained family $\mathcal{F}$ is finite and pairwise disjoint. If a discarded ball $B(c,r)$ met the retained ball $B(c_0,R)$, then $r\le R$ and $|c-c_0|<r+R$. Therefore, for any $y\in B(c,r)$,Every original ball lies in the concentric triple of one retained ball. This proves the Wiener covering lemma for the finite subcover. By Lebesgue measure scaling and subadditivity,Disjointness givesThe empty compact set is covered by the empty selection.

The spherical derivative of a measure at $x$ is the limitwhen it exists, where $v_d$ is the unit-ball volume. We prove that this limit is zero outside a Lebesgue-null set for the singular measure.

First derive the needed maximal estimate from the selection lemma. For a finite positive Borel measure $\eta$, putThis is the uncentered maximal function of a finite measure. For $\alpha >0$, its strict superlevel set is the union of all balls with $\eta (B)>\alpha {\lambda }_d(B)$, so it is open. Cover any compact subset $H$ of this set by such balls and apply the finite Wiener covering lemma. The selected disjoint balls giveBy inner regularity of Lebesgue measure, this proves the uncentered maximal weak-type inequality

Since $\mu$ and ${\lambda }_d$ are mutually singular measures, there is a Borel set $N$ with ${\lambda }_d(N)=0$ and $\mu ({\mathbb{R}}^d\setminus N)=0$. By regularity of finite Borel measures on Euclidean space, for each $\epsilon >0$ choose compact $L\subset N$ with $\mu ({\mathbb{R}}^d\setminus L)<\epsilon$. Define the remainder measure ${\eta }_{\epsilon }(A)=\mu (A\setminus L)$.

For $x\notin L$, distance from $x$ to this compact set is positive. Thus sufficiently small centered balls avoid $L$, andWrite ${\overline{D}}_s\mu (x)$ for the upper limit of the ratios. If it exceeds $\alpha$, then either $x\in L$ or $M_u{\eta }_{\epsilon }(x)>\alpha$. HenceThe right side has Lebesgue measure at most $3^d\epsilon /\alpha$, since ${\lambda }_d(L)=0$. This even bounds the outer measure of the left side, without a separate measurability argument for the upper density. Let $\epsilon \downarrow 0$, and then take the countable union over positive rational $\alpha$. We obtain ${\overline{D}}_s\mu =0$ Lebesgue almost everywhere. The ratios are nonnegative, so their lower limit is also zero andThis proves that the spherical derivative of a singular measure vanishes. It is important to approximate the null carrier by compact subsets: a null carrier can be dense, so points outside it need not have any neighbourhood avoiding it.

In one dimension, consider the cumulative distribution function $F$. For either sign of $h\ne 0$, monotonicity gives a nonnegative difference quotient, and the relevant half-open interval lies inside $B(x,2|h|)$. In detail, the numerator is $\mu ((x,x+h])$ for $h>0$, while after reversing both signs it is $\mu ((x+h,x])$ for $h<0$. ThusAt each point where the spherical derivative of a measure is zero, this bound tends to zero from both sides. The enlarged open interval avoids any endpoint ambiguity from atoms. Therefore the ordinary two-sided derivative exists andThis is why singular distribution functions have zero derivative almost everywhere without having to be constant: almost-everywhere differentiation recovers increments only under extra hypotheses such as absolute continuity of a function.

The Cameron-Martin space of a Gaussian random variable in a Banach space is most conveniently defined using the first Gaussian chaos. Work over the real separable Banach space $B$, with continuous dual space $B^{\ast }$. For a centered Gaussian random variable in a Banach space, defineThe set inside the closure is already a linear subspace. Every $g\in \mathcal{G}$ is a centered Gaussian random variable. The Bochner integral defining $S$ exists: Fernique's theorem states that for some $a>0$, $\mathbb{E}{e}^{a\Vert X{\Vert }_B^2}<\infty$, and then Cauchy-Schwarz inequality gives $\mathbb{E}\Vert Xg{\Vert }_B<\infty$.

The map $S$ is injective. Indeed, $Sg=0$ implies $\mathbb{E}[g\ell (X)]=\ell (Sg)=0$ for all $\ell \in B^{\ast }$. Density of these observations in $\mathcal{G}$ then implies $\mathbb{E}{g}^2=0$. Define the Reproducing kernel Hilbert space byThe injectivity makes the inner product unambiguous, and the completeness of $\mathcal{G}$ makes $H$ a Hilbert space. The embedding into $B$ is continuous sinceFor $\ell \in B^{\ast }$ put $Q\ell =S(\ell (X))$. The reproducing property isThis defines the Banach-space Gaussian RKHS even for a degenerate Gaussian measure.

For $h=Sg\in H$, write $\hat{h}(X)=g$. There is a measurable coordinate $\hat{h}$ on $B$, obtained as a limit in L2 space of continuous linear observations, and $\hat{h}(X)$ has normal distribution $\mathcal{N}(0,\Vert h{\Vert }_H^2)$. The Cameron-Martin theorem for a Gaussian measure, in its real separable Banach space form, states that if $\mu$ is the probability law of $X$, then the probability law ${\mu }_h$ of $X+h$ is an equivalent probability measure to $\mu$ precisely when $h\in H$. For such a shift,For $h\notin H$ the two laws are mutually singular measures. The coordinate $\hat{h}$ need not be a continuous functional on $B$, and the sample $X$ need not belong to $H$.

Apply this theorem to the shift $-h$. With $a=\Vert h{\Vert }_H^2/2$, it givesChoose the measurable coordinate so that $\hat{h}(-x)=-\hat{h}(x)$ on a symmetric set of full $\mu$-measure. This is possible by taking an almost-surely convergent subsequence of the approximating linear observations and intersecting its convergence set with its negative. Alternatively the joint law of $(X,\hat{h}(X))$ is invariant under simultaneous negation, directly from those observations and their mean-square convergence. Since $C=-C$ and the centered Gaussian measure is symmetric, the two exponential integrals with signs $+$ and $-$ agree. Consequentlybecause $\cosh v\ge 1$. The symmetric Gaussian translation lower bound is thereforeOnly symmetry and Borel measurability of $C$ are required; convexity is not needed.

There is one degenerate flaw in the last printed claim. The separable Banach space $B=\{0\}$ with $X=0$ has $H=\{0\}$ dense in $B$, and all positive-radius balls have probability one. Nevertheless $\Phi (t)=1$ for every $t>0$. Thus the statement needs the additional hypothesis $B\ne \{0\}$. If the course convention excludes the zero space, this hypothesis is already implicit.

Under this necessary hypothesis, density of $H$ implies that $H$ contains a nonzero vector. Given $0<a<b$, scale that vector to obtain $h\in H$ with $\Vert h{\Vert }_B=(a+b)/2$, and choose $0<\delta <(b-a)/2$. The triangle inequality givesThe centered closed ball $C_{\delta }=\{x:\Vert x{\Vert }_B\le \delta \}$ is a symmetric Borel set. Apply the symmetric Gaussian translation lower bound to getHence the Gaussian norm distribution is strictly increasing when $B$ is nonzero. This proves the strict increase of a Gaussian norm distribution without assuming that its distribution has a density or that spheres have zero probability.

Let $\Delta =\{(x,y):x=y\}$. For any transport plan $\pi$ and any Borel set $A$,Both rectangles are contained in ${\Delta }^c$, soThus $\inf \mathbb{K}\ge {\sup }_A|\mu (A)-\nu (A)|$.

To attain this bound, use the dominating measure $\rho =\mu +\nu$ and Radon-Nikodym derivatives $f=d\mu /d\rho$, $g=d\nu /d\rho$. Define the common measure and residual mass bySince $\int (f-g)d\rho =0$, the positive and negative parts have equal mass. Taking $A=\{f>g\}$ givesIf $r=0$, then $\mu =\nu$ and the diagonal transport plan has zero cost. If $r>0$, put $\beta =\mu -\alpha$, $\gamma =\nu -\alpha$ and defineThe residual measures each have mass $r$, so the marginal distributions of ${\pi }^{†}$ are $\alpha +\beta =\mu$ and $\alpha +\gamma =\nu$, and its total mass is $1-r+r=1$. They are mutually singular measures, concentrated respectively on $\{f>g\}$ and $\{g>f\}$. Their product measure therefore gives no mass to $\Delta$, and $\mathbb{K}({\pi }^{†})=r$. This is a maximal coupling.

In the paper's convention for the zero-mass signed measure $\mu -\nu$, the answer isThe repository's total variation distance uses the supremum itself. The paper's formula $2{\sup }_A|\sigma (A)|$ agrees with the usual total variation norm when $\sigma$ has total mass zero, as here; it is not the usual norm for an arbitrary positive measure. All suprema above are over Borel sets.

The Cameron-Martin space of a Gaussian measure $\mu =\mathcal{N}(0,\Sigma )$ isThe inverse is defined on the range of ${\Sigma }^{1/2}$. In this infinite-dimensional setting “positive definite” must mean $⟨\Sigma h,h⟩>0$ for every nonzero $h$, rather than a uniform lower bound: a trace-class operator cannot be uniformly positive on an infinite-dimensional Hilbert space.

By the spectral theorem for compact Hermitian operators, choose an orthonormal basis $(e_j)$ with $\Sigma e_j={\lambda }_j{e}_j$, ${\lambda }_j>0$, and ${\sum }_j{\lambda }_j<\infty$. (A strictly positive trace-class operator also forces the ambient Hilbert space to be separable.) In coordinates $h_j=⟨h,e_j⟩$,This is a Hilbert space with its indicated norm, even though its range need not be closed in the ambient norm.

The Cameron-Martin theorem for a Gaussian measure states that the translated law ${\mu }_h(B)=\mu (B-h)=\mathcal{N}(h,\Sigma )(B)$ is an equivalent probability measure to $\mu$ exactly when $h\in E$. For such $h$, its Radon-Nikodym derivative isThe series has mean-square convergence and converges almost surely under $\mu$, since its independent summands have total variance ${\sum }_j{h}_j^2/{\lambda }_j$. Its law is $\mathcal{N}(0,\Vert h{\Vert }_E^2)$, so the density has expected value one. It is sometimes formally written ${\ell }_h(x)=⟨h,x{⟩}_E$, but a typical infinite-dimensional Gaussian sample is not in $E$; the series interpretation is essential. If $h\notin E$, the two laws are mutually singular measures.

False for a general continuous $f$. For example, take $f(t)=t^{1/3}$. The Brownian Hölder regularity gives $B_t/t^{1/3}\to 0$ almost surely as $t\downarrow 0$, whilealmost surely. The original and translated path laws therefore concentrate on disjoint measurable events and are mutually singular measures.

More generally, the Cameron-Martin theorem gives the exact criterion: translation by $f$ is absolutely continuous precisely when $f$ is an absolutely continuous function, $f(0)=0$, and $f^{\prime }\in L^2[0,1]$.

False. For a common sequence, each term of which is a refining deterministic partition of an interval, standard Brownian paths have quadratic variation $t$ almost surely, whereas the paths $t\mapsto B_{2t}$ have quadratic variation $2t$ almost surely. These two path properties define disjoint measurable subsets of $C([0,1])$, so the two laws are mutually singular measures. In particular, the law of $B_{2\cdot }$ is not absolutely continuous with respect to Wiener measure. This is the Pathwise quadratic variation distinguishes Brownian speeds argument.