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A function $f:\mathbb{R}\to \mathbb{R}$ is Borel measurable whenis a Borel set for every Borel set $B\subseteq \mathbb{R}$. It is enough to require this for every open set, or for every interval of the form $(-\infty ,a)$.

Yes. In fact, measurability of $f$ is unnecessary. Define the oscillationFor every $\epsilon >0$, the set $\{x:{\omega }_f(x)<\epsilon \}$ is open, and $f$ is continuous at $x$ exactly when ${\omega }_f(x)=0$. Hence the continuity set of a function isa $G_{\delta }$ set and therefore a Borel set.

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.

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

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.

For every Borel set $A\subseteq {\mathbb{R}}^d$, letbe its random point count. A Non-homogeneous Poisson point process with intensity function of a point process $\lambda$ is characterized by:

Because $\Lambda$ has a density with respect to Lebesgue measure, it is atomless, so the process is simple: almost surely no location occurs more than once.

For a Borel set $A\subseteq [0,1]$, the two inverse branches of the tent map are $y\mapsto y/2$ and $y\mapsto 1-y/2$. Apart from their immaterial common endpoint,Each branch scales Lebesgue measure by $1/2$, and their images lie in opposite half-intervals. HenceAlso $m([0,1])=1$, so $m$ is an invariant probability measure.

Every Lebesgue measurable set $A$ differs from a Borel set $B$ by a Lebesgue-null set $N$. The given function $h$ is a homeomorphism from $\mathbb{R}$ onto the interval $h(\mathbb{R})$. Hence $h(B)$ is Borel relative to $h(\mathbb{R})$ and therefore Lebesgue measurable in $\mathbb{R}$. Moreover,and $h(N)$ is null by Lusin condition N. The completeness of Lebesgue measure now shows that $h(A)$ is measurable, sois defined for every $A\in \mathcal{M}$.

Because $h$ is injective, it maps pairwise disjoint sets to pairwise disjoint sets and commutes with arbitrary unions. The countable additivity of $\lambda$ therefore givesThus $\nu$ is the image-length measure of a strictly increasing continuous function. If $\lambda (A)=0$, the null-set hypothesis gives $\nu (A)=\lambda (h(A))=0$, so

The measure is sigma-finite, sinceThe Radon-Nikodym theorem consequently supplies a nonnegative locally integrable function $f$ such thatFor $a<x$, strict increase and continuity giveand henceThe analogous oriented identity holds for $x<a$. By differentiation of an indefinite Lebesgue integral,

Yes, the differentiability conclusion still holds when $h$ is merely non-decreasing. In fact, the Lebesgue theorem on differentiability of monotone functions says that every monotone function is differentiable almost everywhere; neither continuity nor the null-set hypothesis is needed for that conclusion. Flat intervals mean that the preceding image-set argument no longer gives disjoint images, so its natural replacement is the Lebesgue-Stieltjes measure determined by

A projection-valued measure on the Borel sets of $\mathbb{C}$ is a map $E$ into the orthogonal projections on a separable Hilbert space $\mathcal{H}$ such thatand for pairwise disjoint $B_j$,for every $v\in \mathcal{H}$, with convergence in norm.

The spectral theorem for normal operators on a separable Hilbert space states that a bounded normal operator $A$ has a unique projection-valued measure supported on $\mathrm{Sp}(A)$ for whichMore generally, the Borel functional calculus for a normal operator is

For $v,w\in \mathcal{H}$, the scalar spectral measures areIf $A$ is self-adjoint, its spectrum and hence the support of $E$ lie in $\mathbb{R}$. Moreover,so ${\mu }_v$ is a positive measure, andThe paper prints total mass $\Vert v\Vert$; with the standard definition it is $\Vert v{\Vert }^2$, so the unsquared norm is a typographical error.

The map $F$ is a measure-preserving transformation whenfor every Borel set $B$, equivalently $F_{\ast }\omega =\omega$. In this case its Koopman operator is an isometry on $L^2(\omega )$.

Write $P=\omega {\omega }^T$. Since $P^T=P$ and $P^2=P$, the orthogonal reflection $R_{\omega }=I-2P$ satisfies $R_{\omega }^T{R}_{\omega }=I$ and $R_{\omega }^{-1}=R_{\omega }$. Thus, for every Borel set $E$, invariance under the orthogonal group gives

The spherical average must use normalized surface measure without an additional area factor. Let $d\sigma$ be the normalized surface measure on a sphere, and setThe correct radial disintegration of an orthogonally invariant measure isIndeed, take normalized Haar measure $H$ on the compact orthogonal group $O(d)$. Invariance and Fubini's theorem giveRight invariance of Haar measure and transitivity of the orthogonal group on each sphere show that the inner average depends only on $r=|x|$. Averaging it once more over $x=r\theta$, using invariance of surface measure on a sphere and Fubini's theorem, identifies its value as $A\eta (r)$. At $x=0$ the same identity reads $\eta (0)=\eta (0)$. The definition of pushforward measure now proves the boxed formula. This argument also works for invariant probability measures with an atom of a measure at the origin.

For the printed normalization, choose a smooth cutoff function $\eta$ equal to one on a ball containing the support of a measure $\mu$. Its left integral is one, whereas the printed right integral is ${\sigma }_d$. Thus the extra ${\sigma }_d$ cannot be retained. Equivalently, with unnormalized area measure $dS$, the average is ${\sigma }_d^{-1}\int \eta (r\theta )dS(\theta )$.

The reflection-averaging hint also fails in general dimension. For $x=e_1$,The first equality uses $\int {\omega }_i{\omega }_{j}d\sigma ={\delta }_{ij}/d$, which follows from coordinate reflections, coordinate permutations, and ${\sum }_i{\omega }_i^2=1$. Taking a test function equal to the first coordinate near the unit sphere disproves the hint when $d\ne 2$. Averaging over the full orthogonal group, as above, proves the required corrected result without that hint.

A real measurable function is one whose inverse image of every open subset of $\mathbb{R}$ is Lebesgue measurable. Equivalently,These upper rays generate the Borel sigma-algebra, making this equivalent to requiring measurable inverse images of all Borel sets.

Independence means $\mathrm{Pr}(X\in A,Y\in B)=\mathrm{Pr}(X\in A)\mathrm{Pr}(Y\in B)$ for all Borel sets $A,B$. For variables with a joint density this is equivalent to factorization of that density into the two marginal densities almost everywhere.

The inverse transformation is $x=uv$, $y=(1-u)v$, on $0<u<1$, $v>0$. Its absolute Jacobian determinant is $v$. Multiplying the independent gamma densities by it givesThe density is zero outside that support. Integrating in $v$ with the gamma integral givesIt is a beta density. Its normalization also proves the beta integral ${\int }_0^1{u}^{\alpha -1}(1-u{)}^{\beta -1}du=\Gamma (\alpha )\Gamma (\beta )/\Gamma (\alpha +\beta )$. The joint density consequently factors as this normalized beta density timesThus $V$ is gamma with shape $\alpha +\beta$ and rate $\lambda$, and $U,V$ are independent.

Finally $X=UV$. All relevant first moments are finite, and independence gives $\mathbb{E}X=\mathbb{E}U\mathbb{E}V$, while $\mathbb{E}V=\mathbb{E}X+\mathbb{E}Y>0$. HenceThis identity follows from the particular beta-gamma independence here; a general ratio of random variables would not permit moving expectation through division.

For a Borel set $A\subseteq \mathbb{R}$, the joint probability density givesFor each $x\ne 0$, substitute $z=xy$. The change of variables for probability densities contributes $dy=dz/|x|$, with the absolute value accounting for the reversed orientation when $x<0$. The line $x=0$ has two-dimensional Lebesgue measure zero, so omitting it changes no probability. By the Tonelli theorem,The bracket is nonnegative and its integral over $\mathbb{R}$ is one. This proves the probability density of a product of random variables formula:As usual, a probability density function is determined only almost everywhere; exceptional infinite values are harmless.

For the independent random variables with the specified uniform distribution, $f(x,y)=b^{-2}$ on $(0,b{)}^2$ and zero elsewhere. When $0<z<b^2$, the conditions $0<x<b$ and $0<z/x<b$ are equivalent to $z/b<x<b$. ThereforeThe value assigned at zero is arbitrary. Substitution $s=z/b^2$ checks normalization of a probability distribution: ${\int }_0^1-\log sds=1$. This is the product of independent uniform random variables law.

For every $n$, ${\mathcal{F}}_n$ is a sigma-algebra contained in $\mathcal{F}$, and including one more random variable among its generators can only enlarge it. Hence ${\mathcal{F}}_n\subseteq {\mathcal{F}}_{n+1}$; the trivial ${\mathcal{F}}_0$ is contained in every sigma-algebra. These are exactly the requirements for a filtration.

The constant $X_0=x$ is ${\mathcal{F}}_0$-measurable. Suppose $X_n$ is ${\mathcal{F}}_n$-measurable. The pair $(X_n,U_{n+1})$ is then ${\mathcal{F}}_{n+1}$-measurable, and the event $\{U_{n+1}\le X_n\}$ is ${\mathcal{F}}_{n+1}$-measurable because comparison of two real coordinates defines a Borel set. Define its indicator function by $B_{n+1}$. The recursion becomesIt is a measurable function of that pair. Mathematical induction proves $X_n$ is ${\mathcal{F}}_n$-measurable for every $n$, so $(X_n)$ is an adapted process. The same recursion also proves $0<X_n<1$ at every finite time, a useful bound for the later integrability arguments.

Consider the Borel setIts indicator function is nonnegative and measurable, so the Tonelli theorem applies to the product measure, without needing any separate absolute-integrability argument. At a fixed $y\in (a,b]$, its section in the first coordinate is $(a,y]$; outside $(a,b]$ that section is empty. ThereforeThe last equality uses the definition of a cumulative distribution function on the half-closed interval. It retains all point atoms of a measure on the diagonal $x=y$, excludes a possible point atom of a measure at $x=a$, and includes $y=b$, exactly as required by the set $D$. If $a=b$, both sides vanish. This verifies the identity even when the probability measures are not non-atomic measures.^[ref]

The product measure $\mu \otimes \nu$ is the unique probability measure on the sigma-algebra generated by measurable rectangles for whichHere the printed product notation $\mathcal{B}(\mathbb{R})\times \mathcal{B}(\mathbb{R})$ denotes the generated product sigma-algebra, conventionally written $\mathcal{B}(\mathbb{R})\otimes \mathcal{B}(\mathbb{R})$; it is not just the collection of rectangles. On ${\mathbb{R}}^2$ it equals $\mathcal{B}({\mathbb{R}}^2)$: coordinate projections show that rectangles are Borel sets, while open rectangles with rational endpoints give every open set as a countable union and hence generate the Borel sigma-algebra.

Existence is the product measure theorem for finite measures, which are in particular sigma-finite measures. Uniqueness follows because the rectangles form a pi-system generating the product sigma-algebra, and two finite measures agreeing there agree everywhere by the Dynkin lemma.^[ref]

For each integer $m\ge 1$, letIf $D_m$ had $m+1$ distinct members, finite additivity on their disjoint singleton Borel sets would give total measure at least $(m+1)/m>1$, contradicting $\mu (\mathbb{R})=1$. Thus $|D_m|\le m$.

Every positive singleton measure is at least $1/m$ for some sufficiently large $m$. Consequently the set of point atoms of a measure isA countable union of finite sets is at most countable. This proves the claim, including the possibility that the set is empty. The same argument gives the countability of point atoms of a finite measure, with $|D_m|\le m\mu (\mathbb{R})$ in place of $m$.

We prove the Doob-Dynkin lemma here by simple function approximation. Choose $C<\infty$ such that $|Z|\le C$ everywhere, and letEach $Z_m$ is a finite-valued $\sigma (X)$-measurable function and $Z_m\to Z$ pointwise. Write its level sets as $A_{m,j}$, with associated values $z_{m,j}$. By part (i), choose Borel sets $D_{m,j}$ with $A_{m,j}=X^{-1}(D_{m,j})$. Define the Borel measurable functionsAlthough the chosen $D_{m,j}$ need not partition all of $\mathbb{R}$, their inverse images do partition $\Omega$, so $g_m(X)=Z_m$ exactly.

Apply scalar clipping to each $g_m$ with bounds $[-C-1,C+1]$, preserving this equality, and define $g(x)={\mathrm{limsup}}_m{g}_m(x)$. A bounded limit superior of Borel measurable functions is a finite Borel measurable function. At every $\omega \in \Omega$ the limit of a sequence exists and givesThus the factorization is exact for a bounded, exactly $\sigma (X)$-measurable function. If hypotheses are instead given only up to null sets, the corresponding conclusion is a factorization holding almost surely after choosing a representative.

The sigma-algebra generated by a random variable $X$ is the smallest sigma-algebra on $\Omega$ making $X$ a measurable function into $(\mathbb{R},\mathcal{B}(\mathbb{R}))$. Equivalently, it is the intersection of all sigma-algebras containing the inverse images $X^{-1}(C)$ of Borel sets $C$.

Let $\mathcal{D}=\{X^{-1}(C):C\in \mathcal{B}(\mathbb{R})\}$. This collection is already a sigma-algebra: $X^{-1}(\mathbb{R})=\Omega$,The right-hand sets are in $\mathcal{D}$, since the Borel sigma-algebra is closed under the indicated operations. Any sigma-algebra making $X$ measurable must contain every member of $\mathcal{D}$. Conversely $\mathcal{D}$ makes $X$ measurable by definition. HenceThis equality holds exactly as a collection of sets, without completing the sigma-algebra.