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The integrand $\mathrm{sign}({\beta }_s)$ is a bounded previsible process, so $B$ is a continuous local martingale. The quadratic variation of a stochastic integral isbecause the Brownian zero set has zero Lebesgue measure. Since $B_0=0$, the Lévy characterization of Brownian motion shows that $B$ is a standard Brownian motion.

LetThe clock $A$ is an absolutely continuous function and is strictly increasing: its derivative is positive away from the Brownian zero set, which has zero Lebesgue measure. It also tends to infinity. This is immediate for $\delta =1$; for $\delta >1$, recurrence and the Strong Markov property imply that the Brownian occupation time of, for example, $\{x:1<|x|<2\}$ is unbounded, while the integrand is bounded below there by ${\delta }^2$.

Let $Z=\{t:B_t=0\}$. Since $A$ is the clock from part (b) and $\tau =A^{-1}$,For $\delta >1$, the absolutely continuous function $A$ has derivative $A^{\prime }(t)={\delta }^2|B_t{|}^{2\delta -2}=0$ on $Z$. The one-dimensional area bound for an absolutely continuous function therefore givesFor $\delta =1$, $A_t=t$ and the conclusion follows directly because the Brownian zero set has zero Lebesgue measure. Hence the zero set of $X$ has zero Lebesgue measure almost surely.

Fix $x\in \mathbb{R}$ and define ${\mu }_x(B)=\lambda (B+x)$. This is a $\sigma$-finite measure on the Borel sets. For every half-open interval,The half-open intervals form a generating $\pi$-system, so the $\sigma$-finite uniqueness theorem gives ${\mu }_x=\lambda$. Thus Lebesgue measure is translation invariant.

Let $D>0$ be the diameter of the bounded set $S$, and choose $x_0\in S$. Then $S\subseteq B(x_0,D)$. It is enough to prove a volumetric covering bound for the unit ball.

Choose a maximal $1/4$-separated set $N$ in $B(0,1)$. The balls of radius $1/8$ centred at points of $N$ are disjoint and lie in $B(0,9/8)$. Comparing volumes givesMaximality means that the balls of radius $1/4$ centred at $N$ cover the unit ball.

After translating and scaling, at most $9^n$ balls of radius $D/4$ cover $S$. Assign each point of $S$ to one covering ball containing it. This gives at most $9^n$ disjoint pieces, each of diameter at most $D/2<D$. Thus the claim holds with the absolute constant $C=9$.

The discrete base factors in the displayed construction inherit their variational parameters. Accordingly write $s_{\lambda }(\epsilon )$, not a parameter-independent $s(\epsilon )$, for the joint base probability mass function or probability density function. The continuous noise factors can be parameter independent while the discrete factors are $\lambda$ dependent. Calling the entire base law parameterless while retaining its nonzero score function would be inconsistent.

Let $t_{\lambda }$ push the base law forward to $q_{\lambda }$, and defineUse a fixed base measure $\nu$, with counting measure for discrete coordinates and Lebesgue measure for continuous coordinates. Assume locally parameter-independent support, differentiability and an integrable domination permitting differentiation under the integral sign. A valid reparameterization gradient givesApply the product rule before simplifying:This is the required combined score and pathwise gradient estimator. A sample from $s_{\lambda }$ and the integrand on the last line give an unbiased stochastic gradient under the stated regularity conditions. Its first term is a score-function gradient estimator contribution; the remaining terms differentiate the sampled objective along the continuous transformation. The same separation appears in the general stochastic computation graph construction.^[ref]

All derivatives inside the expectation hold $\epsilon$ fixed. They still include both explicit parameter dependence and dependence through $t_{\lambda }$. For a continuous coordinate $z=t_{\lambda }(\epsilon )$,The model joint density $p(z,x)$ is fixed as $\lambda$ changes, but its composition with $t_{\lambda }$ need not be constant. Discrete coordinates are not differentiated as real-valued paths; their changing probabilities enter through the score function. Moving supports or parameter-dependent branch boundaries require additional care and are not justified merely by formally interchanging a derivative and an integral.

Compact support of a measure gives finite second absolute moments for both probability measures. Since the source satisfies absolute continuity of measures with respect to Lebesgue measure, Brenier's theorem identifies the quadratic Kantorovich optimal transport problem with the Monge optimal transport problem:Here $S_{\#}\mu$ is the pushforward measure. The optimal transport plan is unique and equals $(\mathrm{Id},T{)}_{\#}\mu$. The optimal transport map is unique up to a $\mu$-null set, withThe gradient of this convex function is c-cyclically monotone for $c(x,y)=|x-y{|}^2$ on a full-measure set. In particular, summing the two supporting inequalities for the convex function gives $(T(x)-T(x^{\prime }))\cdot (x-x^{\prime })\ge 0$ wherever both gradients exist. No regularity of the target beyond the stated probability measure assumptions is needed.

A null set has Lebesgue outer measure zero. Equivalently, for every $\epsilon >0$ it can be covered by countably many open intervals whose total length is less than $\epsilon$. Such a set is automatically Lebesgue measurable and has Lebesgue measure zero.

Let $d=h-g$. It vanishes outside the null set $E$, so part (e) proves $d$ is a measurable function, regardless of its values on $E$. Since the sum of two measurable functions is measurable, $h=g+d$ is measurable. This is the completeness of Lebesgue measure under changes on null sets.

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.

The function is measurable and $0\le {\cos }^2(1/x)\le 1$. Since $(0,1)$ has finite Lebesgue measure, comparison gives ${\int }_0^{1}g\le 1$. Therefore $g$ is integrable on $(0,1)$, despite oscillating infinitely often near zero.

For $x\ge 1$, the bound $|\sin u|\le |u|$ gives $|g(x)|\le x^{-3/2}$, whose tail integral is finite. Thus $g{1}_{[1,\infty )}$ is integrable.

Near one, $h(x)=[(x-1)(x+1){]}^{-1/2}$ has an integrable inverse-square-root singularity. At infinity, however, $h(x)\ge 1/x$ for $x>1$, and the harmonic tail diverges. Thus $h{1}_{[1,\infty )}$ is not integrable. Assigning $h(1)=0$ changes neither conclusion, since one point has Lebesgue measure zero.

A null set has Lebesgue outer measure zero. Equivalently, for every $\epsilon >0$ it can be covered by countably many open intervals whose total length is less than $\epsilon$. Such a set is automatically Lebesgue measurable and has Lebesgue measure zero.

Let $d=h-g$. It vanishes outside the null set $E$, so part (e) proves $d$ is a measurable function, regardless of its values on $E$. Since the sum of two measurable functions is measurable, $h=g+d$ is measurable. This is the completeness of Lebesgue measure under changes on null sets.

The squared norm of $p_{\ell }(x)e^{-x^2/4}$ in $L^2(\mathbb{R},dx)$ is $\sqrt{2\pi }{h}_{\ell }$, where $(h_0,h_1,h_2)=(1,1,2)$. Choose positive normalizing constantsThen ${\varphi }_0,{\varphi }_1,{\varphi }_2$ are an orthonormal set for Lebesgue measure. Expanding the finite sums givesThus this is an orthogonal polynomial projection kernel, explicitlyThe positive constants above are forced by the reproducing identity: before normalization, its convolution multiplies each independent term ${\varphi }_{\ell }(x){\varphi }_{\ell }(z)$ by $\sqrt{2\pi }{h}_{\ell }/{\kappa }_{\ell }^2$. Each multiplier must equal one. Changing the sign of a real ${\kappa }_{\ell }$ leaves the kernel unchanged.

Evaluate the orthogonal polynomial projection kernel on its diagonal:The ${\lambda }^2$ terms cancel. Integrating against Lebesgue measure gives $(\mathbb{E}{Z}^4+3)/2=3$ for a standard normal random variable $Z$, checking the correlation function of a point process normalization.

The absolute continuity of measures condition isThe Radon-Nikodym theorem says that if both positive measures are sigma-finite measures, then this condition is equivalent to the existence of a nonnegative measurable function $h$, unique up to equality $\kappa$-almost everywhere, such thatOne writes $h=d\nu /d\kappa$, the Radon-Nikodym derivative. Under these hypotheses $h$ is finite $\kappa$-almost everywhere. If $\nu$ is finite, then $h\in L^1(\kappa )$. A finite signed measure absolutely continuous with respect to a sigma-finite positive measure likewise has an integrable signed Radon-Nikodym derivative.

The sigma-finiteness hypothesis on the dominating measure cannot simply be omitted. On the Borel subsets of $[0,1]$, let $\kappa$ be counting measure and $\nu$ Lebesgue measure. Absolute continuity holds because the only $\kappa$-null set is empty. A Radon-Nikodym derivative would satisfy $h(x)=\nu (\{x\})=0$ at every point, contradicting $\nu ([0,1])=1$. Here counting measure is not sigma-finite.

For $E\subseteq {\mathbb{R}}^n$, define the Lebesgue outer measure bywhere the boxes are bounded and $\mathrm{vol}(Q_j)={\prod }_r(b_{jr}-a_{jr})$. The empty covering handles the empty set. Monotonicity follows by inclusion of admissible covers. Taking covers within $\epsilon 2^{-j}$ of the outer measures of a sequence of sets proves countable subadditivity; when the sum is infinite the inequality is automatic. Lebesgue measure is the restriction of ${\mathcal{L}}^{\ast }$ to its Caratheodory-measurable sets.

If $d(A,B)=\delta >0$, subdivide each box of any countable cover of $A\cup B$ into finitely many half-open boxes of diameter smaller than $\delta$, preserving its total volume. No small box can meet both $A$ and $B$. The boxes meeting $A$ cover $A$, and the disjoint family meeting $B$ covers $B$, so the original cover costs at least ${\mathcal{L}}^{\ast }(A)+{\mathcal{L}}^{\ast }(B)$. Taking the infimum and then using subadditivity givesThus the outer measure underlying Lebesgue measure is a metric outer measure, and in particular every Borel set is Lebesgue measurable.

The Lebesgue outer measure iswhere $|Q|$ is the product of its side lengths. The infimum ranges over countable covers; allowing other standard endpoint conventions gives the same outer measure by arbitrarily small enlargement. Lebesgue measure is its restriction to Lebesgue measurable sets.

Enlarged unit cubes give ${\mathcal{L}}^{\ast }([0,1{]}^n)\le 1$. For any countable open-box cover of this compact cube choose a finite subcover. A finite grid formed from its box endpoints and the cube endpoints partitions the cube into rectangular cells. The sum of the volumes of covering boxes is at least the sum of these cell volumes, which is one. Boundary faces have elementary zero volume and do not affect the finite volume comparison. Thus every covering sum is at least one, proving

The existence assertion is false under the hypotheses printed in the source. No measurability of $\theta$ is specified. Even requiring $\theta$ to be a Borel measurable function does not suffice: a coefficient can be positive at every point while approaching zero too rapidly near the starting point. This is the obstruction described by pointwise positivity does not ensure weak existence.

Here is a counterexample in dimension one, with $\Theta =1$:Suppose that a solution to the martingale problem with domain $C_c^{\infty }(\mathbb{R})$ existed. The cutoff argument from part (a)(ii), now with $\mathcal{L}=\theta \Delta$, makes $X$ a continuous local martingale withThe criterion for a continuous local martingale with bounded quadratic variation makes it a square-integrable martingale, and the Itô isometry gives $\mathbb{E}{X}_t^2=\mathbb{E}[X{]}_t$.

Apply the occupation-times formula with the indicator of the singleton $\{0\}$. Since that singleton has zero Lebesgue measure,Thus $X$ spends zero ordinary time at zero. Away from zero, $\theta (x)\le x^2$, soThe Gronwall inequality forces $\mathbb{E}{X}_t^2=0$ for every $t$. Continuity then forces $X\equiv 0$ almost surely, contradicting either the zero occupation time just proved or $[X{]}_t=2t$ for that constant path. Consequently

For completeness, the intended construction works if $\theta$ is a Borel measurable function and bounded away from zero on every compact set. In particular, it works for continuous strictly positive $\theta$. Let $B$ be standard $d$-dimensional Brownian motion, and define a clock and its inverse byThe Brownian path on each finite interval has compact range, so the added hypothesis makes $A_u$ finite. The clock is continuous and strictly increasing, and $A_u\ge u/(2\Theta )$, so $A_u\to \infty$. Its inverse is continuous and satisfies ${\tau }_t\le 2\Theta t$. Each ${\tau }_t$ is a stopping time for the Brownian filtration.

For $f\in C_c^{\infty }({\mathbb{R}}^d)$, the optional sampling theorem applied at these bounded random times shows thatis a true martingale in the filtration ${\mathcal{G}}_t={\mathcal{F}}_{{\tau }_t}^B$. The inverse-clock substitution $d{\tau }_t=2\theta (X_t)dt$, valid almost everywhere, rewrites it asThis proves the Brownian time change for a variable diffusion speed:The factor $2$ in the clock is necessary because standard Brownian motion has infinitesimal generator $\Delta /2$.