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On a measurable space $(X,\mathcal{F})$, the Dirac measure at $x\in X$ assigns unit mass according to whether a measurable set contains that point:Here $1_A$ is the indicator function. This is a probability measure, since ${\delta }_x(X)=1$. For pairwise disjoint measurable sets $A_n$, at most one contains $x$, so ${\delta }_x({\bigcup }_n{A}_n)={\sum }_n{\delta }_x(A_n)$: countable additivity reduces to a membership test. The definition works even when the singleton $\{x\}$ is not a measurable set. ^[ref]

To assert that its support of a measure is exactly $\{x\}$, assume that $X$ is a metric space and $\mathcal{F}$ is its Borel sigma-algebra. The singleton $\{x\}$ is then a closed set and a Borel set, with ${\delta }_x(\{x\})=1$. Every open neighbourhood of $x$ has mass one; every other point $y$ has an open ball of radius less than its distance from $x$, with mass zero. Thus $\mathrm{supp}{\delta }_x=\{x\}$. An abstract measurable space has no specified topology, so this topological conclusion is not part of the bare definition. Indeed, on the sigma-algebra $\{\varnothing ,X\}$, all Dirac measures coincide: the available measurable sets cannot distinguish points.

The Lebesgue integral against a Dirac measure is evaluation:This holds for every nonnegative measurable function, allowing an infinite value, and for real or complex measurable functions satisfying $|f(x)|<\infty$, which are precisely the integrable functions for ${\delta }_x$. For a nonnegative simple function, only the cell containing $x$ contributes to its integral; increasing simple functions and the monotone convergence theorem extend the identity. Splitting into positive and negative parts, and then real and imaginary parts, gives the remaining cases. This is why a Dirac measure turns integration into a point query.

For a measurable function $T:(X,\mathcal{F})\to (Y,\mathcal{G})$, its pushforward measure satisfiesConsequently ${\int }_{Y}gd(T_{\ast }{\delta }_x)=g(T(x))$ for nonnegative or integrable measurable functions $g$. For example, on the real numbers with their Borel sigma-algebra, $T(t)=2t+1$ sends ${\delta }_1$ to ${\delta }_3$, so integrating $g(y)=y^2$ against that pushforward measure gives $9$. ^[ref]

Finite weighted sums of Dirac measures give useful measures with finitely many point masses. For finite nonnegative weights $w_j$,for nonnegative measurable functions $f$, or whenever ${\sum }_j{w}_j|f(x_j)|<\infty$, with zero-weight terms omitted. Repeated locations have their weights added. On the real numbers, this is the finite version of the measure underlying the left-continuous cumulative function of an atomic measure; a positive-weight singleton is an atom of a measure. Unit weights at the distinct points of a finite set recover counting measure on that set. ^[ref]

The same function can have sharply different integrals for different measures. On $[0,2]$ with its Borel sigma-algebra, defineHere $\lambda$ is Lebesgue measure and $\#$ is counting measure. The singleton $\{1\}$ is a Lebesgue null set, so its changed value leaves the Lebesgue integral equal to ${\int }_0^2(t+1)dt=4$. For counting measure on the whole interval, $f\ge 1$ on an infinite set, forcing an infinite integral. On the smaller measure space $D=\{0,1,2\}$ with its power set and counting measure, the restricted function instead has integral $1+5+3=9$.

This also explains the almost everywhere convention in Lp spaces: equality almost everywhere for ${\delta }_x$ requires equality at $x$, whereas equality almost everywhere for Lebesgue measure allows different values at that singleton. A point value therefore cannot be read unambiguously from an equivalence class in an Lp space formed using Lebesgue measure.

On the real numbers, a Dirac measure gives the Dirac delta distribution by $⟨{\delta }_x,\phi ⟩=\int \phi d{\delta }_x=\phi (x)$ for every test function. The name Dirac delta function does not mean an ordinary function that is zero away from $x$ and infinitely high at $x$. Such a nonnegative measurable function still has Lebesgue integral zero, because its nonzero values occur on a Lebesgue null set. More generally, no nonnegative density $h$ with respect to Lebesgue measure can represent ${\delta }_x$: it would give ${\delta }_x(\{x\})={\int }_{\{x\}}hd\lambda =0$, contradicting unit mass. The measure and its associated distribution act on different kinds of inputs, namely measurable sets and test functions, but express the same point evaluation. ^[ref]

Let $H_0^1$ be the Cameron-Martin space of Wiener measure on $[0,T]$: its elements start at zero, are absolutely continuous, and have square-integrable derivatives. For $h\in H_0^1$, let $y^h$ solve the controlled ordinary differential equationThe Stroock-Varadhan support theorem in the uniform topology isOne may equivalently take piecewise linear or smooth controls starting at zero. The skeleton uses the displayed vector fields, as appropriate for the Stratonovich equation.

Choose $2<p<3$ and work in the complete separable space of anchored geometric p-rough paths, with its rough path metric $d_p$. The enhanced Brownian motion is a random element of this space. Its support of a measure isWe explain this Brownian support step using the permitted results about the enhancement. The almost-sure polygonal convergence proved in Question 3 gives the inclusion of its support in $K$. Indeed, polygonal controls belong to $H_0^1$, and the enhancement belongs to their closed signature closure almost surely. Hölder rough-path convergence at an exponent greater than $1/p$ implies $p$-variation convergence: for the level-$k$ differences, sum the bound proportional to $|t-s{|}^{kp\alpha /k}=|t-s{|}^{p\alpha }$ over a partition. Since $p\alpha >1$, these sums are uniformly bounded and tend to zero with the Hölder distance.

For the reverse inclusion, use Brownian rough path small-ball positivity, namely $\mathbb{P}(d_p(\mathbf{B},e)<r)>0$ for every $r>0$, and Cameron-Martin quasi-invariance of enhanced Brownian motion. These are the Brownian input results used without proof here. For $h\in H_0^1$, the rough path translation $T_h$ is continuous and sends $e$ to $S_2(h)$. The law of $T_h\mathbf{B}$ is equivalent to that of $\mathbf{B}$. Given an open neighborhood $U$ of $S_2(h)$, continuity supplies an identity neighborhood mapped into $U$. Small-ball positivity gives positive probability that $T_h\mathbf{B}\in U$; equivalence of the two laws gives $\mathbb{P}(\mathbf{B}\in U)>0$. Hence $S_2(h)$ belongs to the Brownian support for every $h$. Closedness of support proves the equality with $K$.

Let $\Phi$ be the state-path solution map of the rough differential equation driven by these vector fields, with the fixed initial value $y_0$. The assumed identification of the Stratonovich solution says $Y=\Phi (\mathbf{B})$. Boundedness of the fields and all their derivatives ensures global existence and the regularity needed by the universal limit theorem. That theorem makes $\Phi$ continuous from the rough-path space into the uniform state-path space. For a control signature, consistency with ordinary equations gives $\Phi (S_2(h))=y^h$.

We use the elementary support under a continuous map identityTo prove it, a neighborhood of $f(x)$ with $x\in \mathrm{supp}\mu$ has an open inverse image containing $x$, and therefore positive measure. This puts $f(\mathrm{supp}\mu )$, and its closure, in the image support. Conversely, the image closure is closed and has full image measure, since the original support has full measure. The minimal closed full-measure characterization from Question 4(i) gives the other inclusion.

Apply this identity to $f=\Phi$ and the Brownian rough-path law. We obtainContinuity and the definition of $K$ imply $\Phi (K)\subset \overline{\{\Phi (S_2(h)):h\in H_0^1\}}$. The reverse inclusion after taking closures follows from $S_2(h)\in K$. Therefore the last displayed support is exactly the uniform closure of the skeleton paths, proving the theorem.

Finally, piecewise constant and smooth approximations of $\dot{h}$ converge in $L^2$, hence in $L^1$ on a finite interval. Their controls converge in total variation, their signatures converge in the rough-path metric, and the universal limit theorem gives uniform convergence of the corresponding skeleton solutions. This proves the stated equivalence of the control classes.

For a Borel probability measure $\mu$ on a Polish space $E$, its support of a measure isEquivalently, for a compatible metric, every open ball about $x$ has positive measure. The complement is open: any zero-measure open neighborhood of a point is a zero-measure neighborhood of all its points. Hence the support is closed.

A Polish space has a countable base. Its support complement is therefore a countable union of zero-measure basic open sets, so $\mu (\mathrm{supp}\mu )=1$. If $F$ is any closed full-measure set, a point outside $F$ has the zero-measure open neighborhood $E\setminus F$ and is outside the support. Thus the support is also the smallest closed set of full measure. Countability is what justifies the full-measure assertion; an arbitrary uncountable union of null sets would not suffice.

Work on a finite interval $[0,T]$ with a fixed initial point $y_0\in {\mathbb{R}}^m$. To give a clean global statement, assume $V_0,V_1,\dots ,V_d$ are bounded vector fields with bounded continuous derivatives through order three. Consider the Stratonovich integral form of the stochastic differential equationFor a control in the Cameron-Martin space of Wiener measure,let $y^h$ solve the controlled ordinary differential equationThe Stroock-Varadhan support theorem says, in the uniform metric on continuous paths,Thus a path is in the support of a measure of the diffusion precisely when it can be approximated by these deterministic controlled trajectories. There is no ellipticity assumption: the allowed controls and vector fields already encode any restrictions on reachability.

The central point of the rough-path proof is that ordinary uniform approximation of the driver does not make the solution map continuous. Small planar loops can have small amplitude but accumulated area of fixed size, and vector fields with a nonzero Lie bracket detect that area. The correct driver is enhanced Brownian motion $\mathbf{B}$, including Lévy area. Fix $2<p<3$. The universal limit theorem gives a continuous mapfrom the geometric $p$-variation rough-path space to solution paths. It agrees with the controlled ordinary differential equation when $\mathbf{X}=S_2(h)$. Adding the time coordinate, with its mixed integrals defined by integration against $dt$ and integration by parts, incorporates $V_0$ without changing the continuity argument. This is the role of the universal limit theorem: it turns a support problem for all these nonlinear diffusions into a support problem for one enhanced driving process.

For $\mathbf{X}=\mathbf{B}$, the rough differential equation is the stated Stratonovich equation. Its second-order increment includesReplacing the Stratonovich integral in the second level by its Itô integral expression contributes $\frac{1}{2}{\sum }_{i}D{V}_i{V}_{i}dt$, which is exactly the Stratonovich-to-Itô correction. Polygonal convergence from Question 3 and the universal limit theorem therefore give the Wong-Zakai approximation: the ordinary equations driven by $B^n$ converge to this Stratonovich diffusion.

It remains to establish the support of enhanced Brownian motion:One inclusion follows immediately from Question 3. Every polygonal $B^n$ belongs to $H$, and $S_2(B^n)\to \mathbf{B}$ almost surely in the rough path metric. Since $K$ is closed, $\mathbf{B}\in K$ almost surely, whence its support is contained in $K$. This alone does not prove the reverse inclusion: convergence of random approximations does not automatically give positive mass near every control.

For that inclusion, take dyadic mesh ${\delta }_n=T{2}^{-n}$ and writeOn each grid interval $R^n$ is a pinned Brownian bridge. The bridges are mutually independent and independent of the grid values defining $B^n$, by the Gaussian projection formula. Their canonical Stratonovich enhancement ${\mathbf{R}}^n$ is therefore independent of $B^n$. It is important to enhance the bridge residual itself, including its area, rather than infer area smallness from its uniform size.

The bridge enhancement tends to the zero rough path in probability. Here are the estimates behind this assertion. A bridge on an interval of length ${\delta }_n$ scales as $\sqrt{{\delta }_n}$ times a unit bridge; its full antisymmetric area has mean zero, moments of order ${\delta }_n$, and second moment of order ${\delta }_n^2$. Since its endpoint displacement is zero, at grid points the cumulative second level is just the sum of the independent bridge areas. The Doob L2 maximal inequality givesPartial intervals at the two ends contribute bridge increments and at most two local areas. Bridge scaling, Gaussian moment bounds and the corresponding iterated-integral estimates yield, uniformly in $n$ and for arbitrarily large $q$,For an interval crossing several grid cells, the full-cell second levels are independent centered areas, whose sum has moments bounded by $C_q((t-s){\delta }_n{)}^{q/2}$; the two boundary pieces and their cross product satisfy the displayed bound as well. For an interval inside one cell these are the usual scaled bridge increment and area bounds. At each fixed pair $(s,t)$ both levels converge to zero in probability: the first-level variance is $O({\delta }_n)$, and the second-level second moment is $O((t-s){\delta }_n+{\delta }_n^2)$ for sufficiently fine grids. The Kolmogorov continuity theorem gives uniformly bounded moments of a Hölder rough-path norm at any exponent $\beta <1/2$. Tightness, convergence on a dense set of pairs and interpolation to $\alpha <\beta$ now give ${\rho }_{\alpha }({\mathbf{R}}^n,e)\to 0$ in probability. Taking $1/p<\alpha <1/2$ gives $d_p({\mathbf{R}}^n,e)\to 0$ in probability.

Fix $h\in H$. Its polygonal interpolations $h^n$ converge to $h$ in the total-variation norm: their derivatives are the cell averages of $\dot{h}$, which converge in $L^2$, hence in $L^1$. Moreover $|h^n{|}_{1\text{-var}}\le |h{|}_{1\text{-var}}$. Thus $S_2(h^n)\to S_2(h)$ in the rough path metric.

The rough path translation operation satisfiesFor a bounded variation path $b$, translation adds $b$ at level one and, at level two, the mixed termsThe second mixed term is defined by integration by parts. Their variation bounds show that translation is continuous, uniformly when $|b{|}_{1\text{-var}}$ is bounded, near the zero rough path. More explicitly the mixed second-level norms are bounded by a constant times $\Vert R{\Vert }_{p\text{-var}}|b{|}_{1\text{-var}}$; the other new second-level term is the path signature of $b$.

Given a ball of radius $\epsilon$ around $S_2(h)$, choose a sufficiently large $n$ so that $S_2(h^n)$ is close andfor the small $\eta$ supplied by translation continuity. The relevant variation bounds can be chosen uniformly in $n$, since the variations of the $h^n$ are uniformly bounded. For this fixed $n$, the event $|B^n-h^n{|}_{1\text{-var}}<\eta$ has positive probability: the finite vector of Brownian increments has an everywhere positive Gaussian density, and sufficiently small deviations of its vertices imply this variation bound. These two events are independent. On their intersection, translation continuity gives $d_p(\mathbf{B},S_2(h))<\epsilon$. Every ball around $S_2(h)$ therefore has positive probability. This proves $S_2(h)$ lies in the support, and taking closures gives $K$.

Finally, a continuous map between the separable path spaces in use satisfiesFor one direction the measure is concentrated on its support. For the other, an open neighborhood of $\Phi (x)$ pulls back to an open neighborhood of a support point $x$, which has positive measure. Using continuity of $\Phi$ and the identity $\Phi (S_2(h))=y^h$ proves the boxed Stroock-Varadhan support theorem. This separates the two ingredients: the probabilistic support calculation for Brownian motion and area, and the deterministic continuity supplied by the universal limit theorem.

If the diffusion is instead presented as an Itô integral equation with drift $b$, the controlled equation must useUsing $b$ unchanged in the controlled equation would generally give the wrong support description. The result above states explicit sufficient smoothness and nonexplosion assumptions; extensions to more general coefficients require the usual localization, not an unqualified assertion for arbitrary vector fields.

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.

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.

Write $d\mu =f(x)dx$, with $f\ge 0$. Applying polar coordinates and Fubini's theorem gives, for bounded nonnegative measurable $h$,Also $\tilde{\mu }(\{0\})=\mu (\{0\})=0$ by absolute continuity of measures. Extending $g$ by zero on the negative half-line proves

Both radial pushforward measures have compact support of a measure, so Brenier's theorem on the line supplies a unique optimal transport plan and a transport map $\tau$, unique $\tilde{\mu }$-almost everywhere, withIn one dimension, the gradient of a convex function is a nondecreasing function. Nonnegativity holds $\tilde{\mu }$-almost everywhere because ${\tau }_{\#}\tilde{\mu }=\tilde{\nu }$ is carried by $[0,\infty )$. To choose a nonnegative, nondecreasing representative everywhere, let $F,G$ be the radial cumulative distribution functions and use the monotone rearrangementAt $F(r)=0$ set $\tau (r)=0$, and at $F(r)=1$ use the largest point of the support of a measure $\tilde{\nu }$. These choices preserve monotonicity and boundedness. Since $F$ is continuous, $F(R)$ has the uniform distribution on $(0,1)$ when $R$ has law $\tilde{\mu }$, and the quantile function $G^{-1}$ sends this uniform distribution to $\tilde{\nu }$. Thus this representative has the required pushforward measure, even if the target has atoms.

For $2<p<3$, the support of a measure of enhanced Brownian motion is the rough-path closure of path signatures of paths in the Cameron-Martin space of Wiener measure. Polygonal convergence gives one inclusion. For the reverse inclusion, decompose Brownian motion into its polygonal interpolation and independent pinned bridges. The bridge enhancement becomes small in probability as the mesh decreases, while every finite-dimensional neighborhood of the polygonal vertices has positive Gaussian probability. Rough path translation then gives positive probability in every ball around a prescribed control path signature.