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Put ${\mathcal{F}}_n=\sigma (B_{k{2}^{-n}}:0\le k\le 2^n)$. These sigma-fields increase, and their union generates the continuous Brownian path. Let ${\mathbf{B}}^{(n)}=S_2(B^{(n)})$ be the polygonal enhancement. Its coordinates are determined by its displacement $b_{s,t}^{(n)}$ and antisymmetric area $a_{s,t}^{(n)}$; the symmetric second level is $\frac{1}{2}{b}_{s,t}^{(n)}\otimes b_{s,t}^{(n)}$.

The given conditional-expectation identities implyFor the second assertion it is important to use area coordinates rather than asserting that the whole tensor signature is a conditional expectation. Write $I_t={\int }_0^t\beta d\tilde{\beta }$. The local off-diagonal second integral isConditional on their separate dyadic values, the two independent Brownian components remain independent. Thus the conditional expectation of the product in the last term is ${\beta }_s^{(n)}({\tilde{\beta }}_t^{(n)}-{\tilde{\beta }}_s^{(n)})$. The assumed identity for $I$ gives the polygonal $12$ entry. The reverse entry follows by integration by parts, since the product of the two component increments also has the product of their conditional means. Taking the antisymmetric part proves the area identity. Diagonal second levels are recovered from the geometric identity, rather than from a conditional-expectation claim about squares.

For each fixed dyadic time $t$, the first-level and area conditional expectations are uniformly integrable martingales in $n$. Their targets are measurable with respect to ${\mathcal{F}}_{\infty }=\sigma ({\bigcup }_n{\mathcal{F}}_n)$: Brownian area is a limit of stochastic integral sums determined by the path. The martingale convergence theorem gives convergence to $B_t$ and $A_{0,t}$, respectively. The continuous group coordinate formulas then giveat all dyadic times on a common event of probability one.

For the uniform Hölder estimate, fix $\alpha <1/2$ and put $K=\Vert \mathbf{B}{\Vert }_{\alpha \text{-Ho¨lder};[0,1]}$. The given all-order integrability and the group norm comparison implyUse the conditional-expectation formulas, the conditional Jensen inequality, and $\mathbb{E}[K\mid {\mathcal{F}}_n]\le \sqrt{\mathbb{E}[K^2\mid {\mathcal{F}}_n]}$. The group norm comparison yieldsFirst take all dyadic pairs, a countable collection, and then extend the inequality to all $s,t$ by continuity. Hence the Brownian rough path martingale Hölder bound isFor any $q>1$, the Doob Lp maximal inequality givesTherefore

To upgrade convergence from dyadic times to every time simultaneously, apply this bound at one positive exponent $\beta <1/2$. The polygonal enhancements are then equicontinuous in the group metric, with a common finite random constant. For a finite dyadic time net, convergence holds at every net point. The triangle inequality bounds the errors at the remaining times by the two Hölder errors to the nearest net point and the error at that point. First let $n\to \infty$, and then let the net mesh tend to zero. This proves uniform convergence, and in particular convergence for every $t\in [0,1]$ on one event of probability one.

There is also convergence in each weaker Hölder rough path metric. Uniform convergence of the group coordinates gives uniform convergence of their levelwise increments through the Chen formulas. If $\alpha <\beta <1/2$, the uniform $\beta$ bounds bound the level-$k$ increment differences by a random constant times $|t-s{|}^{k\beta }$, while their unweighted suprema tend to zero. Interpolating these two bounds proves convergence with denominators $|t-s{|}^{k\alpha }$, for $k=1,2$. This establishes the nested polygonal approximation of Brownian rough paths in the topology needed for the support argument.

Fix $T<\infty$ and a standard $d$-dimensional Brownian motion $B$ starting at zero. Let $D_n$ be deterministic nested finite partitions of $[0,T]$ with mesh tending to zero; dyadic partitions are the basic example. Let $B^n$ be the polygonal interpolation through the partition values and put ${\mathbf{B}}^n=S_2(B^n)$. The limit is enhanced Brownian motion, with incrementsHere the integral is a Stratonovich integral. In terms of Itô integrals,Its symmetric part is $\frac{1}{2}{B}_{s,t}^{\otimes 2}$, while its antisymmetric part is Lévy area,These identities and splitting the integral at an intermediate time give the Chen identity, so $\mathbf{B}$ has values in $G^2({\mathbb{R}}^d)$.

A precise nested polygonal approximation of Brownian rough paths statement is: for every $0<\alpha <1/2$,where the rough path metric used here isIn particular, for each $2<p<3$, choosing $1/p<\alpha <1/2$ gives convergence in the inhomogeneous $p$-variation rough path metric. The limiting signed area is the Lévy area, and the limiting full second level is Stratonovich, not Itô. This distinction is necessary: a polygonal lift has the prescribed symmetric second level, whereas an Itô lift would lack half the quadratic-variation correction.

Here is a proof sketch emphasizing the nested martingale structure. SetA Brownian bridge between two prescribed endpoints has the affine interpolation as conditional mean. ConsequentlyFor $i\ne j$ the coordinate processes remain independent when their respective partition values are conditioned on. Approximating the Itô integral by its left sums, conditioning each product, and then taking its $L^2$ limit givesIndeed each conditioned factor is its polygonal interpolation, and their sums converge to the ordinary integral of those piecewise linear functions. For a fixed pair $s,t$, the right side is therefore a martingale in the refinement index $n$. It is not being asserted to be a martingale in physical time; polygonal interpolation uses the right endpoint of its current cell.

The union of the partitions is dense. Continuity of $B$ implies that ${\mathcal{G}}_{\infty }=\sigma ({\bigcup }_n{\mathcal{G}}_n)$ contains the full Brownian path and its stochastic integrals, up to completion. The martingale convergence theorem for conditional expectations therefore yields convergence of $B_t^n$ and all off-diagonal ${\mathbb{B}}_{s,t}^{n,ij}$ at every fixed pair, almost surely and in $L^q$ for finite $q$. Simultaneously it holds for a countable dense collection of pairs. The diagonal terms must be handled separately:They are not conditional expectations of the diagonal Brownian second level: the latter would also contain a conditional variance.

Pointwise convergence alone is insufficient for the rough path metric. The Burkholder-Davis-Gundy inequalities, Brownian scaling and Gaussian increment moments give, for arbitrarily large $q$,Equivalently, using the homogeneous norm on $G^2$, the $q$-th moment of the Carnot-Carathéodory distance of an increment is bounded by $C_q|t-s{|}^{q/2}$. The metric form of the Kolmogorov continuity theorem, with $q$ chosen large, gives for every $\beta <1/2$ a random $K_{\beta }$ having all finite moments such thatConditioning these bounds givesThe diagonal bounds follow by squaring the first bound. The two conditional-expectation processes are nonnegative martingales; the Doob Lp maximal inequality makes their suprema over $n$ finite almost surely, with all needed moments. Thus all the polygonal lifts have a common random bound at exponent $\beta$.

These bounds and the Chen identity give joint equicontinuity of the increments as functions of $(s,t)$. Convergence on the dense collection of pairs consequently gives uniform convergence of both levels on the time triangle. Take $\alpha <\beta <1/2$. For either level $k=1,2$, with $M$ a common bound and ${\epsilon }_n$ its uniform difference, interpolation givesThis proves the stated Hölder rough path metric convergence. Finally, for a partition $D$, the $k$-th level difference satisfiesbecause $\alpha p>1$. Taking the appropriate powers and suprema proves $p$-variation convergence too. Nesting is what permits the conditional-expectation martingale proof; without nesting one instead needs direct approximation estimates.