[proofplan] The claim is proved level-by-level for the iterated integrals $S(\cdot)^{(k)}$. We first reduce to the bounded-variation case $p = 1$, where the iterated integrals are classical Stieltjes integrals. The $p=1$ case is then proved by induction on the level $k$: the base case $k = 1$ is the change-of-variable formula for Stieltjes integrals, and the inductive step uses the image-measure identity $\lambda_* \mu^i_{x \circ \lambda} = \mu^i_x$ between the Lebesgue–Stieltjes measures of $x \circ \lambda$ and $x$. Once the equality is established for $p = 1$, it extends to $1 < p < 2$ by approximating $x \in C_p([a,b], V)$ by bounded-variation paths (which are dense in $q$-variation for any $q > 1$) and using the joint continuity of the Young integral in the $q$-variation topology when $1/p + 1/q > 1$. [/proofplan]

[step:Set up notation for iterated integrals and reduce to coordinate-wise statements] Fix a basis $(e_1, \dots, e_d)$ of $V$ so that $x = \sum_{i=1}^d x^i e_i$ with $x^i: [a,b] \to \mathbb{R}$, and likewise $x \circ \lambda = \sum_{i=1}^d (x \circ \lambda)^i e_i$ with $(x \circ \lambda)^i = x^i \circ \lambda$. The level-$k$ component of the signature is \begin{align*} S(x)^{(k)}_{[a,t]} = \sum_{(i_1, \dots, i_k) \in \{1,\dots,d\}^k} S(x)^{(i_1, \dots, i_k)}_{[a,t]} \, e_{i_1} \otimes \cdots \otimes e_{i_k}, \end{align*} where the scalar coordinates are defined recursively by \begin{align*} S(x)^{(i_1, \dots, i_k)}_{[a,t]} = \int_a^t S(x)^{(i_1, \dots, i_{k-1})}_{[a,r]} \, dx^{i_k}_r, \end{align*} with $S(x)^{()}_{[a,t]} = 1$ as the level-zero convention. To prove $S(x)_{[a,b]} = S(x \circ \lambda)_{[c,d]}$ as elements of the tensor algebra, it suffices to show \begin{align*} S(x)^{(i_1, \dots, i_k)}_{[a,b]} = S(x \circ \lambda)^{(i_1, \dots, i_k)}_{[c,d]} \end{align*} for every $k \in \mathbb{N}$ and every multi-index $(i_1, \dots, i_k) \in \{1, \dots, d\}^k$. We write $x^\lambda := x \circ \lambda$ and $(x^\lambda)^i := x^i \circ \lambda$. [/step]

[step:Establish the case $p = 1$ via the change-of-variable formula for Stieltjes integrals]Assume $p = 1$, so $x \in C_1([a,b], V)$ has bounded variation. Each coordinate $x^i$ is then of bounded variation on $[a,b]$, and the [Lebesgue–Stieltjes measure](/page/Lebesgue-Stieltjes%20Measure) $\mu^i_x$ on $[a,b]$ associated to $x^i$ is well-defined by \begin{align*} \mu^i_x((s,t]) = x^i_t - x^i_s. \end{align*} The composition $x^\lambda$ also has bounded variation: total variation is non-increasing under composition with a continuous non-decreasing surjection, since for any partition $c = s_0 < s_1 < \dots < s_n = d$ the points $\lambda(s_j) \in [a,b]$ form a non-decreasing sequence and \begin{align*} \sum_{j=1}^n |x^\lambda_{s_j} - x^\lambda_{s_{j-1}}| = \sum_{j=1}^n |x^i_{\lambda(s_j)} - x^i_{\lambda(s_{j-1})}| \le \|x^i\|_{1;[a,b]}. \end{align*} Hence $\mu^i_{x^\lambda}$ on $[c,d]$ is well-defined. We claim the **image-measure identity** \begin{align*} \lambda_* \mu^i_{x^\lambda} = \mu^i_x \qquad \text{on } \mathcal{B}([a,b]), \end{align*} where $\lambda_* \mu := \mu \circ \lambda^{-1}$ denotes the pushforward. To verify, it suffices (by the uniqueness portion of the [Carathéodory extension](/theorems/???)) to check the identity on the half-open interval $\pi$-system. For $a \le u < v \le b$, since $\lambda$ is continuous and non-decreasing the preimage $\lambda^{-1}((u, v])$ is a (possibly empty) interval $(s, t] \subseteq [c, d]$ with $\lambda(s) = u$ and $\lambda(t) = v$ when non-empty. Then \begin{align*} (\lambda_* \mu^i_{x^\lambda})((u,v]) = \mu^i_{x^\lambda}(\lambda^{-1}((u,v])) = \mu^i_{x^\lambda}((s,t]) = x^\lambda_t - x^\lambda_s = x^i_{\lambda(t)} - x^i_{\lambda(s)} = x^i_v - x^i_u = \mu^i_x((u,v]). \end{align*} The two finite measures agree on the $\pi$-system of half-open intervals generating $\mathcal{B}([a,b])$, hence on all of $\mathcal{B}([a,b])$. The image-measure identity gives the **change-of-variable formula**: for any bounded Borel $f: [a,b] \to \mathbb{R}$ and $a \le u \le v \le b$ with $\lambda(s) = u$, $\lambda(t) = v$, \begin{align*} \int_{(s,t]} f(\lambda(r)) \, d\mu^i_{x^\lambda}(r) = \int_{(u,v]} f(z) \, d\mu^i_x(z), \end{align*} which is the [pushforward integration formula](/theorems/???) ($\int g \, d(\lambda_* \mu) = \int (g \circ \lambda) \, d\mu$) applied to $g = f \cdot \mathbb{1}_{(u,v]}$.[/step]

[guided]The setup. We are working in the bounded-variation regime $p = 1$, where every coordinate $x^i: [a,b] \to \mathbb{R}$ has finite total variation, and the [Lebesgue–Stieltjes measure](/page/Lebesgue-Stieltjes%20Measure) $\mu^i_x$ on $[a,b]$ is the unique finite signed Borel measure satisfying $\mu^i_x((s,t]) = x^i_t - x^i_s$. The reparameterized path $x^\lambda = x \circ \lambda: [c,d] \to V$ inherits bounded variation: any partition $c = s_0 < \cdots < s_n = d$ pushes forward to a non-decreasing sequence $\lambda(s_0) \le \cdots \le \lambda(s_n)$ in $[a,b]$, so \begin{align*} \sum_{j=1}^n |x^\lambda_{s_j} - x^\lambda_{s_{j-1}}| = \sum_{j=1}^n |x^i_{\lambda(s_j)} - x^i_{\lambda(s_{j-1})}| \le \|x^i\|_{1;[a,b]}, \end{align*} and taking the supremum gives $\|x^\lambda\|_{1;[c,d]} \le \|x\|_{1;[a,b]} < \infty$. So $\mu^i_{x^\lambda}$ on $[c,d]$ is well-defined. The key identity. We claim the pushforward $\lambda_* \mu^i_{x^\lambda}$ on $[a,b]$ equals $\mu^i_x$. Why is this the right identity to prove? Because once we have it, the [pushforward integration formula](/theorems/???) (which says $\int g \, d(\lambda_* \mu) = \int (g \circ \lambda) \, d\mu$ for any bounded Borel $g$) immediately translates Stieltjes integrals on $[c,d]$ against $\mu^i_{x^\lambda}$ into Stieltjes integrals on $[a,b]$ against $\mu^i_x$. This is the fundamental change-of-variable mechanism we will use throughout the inductive proof. Verifying the identity. Two finite Borel measures on $[a,b]$ are equal iff they agree on a $\pi$-system generating $\mathcal{B}([a,b])$. The half-open intervals $\{(u,v] : a \le u < v \le b\}$ form such a $\pi$-system (closed under finite intersection, and they generate the Borel $\sigma$-algebra). So it suffices to compute both measures on $(u,v]$. For $\mu^i_x$, by definition, $\mu^i_x((u,v]) = x^i_v - x^i_u$. For the pushforward, we compute $(\lambda_* \mu^i_{x^\lambda})((u,v]) = \mu^i_{x^\lambda}(\lambda^{-1}((u,v]))$. The continuity and monotonicity of $\lambda$ imply that $\lambda^{-1}((u,v])$ is itself a (possibly empty) interval — specifically, if $u, v$ are both in the image $\lambda([c,d])$, write $\lambda(s) = u$ and $\lambda(t) = v$ for some $s \le t$ in $[c,d]$, and then $\lambda^{-1}((u,v]) = (s, t]$. (If $u$ or $v$ falls in a "plateau" where $\lambda$ is constant, the preimage shape is the same up to choice of endpoint, and the measure is unaffected because plateaus contribute zero to $\mu^i_{x^\lambda}$.) Then \begin{align*} (\lambda_* \mu^i_{x^\lambda})((u,v]) = \mu^i_{x^\lambda}((s,t]) = x^\lambda_t - x^\lambda_s = x^i_{\lambda(t)} - x^i_{\lambda(s)} = x^i_v - x^i_u = \mu^i_x((u,v]). \end{align*} The two measures agree on the $\pi$-system, so by [Carathéodory uniqueness](/theorems/???) they agree on all of $\mathcal{B}([a,b])$, establishing $\lambda_* \mu^i_{x^\lambda} = \mu^i_x$. The change-of-variable formula. With the pushforward identity in hand, the [pushforward integration formula](/theorems/???) applied to $g = f \cdot \mathbb{1}_{(u,v]}$ (bounded and Borel) gives \begin{align*} \int_{(s,t]} f(\lambda(r)) \, d\mu^i_{x^\lambda}(r) = \int_{[c,d]} (f \cdot \mathbb{1}_{(u,v]})(\lambda(r)) \, d\mu^i_{x^\lambda}(r) = \int_{[a,b]} f \cdot \mathbb{1}_{(u,v]} \, d\mu^i_x = \int_{(u,v]} f(z) \, d\mu^i_x(z). \end{align*} This is the operational form we will plug into the inductive computation. The same formula extends from $f \cdot \mathbb{1}_{(u,v]}$ to any bounded Borel function on $[a,b]$ by the standard $\pi$-$\lambda$ / monotone class argument, but we will only need the half-open-interval case. A subtle point: $\lambda$ is allowed to have flat plateaus (it is non-decreasing, not strictly increasing). If $\lambda$ is constant on $[s, s']$, the path $x^\lambda$ is also constant on $[s, s']$, so $\mu^i_{x^\lambda}$ assigns zero mass to that plateau. The pushforward identity therefore goes through whether $\lambda$ is strictly increasing or merely non-decreasing.[/guided]

[step:Prove the level-1 base case using the change-of-variable identity] Fix any single index $i \in \{1, \dots, d\}$ and any $t \in [c, d]$. By definition of the level-1 signature and the Stieltjes-integral identity above (taking $f \equiv 1$, $s = c$, $t = t$, $u = a$, $v = \lambda(t)$), \begin{align*} S(x^\lambda)^{(i)}_{[c,t]} = \int_{(c,t]} d\mu^i_{x^\lambda}(r) = \mu^i_{x^\lambda}((c,t]) = x^i_{\lambda(t)} - x^i_a = \int_{(a,\lambda(t)]} d\mu^i_x(z) = S(x)^{(i)}_{[a, \lambda(t)]}. \end{align*} In particular, taking $t = d$ so that $\lambda(d) = b$, \begin{align*} S(x^\lambda)^{(i)}_{[c,d]} = S(x)^{(i)}_{[a,b]}, \end{align*} proving the equality at level $k = 1$ for each coordinate. [/step]

[step:Inductive step: lift the equality from level $k$ to level $k+1$]Fix $k \ge 1$ and a multi-index $(i_1, \dots, i_k, i_{k+1})$. Assume the inductive hypothesis \begin{align*} \text{(IH)} \qquad S(x^\lambda)^{(i_1, \dots, i_k)}_{[c,t]} = S(x)^{(i_1, \dots, i_k)}_{[a, \lambda(t)]} \qquad \text{for all } t \in [c,d]. \end{align*} We prove the same identity holds at level $k+1$ when evaluated at $t = d$ (so $\lambda(t) = b$); the same argument with arbitrary $t$ and $\lambda(t)$ gives the version needed for further induction. Write $F(z) := S(x)^{(i_1, \dots, i_k)}_{[a,z]}$ for $z \in [a,b]$. By definition of the level-$(k+1)$ iterated integral and (IH), \begin{align*} S(x^\lambda)^{(i_1, \dots, i_k, i_{k+1})}_{[c,d]} &= \int_{(c,d]} S(x^\lambda)^{(i_1, \dots, i_k)}_{[c,r]} \, d\mu^{i_{k+1}}_{x^\lambda}(r) \\ &= \int_{(c,d]} F(\lambda(r)) \, d\mu^{i_{k+1}}_{x^\lambda}(r). \end{align*} The function $F: [a,b] \to \mathbb{R}$ is continuous (it is a finite-variation indefinite Stieltjes integral with continuous integrators, hence continuous as a function of its upper limit) and bounded (continuous on a compact interval). Applying the change-of-variable formula from Step 2 with $u = a$, $v = b$, $s = c$, $t = d$, and $f = F$: \begin{align*} \int_{(c,d]} F(\lambda(r)) \, d\mu^{i_{k+1}}_{x^\lambda}(r) = \int_{(a,b]} F(z) \, d\mu^{i_{k+1}}_x(z) = \int_a^b S(x)^{(i_1, \dots, i_k)}_{[a,z]} \, dx^{i_{k+1}}_z = S(x)^{(i_1, \dots, i_k, i_{k+1})}_{[a,b]}. \end{align*} Combining the two displays, \begin{align*} S(x^\lambda)^{(i_1, \dots, i_k, i_{k+1})}_{[c,d]} = S(x)^{(i_1, \dots, i_k, i_{k+1})}_{[a,b]}. \end{align*} The same computation with $d$ replaced by an arbitrary $t \in [c,d]$ (and $b$ correspondingly replaced by $\lambda(t)$) verifies the strengthened identity needed to continue the induction. By induction on $k$ starting from the base case in Step 3, the level-$k$ equality holds for every $k \in \mathbb{N}$ and every multi-index, establishing the theorem when $p = 1$.[/step]

[guided]We are inducting on the level $k$ of the iterated integral. The base case $k = 1$ is Step 3. We now show: assuming the level-$k$ identity \begin{align*} \text{(IH)} \qquad S(x^\lambda)^{(i_1, \dots, i_k)}_{[c,t]} = S(x)^{(i_1, \dots, i_k)}_{[a, \lambda(t)]} \qquad \forall t \in [c,d], \end{align*} the same identity holds at level $k+1$. The strategy. The level-$(k+1)$ iterated integral is, by definition, an integral of the level-$k$ iterated integral against the path $x^{i_{k+1}}$. We exploit (IH) to rewrite the inner integrand and then apply the change-of-variable formula from Step 2 to the outer integral. Setting up the integral. Fix the multi-index $(i_1, \dots, i_k, i_{k+1})$ and write $F(z) := S(x)^{(i_1, \dots, i_k)}_{[a,z]}$ for $z \in [a,b]$. By definition of the level-$(k+1)$ iterated integral and the inductive hypothesis, \begin{align*} S(x^\lambda)^{(i_1, \dots, i_k, i_{k+1})}_{[c,d]} &= \int_{(c,d]} S(x^\lambda)^{(i_1, \dots, i_k)}_{[c,r]} \, d\mu^{i_{k+1}}_{x^\lambda}(r) \\ &\stackrel{\text{(IH)}}{=} \int_{(c,d]} S(x)^{(i_1, \dots, i_k)}_{[a, \lambda(r)]} \, d\mu^{i_{k+1}}_{x^\lambda}(r) = \int_{(c,d]} F(\lambda(r)) \, d\mu^{i_{k+1}}_{x^\lambda}(r). \end{align*} Verifying the hypotheses for change of variable. The change-of-variable formula from Step 2 (with $u = a$, $v = b$, $s = c$, $t = d$) requires the integrand $f$ to be bounded and Borel-measurable on $[a,b]$. We verify both for $f = F$: - **Continuity of $F$**: $F$ is an indefinite Stieltjes integral of a continuous function (the level-$(k-1)$ signature $S(x)^{(i_1, \dots, i_{k-1})}_{[a, \cdot]}$, which is continuous in its upper endpoint by continuity of Stieltjes integration against a continuous integrator) against a continuous bounded-variation integrator $x^{i_k}$. Hence $F$ is continuous on $[a,b]$. - **Boundedness**: continuity of $F$ on the compact interval $[a,b]$ implies $F$ is bounded. - **Borel-measurability**: continuous functions are Borel-measurable. Applying the change of variable. By Step 2 applied to $f = F$, \begin{align*} \int_{(c,d]} F(\lambda(r)) \, d\mu^{i_{k+1}}_{x^\lambda}(r) = \int_{(a,b]} F(z) \, d\mu^{i_{k+1}}_x(z). \end{align*} Now recognise the right-hand side as a level-$(k+1)$ signature component of $x$: \begin{align*} \int_{(a,b]} F(z) \, d\mu^{i_{k+1}}_x(z) = \int_a^b S(x)^{(i_1, \dots, i_k)}_{[a,z]} \, dx^{i_{k+1}}_z = S(x)^{(i_1, \dots, i_k, i_{k+1})}_{[a,b]}. \end{align*} Conclusion. Stringing the equalities together, \begin{align*} S(x^\lambda)^{(i_1, \dots, i_k, i_{k+1})}_{[c,d]} = S(x)^{(i_1, \dots, i_k, i_{k+1})}_{[a,b]}. \end{align*} The same computation with $d$ replaced by an arbitrary $t \in [c,d]$ (and correspondingly $b$ replaced by $\lambda(t)$) gives the strengthened version $S(x^\lambda)^{(i_1, \dots, i_k, i_{k+1})}_{[c,t]} = S(x)^{(i_1, \dots, i_k, i_{k+1})}_{[a, \lambda(t)]}$ needed to continue the induction. By induction starting from Step 3, the level-$k$ equality holds for every $k \in \mathbb{N}$ and every multi-index, establishing $S(x^\lambda)_{[c,d]} = S(x)_{[a,b]}$ in $T((V))$ when $p = 1$. A subtle point: $\lambda$ may have flat plateaus. The plateau-handling is bundled into the pushforward identity established in Step 2 — flat plateaus in $\lambda$ correspond to plateaus in $x^\lambda$, on which $\mu^{i_{k+1}}_{x^\lambda}$ vanishes, so the pushforward computation is unaffected. We do not need a separate plateau argument here.[/guided]

[step:Extend the equality from $p = 1$ to $1 < p < 2$ by density and continuity of Young integration]Now suppose $1 < p < 2$. Choose any $q > 1$ with $1/p + 1/q > 1$ (such $q$ exists: as $q$ runs through $(1, p/(p-1))$, the sum $1/p + 1/q$ exceeds $1$). By the [density of bounded-variation paths in $q$-variation](/theorems/???), there exists a sequence $(x^{(n)})_{n \in \mathbb{N}} \subset C_1([a,b], V)$ with \begin{align*} \|x - x^{(n)}\|_{q;[a,b]} \to 0 \quad \text{as } n \to \infty. \end{align*} Composition with $\lambda$ preserves the $q$-variation in the same way it preserves total variation: for any partition $c = s_0 < \dots < s_n = d$, \begin{align*} \sum_{j=1}^n |x^\lambda_{s_j} - x^\lambda_{s_{j-1}}|^q = \sum_{j=1}^n |x_{\lambda(s_j)} - x_{\lambda(s_{j-1})}|^q, \end{align*} and the points $\lambda(s_j)$ form a non-decreasing sequence in $[a,b]$, so the right-hand side is bounded by $\|x\|_{q;[a,b]}^q$. Taking the supremum over partitions, $\|x^\lambda\|_{q;[c,d]} \le \|x\|_{q;[a,b]}$. The same bound applies to $x - x^{(n)}$, so \begin{align*} \|x^\lambda - (x^{(n)})^\lambda\|_{q;[c,d]} \le \|x - x^{(n)}\|_{q;[a,b]} \to 0. \end{align*} The signature map $S: C_q([a,b], V) \to T((V))$ is continuous in $q$-variation when $1/p + 1/q > 1$; this is the [continuity of the Young integral](/theorems/???) applied iteratively level by level. Concretely, at each level the map $(f, g) \mapsto \int f \, dg$ is jointly continuous on $C_q \times C_q$ in the relevant variations, and the iterated integral is built by composing such bilinear maps. By Step 4, $S(x^{(n)})_{[a,b]} = S((x^{(n)})^\lambda)_{[c,d]}$ for every $n$ since each $x^{(n)}$ lies in $C_1$. Passing to the limit on both sides using the established continuity gives \begin{align*} S(x)_{[a,b]} = \lim_{n \to \infty} S(x^{(n)})_{[a,b]} = \lim_{n \to \infty} S((x^{(n)})^\lambda)_{[c,d]} = S(x^\lambda)_{[c,d]}, \end{align*} where convergence is in the $q$-variation topology on $T((V))$ at every level. This completes the proof.[/step]

[guided]The strategy. Prove the identity on a dense subclass where it is easy ($p = 1$, Step 4), then transport it to the full class by continuity. The dense subclass is $C_1$ (bounded-variation paths), where iterated integrals are honest Stieltjes integrals and the change-of-variable formula from Step 2 applies directly. To make the transport work we need (i) approximating sequences $x^{(n)} \to x$ in a topology in which (ii) the signature map $S$ is continuous. Choosing the approximating topology. Both criteria are satisfied by the $q$-variation topology when $1/p + 1/q > 1$: the [continuity of the Young integral](/theorems/???) makes $(f, g) \mapsto \int f \, dg$ jointly continuous in $q$-variation, and iterating this up the tower of iterated integrals makes $S$ continuous on $C_q$. We need $q > 1$ (so that $C_q$ contains $C_p$ for $p < q$, but here it is the dual condition that matters) with $1/p + 1/q > 1$, i.e., $q < p/(p-1)$. Such $q$ exists because as $q$ approaches $1$ from above, $1/q$ approaches $1$ and $1/p + 1/q$ exceeds $1$ since $1/p > 0$. Constructing the approximation. By the [density of bounded-variation paths in $q$-variation](/theorems/???), there is a sequence $(x^{(n)})_{n \in \mathbb{N}} \subset C_1([a,b], V)$ with \begin{align*} \|x - x^{(n)}\|_{q;[a,b]} \to 0. \end{align*} The reparameterization respects $q$-variation. We need to verify that $(x^{(n)})^\lambda \to x^\lambda$ in $q$-variation on $[c,d]$. The same partition argument as in Step 2 (now raised to the $q$-th power instead of summed in absolute value) gives: for any partition $c = s_0 < \cdots < s_n = d$, the points $\lambda(s_j)$ form a non-decreasing sequence in $[a,b]$, so \begin{align*} \sum_{j=1}^n |x^\lambda_{s_j} - x^\lambda_{s_{j-1}}|^q = \sum_{j=1}^n |x_{\lambda(s_j)} - x_{\lambda(s_{j-1})}|^q \le \|x\|_{q;[a,b]}^q. \end{align*} Taking the supremum over all partitions of $[c,d]$, \begin{align*} \|x^\lambda\|_{q;[c,d]} \le \|x\|_{q;[a,b]}. \end{align*} Applying the same bound to the difference $x - x^{(n)}$, \begin{align*} \|x^\lambda - (x^{(n)})^\lambda\|_{q;[c,d]} \le \|x - x^{(n)}\|_{q;[a,b]} \to 0. \end{align*} So composition with $\lambda$ is non-expansive in $q$-variation, and $(x^{(n)})^\lambda \to x^\lambda$ in $C_q([c,d], V)$. Passing to the limit. By Step 4 applied to each $x^{(n)} \in C_1$, \begin{align*} S(x^{(n)})_{[a,b]} = S((x^{(n)})^\lambda)_{[c,d]} \qquad \forall n. \end{align*} Both sides are continuous functions of their respective inputs in the $q$-variation topology on $C_q$: - The left-hand side $x \mapsto S(x)_{[a,b]}$ is continuous on $C_q([a,b], V)$ by joint continuity of the Young integral, applied iteratively at every level. - The right-hand side $x \mapsto S(x^\lambda)_{[c,d]}$ is the composition of $x \mapsto x^\lambda$ (continuous from $C_q([a,b])$ to $C_q([c,d])$ by the bound just established) with $S(\cdot)_{[c,d]}$ (continuous on $C_q([c,d])$ by the same Young-integral argument). Two continuous functions agreeing on the dense subset $C_1 \subseteq C_q$ must agree on the closure, hence on all of $C_p \subseteq C_q$. Concretely, taking $n \to \infty$ in both sides: \begin{align*} S(x)_{[a,b]} = \lim_{n \to \infty} S(x^{(n)})_{[a,b]} = \lim_{n \to \infty} S((x^{(n)})^\lambda)_{[c,d]} = S(x^\lambda)_{[c,d]}, \end{align*} where convergence is at every level of $T((V))$ in the $q$-variation topology. This completes the proof. Why is $p < 2$ essential? The Young-integration regime requires $1/p + 1/q > 1$ with $q > 1$. Solving: $1/p > 1 - 1/q$, so $p < q/(q-1)$. The right-hand side decreases from $\infty$ (at $q = 1^+$) to $1$ (at $q \to \infty$), passing through $2$ at $q = 2$. So for any $p < 2$ we can find $q > 1$ with $q/(q-1) > p$, but for $p \ge 2$ no such $q > 1$ exists. The threshold $p = 2$ is the boundary of Young integration; for $p \ge 2$, Chen's identity and reparameterization invariance still hold, but their proofs require the extra structure of a rough-path lift.[/guided]