[guided]We claim that the level-$k$ signature has the closed-form representation
\begin{align*}
S(x^\tau)^{(k)} = \int_{\Delta_k(L)} (x^\tau_{t_1})' \otimes \cdots \otimes (x^\tau_{t_k})' \, d\mathcal{L}^k(t_1, \dots, t_k),
\end{align*}
where $\Delta_k(L) := \{(t_1, \dots, t_k) \in [0, L]^k : 0 < t_1 < t_2 < \cdots < t_k < L\}$ is the open $k$-simplex. Two ingredients combine here: (a) **Stieltjes-to-Lebesgue conversion** — replace each $dx^\tau$ by $(x^\tau)' \, d\mathcal{L}^1$ — and (b) **Fubini's theorem** — fold the iterated integral into a single multidimensional integral. We prove the claim by induction on $k$, keeping careful track of the hypotheses being verified at each step.
**Base case $k = 1$.** By definition of the level-$1$ signature,
\begin{align*}
S(x^\tau)^{(1)}_{[0, L]} = x^\tau_L - x^\tau_0.
\end{align*}
By Step 2.2 (the fundamental theorem of calculus for absolutely continuous functions),
\begin{align*}
x^\tau_L - x^\tau_0 = \int_0^L (x^\tau_t)' \, d\mathcal{L}^1(t) = \int_{(0, L)} (x^\tau_t)' \, d\mathcal{L}^1(t),
\end{align*}
where the second equality holds because removing the two endpoints — a $\mathcal{L}^1$-null set — does not change the integral. Since $\Delta_1(L) = (0, L)$, this is exactly the claim at level $k = 1$.
**Inductive hypothesis.** Suppose at level $k$ that for every $t \in [0, L]$,
\begin{align*}
S(x^\tau)^{(k)}_{[0, t]} = \int_{\Delta_k(t)} (x^\tau_{t_1})' \otimes \cdots \otimes (x^\tau_{t_k})' \, d\mathcal{L}^k(t_1, \dots, t_k).
\end{align*}
**Inductive step: $k \to k+1$.** By the recursive definition of iterated integrals,
\begin{align*}
S(x^\tau)^{(k+1)}_{[0, L]} = \int_0^L S(x^\tau)^{(k)}_{[0, t_{k+1}]} \otimes dx^\tau_{t_{k+1}}.
\end{align*}
The outer integral is a Stieltjes integral with respect to the path-measure $dx^\tau$. To convert it to a Lebesgue integral, we use that $x^\tau$ is absolutely continuous (Step 2.2): the Stieltjes measure $dx^\tau$ on $[0, L]$ is absolutely continuous with respect to $\mathcal{L}^1$ with Radon–Nikodym density $(x^\tau)'$. Concretely,
\begin{align*}
\int_0^L f(t) \, dx^\tau_t = \int_0^L f(t) (x^\tau_t)' \, d\mathcal{L}^1(t)
\end{align*}
for every Borel $f: [0, L] \to V^{\otimes k}$ that is integrable against $|dx^\tau|$. Applying this with $f(t_{k+1}) = S(x^\tau)^{(k)}_{[0, t_{k+1}]}$ — which is bounded (by the inductive expression and the bound $\|(x^\tau)'\| \le 1$ a.e.) and continuous in $t_{k+1}$, hence Borel measurable and $|dx^\tau|$-integrable — gives
\begin{align*}
S(x^\tau)^{(k+1)}_{[0, L]} = \int_0^L S(x^\tau)^{(k)}_{[0, t_{k+1}]} \otimes (x^\tau_{t_{k+1}})' \, d\mathcal{L}^1(t_{k+1}).
\end{align*}
Now substitute the inductive expression for the inner factor:
\begin{align*}
S(x^\tau)^{(k+1)} = \int_0^L \!\!\!\int_{\Delta_k(t_{k+1})} (x^\tau_{t_1})' \otimes \cdots \otimes (x^\tau_{t_k})' \otimes (x^\tau_{t_{k+1}})' \, d\mathcal{L}^k(t_1, \dots, t_k) \, d\mathcal{L}^1(t_{k+1}).
\end{align*}
**Verification of Fubini's hypotheses.** We wish to apply [Fubini's Theorem](/theorems/???) on the product space $\mathbb{R}^k \times \mathbb{R}$ with the product Lebesgue measure $\mathcal{L}^k \otimes \mathcal{L}^1 = \mathcal{L}^{k+1}$, swapping the iterated integral into a single integral on $\mathbb{R}^{k+1}$. Fubini requires:
- (F1) The integrand is jointly measurable on the product space. The integrand is a product of measurable factors $(x^\tau_{t_i})'$ projected onto coordinates $t_i$, so it is jointly measurable. (More precisely: $(x^\tau)' \in L^1([0, L], V; \mathcal{L}^1)$ has a Borel-measurable representative, and tensor products of Borel functions on different coordinates remain jointly Borel.)
- (F2) The integrand is in $L^1$ of the product measure (or non-negative). The norm of the integrand satisfies
\begin{align*}
\|(x^\tau_{t_1})' \otimes \cdots \otimes (x^\tau_{t_{k+1}})'\|_{V^{\otimes (k+1)}} \le \prod_{i=1}^{k+1} \|(x^\tau_{t_i})'\|_V \le 1
\end{align*}
$\mathcal{L}^{k+1}$-a.e., by submultiplicativity of the projective tensor norm and the bound $\|(x^\tau)'\| \le 1$ a.e. from Step 2.3. The set on which the integrand is non-zero is contained in $[0, L]^{k+1}$, which has $\mathcal{L}^{k+1}$-measure $L^{k+1} < \infty$. Hence the integrand is bounded a.e. on a finite-measure set, so it lies in $L^\infty(\mathcal{L}^{k+1}) \subset L^1(\mathcal{L}^{k+1})$.
Both hypotheses are verified. Fubini gives
\begin{align*}
S(x^\tau)^{(k+1)} = \int_E (x^\tau_{t_1})' \otimes \cdots \otimes (x^\tau_{t_{k+1}})' \, d\mathcal{L}^{k+1}(t_1, \dots, t_{k+1}),
\end{align*}
where the combined region of integration is
\begin{align*}
E = \{(t_1, \dots, t_{k+1}) \in [0, L]^{k+1} : 0 < t_1 < \cdots < t_k < t_{k+1} \le L\}.
\end{align*}
The set $\{t_{k+1} = L\}$ has $\mathcal{L}^{k+1}$-measure zero, so we may equivalently integrate over the open simplex $\Delta_{k+1}(L) = \{0 < t_1 < \cdots < t_{k+1} < L\}$. This completes the induction.
**Why is the Fubini step non-trivial?** The two iterated orders — first integrate over $\Delta_k(t_{k+1})$ then over $t_{k+1} \in [0, L]$, vs. integrate over $\Delta_{k+1}(L)$ in one go — agree only when an integrability hypothesis is satisfied. The pointwise bound $\|(x^\tau)'\| \le 1$ a.e. (Step 2.3) is doing real work here: without it, the integrand could be unbounded and Fubini would fail. The Lipschitz reduction in Steps 1–2 was precisely the vehicle for getting this bound.
**Why convert Stieltjes to Lebesgue?** The recursive definition of the signature uses Stieltjes integrals against $dx^\tau$. Stieltjes integrals against an absolutely continuous path are Lebesgue integrals against $(x^\tau)' \, d\mathcal{L}^1$. The conversion is essential because Fubini's theorem applies to integrals against a product **measure** — and we need the same measure on each factor of the product. The Stieltjes measures $dx^\tau$ on each axis assemble into the product Stieltjes measure, but this product is opaque; replacing each with $(x^\tau)' \, d\mathcal{L}^1$ exposes a clean product measure $\mathcal{L}^{k+1}$ and pulls all the path data into the integrand.
**The simplex $\Delta_k(L)$.** The chronological constraint $0 < t_1 < \cdots < t_k < L$ identifies the natural geometric object underlying the iterated integral: the $k$-dimensional standard simplex of side $L$. Its $\mathcal{L}^k$-measure is $L^k / k!$, computed in Step 4.[/guided]
