[proofplan] We construct a Brownian motion from a continuous local martingale $M$ by time-changing via the right-continuous inverse $T_s = \inf\{t : \langle M \rangle_t > s\}$ of its quadratic variation. The proof has four stages: (1) verify that $(\mathcal{G}_s) = (\mathcal{F}_{T_s})$ is a filtration satisfying the usual conditions; (2) show $B_s = M_{T_s}$ is continuous and adapted, where the key subtlety is left-continuity across jumps of $s \mapsto T_s$; (3) prove $B$ is a $(\mathcal{G}_s)$-martingale by applying the Optional Stopping Theorem to $M$ and $M^2 - \langle M \rangle$; (4) identify $B$ as Brownian motion via Levy's Characterization, using $\langle B \rangle_s = s$. [/proofplan] [step:Verify that $T_s$ is well-defined, finite, and satisfies $\langle M \rangle_{T_s} = s$] The quadratic variation $\langle M \rangle$ is continuous (by the [Existence and Uniqueness of Quadratic Variation](/theorems/2113)), adapted, non-decreasing with $\langle M \rangle_0 = 0$, and $\langle M \rangle_\infty = \infty$ almost surely by hypothesis. For each $s \geq 0$, define \begin{align*} T_s = \inf\{t \geq 0 : \langle M \rangle_t > s\}. \end{align*} Since $\langle M \rangle_\infty = \infty$ and $\langle M \rangle$ is continuous and non-decreasing, the set $\{t \geq 0 : \langle M \rangle_t > s\}$ is non-empty for every $s \geq 0$, so $T_s < \infty$ almost surely. The process $\langle M \rangle$ is continuous and adapted to $(\mathcal{F}_t)$, and the set $\{t : \langle M \rangle_t > s\}$ is open (by continuity of $\langle M \rangle$). Therefore $T_s = \inf\{t : \langle M \rangle_t > s\}$ is a stopping time with respect to $(\mathcal{F}_t)$. The map $s \mapsto T_s$ is non-decreasing (since $\langle M \rangle$ is non-decreasing, a larger threshold $s$ requires a later hitting time) and right-continuous (since $\langle M \rangle$ is continuous: if $s_n \searrow s$, then $T_{s_n} \searrow T_s$ because $\langle M \rangle_{T_s} \leq s < s_n$ forces $T_{s_n} \geq T_s$, and for any $\varepsilon > 0$, continuity of $\langle M \rangle$ at $T_s$ ensures $\langle M \rangle_{T_s + \varepsilon} > s_n$ for large $n$, giving $T_{s_n} \leq T_s + \varepsilon$). We claim $\langle M \rangle_{T_s} = s$ for all $s \geq 0$. By definition, $\langle M \rangle_t \leq s$ for all $t < T_s$. By right-continuity of $\langle M \rangle$, taking $t \nearrow T_s$ gives $\langle M \rangle_{T_s^-} \leq s$. Since $\langle M \rangle$ is continuous, $\langle M \rangle_{T_s} = \langle M \rangle_{T_s^-} \leq s$. On the other hand, $T_s = \inf\{t : \langle M \rangle_t > s\}$ and $\langle M \rangle_\infty = \infty$ imply that for any $\varepsilon > 0$, $\langle M \rangle_{T_s + \varepsilon} > s$ (otherwise $T_s$ would not be the infimum). By continuity of $\langle M \rangle$, taking $\varepsilon \searrow 0$ gives $\langle M \rangle_{T_s} \geq s$. Therefore $\langle M \rangle_{T_s} = s$. [guided] The identity $\langle M \rangle_{T_s} = s$ is the fundamental property of the time change and relies on the continuity of $\langle M \rangle$. If $\langle M \rangle$ had jumps, we would only have $\langle M \rangle_{T_s^-} \leq s \leq \langle M \rangle_{T_s}$, and the equality could fail. This is why the Dubins-Schwarz theorem requires $M$ to be a continuous local martingale — the continuity of the paths ensures the continuity of $\langle M \rangle$, which in turn ensures the right-continuous inverse $T_s$ is an exact inverse. Geometrically, we are inverting the graph of $t \mapsto \langle M \rangle_t$. Since $\langle M \rangle$ is continuous and increases from $0$ to $\infty$, it is a bijection from $[0, \infty)$ onto $[0, \infty)$ if it is strictly increasing. If $\langle M \rangle$ has flat stretches (intervals where $\langle M \rangle$ is constant), the inverse $T_s$ jumps across those intervals. This is why $s \mapsto T_s$ is only right-continuous in general, not continuous. [/guided] [/step] [step:Show that $(\mathcal{G}_s)_{s \geq 0}$ is a filtration satisfying the usual conditions] Define $\mathcal{G}_s = \mathcal{F}_{T_s}$ for $s \geq 0$. We verify this is a filtration: for $r \leq s$, we have $T_r \leq T_s$ (since $s \mapsto T_s$ is non-decreasing), and by definition of stopped $\sigma$-algebras, $\mathcal{F}_{T_r} \subseteq \mathcal{F}_{T_s}$ whenever $T_r \leq T_s$. To verify this inclusion, take $A \in \mathcal{G}_r = \mathcal{F}_{T_r}$. For any $u \geq 0$: \begin{align*} A \cap \{T_s \leq u\} = A \cap \{T_r \leq u\} \cap \{T_s \leq u\}. \end{align*} Since $T_r \leq T_s$, the event $\{T_s \leq u\}$ is contained in $\{T_r \leq u\}$, so $A \cap \{T_s \leq u\} = A \cap \{T_s \leq u\}$. Since $A \in \mathcal{F}_{T_r}$, we have $A \cap \{T_r \leq u\} \in \mathcal{F}_u$, and since $\{T_s \leq u\} \in \mathcal{F}_u$ (as $T_s$ is a stopping time), the intersection $A \cap \{T_s \leq u\} \in \mathcal{F}_u$. This holds for all $u \geq 0$, so $A \in \mathcal{F}_{T_s} = \mathcal{G}_s$. For right-continuity: $\bigcap_{n \geq 1} \mathcal{G}_{s + 1/n} = \bigcap_{n \geq 1} \mathcal{F}_{T_{s+1/n}} = \mathcal{F}_{\lim_n T_{s+1/n}}$, where the last equality uses right-continuity of $(\mathcal{F}_t)$. Since $s \mapsto T_s$ is right-continuous, $\lim_n T_{s+1/n} = T_s$, so $\bigcap_n \mathcal{G}_{s+1/n} = \mathcal{F}_{T_s} = \mathcal{G}_s$. Therefore $(\mathcal{G}_s)$ is right-continuous. [/step] [step:Show that $B_s = M_{T_s}$ is continuous and $(\mathcal{G}_s)$-adapted] **Adaptedness.** For each $s \geq 0$, $T_s$ is a stopping time for $(\mathcal{F}_t)$ and $M$ is a continuous adapted process. The general fact that $X_T$ is $\mathcal{F}_T$-measurable whenever $X$ is a continuous adapted process and $T$ is a stopping time gives $B_s = M_{T_s} \in \mathcal{F}_{T_s} = \mathcal{G}_s$. **Continuity.** Since $M$ is continuous and $s \mapsto T_s$ is right-continuous, the composition $s \mapsto M_{T_s}$ is right-continuous. Left-continuity requires a separate argument. Define the left limit $T_{s-} = \lim_{r \nearrow s} T_r = \inf\{t \geq 0 : \langle M \rangle_t \geq s\}$. We have $T_{s-} \leq T_s$, with two cases: **Case 1: $T_s = T_{s-}$.** Then $s \mapsto T_s$ is continuous at $s$, so $B_{r} = M_{T_r} \to M_{T_{s-}} = M_{T_s} = B_s$ as $r \nearrow s$. The process $B$ is continuous at $s$. **Case 2: $T_s > T_{s-}$.** Since $\langle M \rangle_{T_{s-}} \leq s$ (by the definition of $T_{s-}$) and $\langle M \rangle_{T_s} = s$ (established above), the quadratic variation $\langle M \rangle$ takes the value $s$ at both $T_{s-}$ and $T_s$. Since $\langle M \rangle$ is non-decreasing, $\langle M \rangle$ is constant on the interval $[T_{s-}, T_s]$, equal to $s$. [claim:$M$ is constant on any interval where $\langle M \rangle$ is constant] Let $[a, b] \subset [0, \infty)$ and suppose $\langle M \rangle_t = \langle M \rangle_a$ for all $t \in [a, b]$. Then $M_t = M_a$ for all $t \in [a, b]$ almost surely. [/claim] [proof] The stopped process $M^b - M^a$ is a continuous local martingale starting at $0$ with quadratic variation $\langle M^b - M^a \rangle_t = \langle M \rangle_{(t \wedge b) \vee a} - \langle M \rangle_a$ for $t \geq a$. For $t \in [a, b]$, this equals $\langle M \rangle_t - \langle M \rangle_a = 0$. By the [Quadratic Variation Vanishes Iff the Martingale Vanishes](/theorems/2084) theorem applied to the continuous local martingale $N_t := M_{(t+a) \wedge b} - M_a$ (which satisfies $N_0 = 0$ and $\langle N \rangle_t = \langle M \rangle_{(t+a) \wedge b} - \langle M \rangle_a = 0$ for $t \in [0, b-a]$), we conclude $N_t = 0$ for all $t \in [0, b-a]$, i.e., $M_t = M_a$ for all $t \in [a, b]$. [/proof] Since $\langle M \rangle$ is constant on $[T_{s-}, T_s]$, the claim gives $M_{T_s} = M_{T_{s-}}$. Therefore \begin{align*} \lim_{r \nearrow s} B_r = \lim_{r \nearrow s} M_{T_r} = M_{T_{s-}} = M_{T_s} = B_s. \end{align*} The process $B$ is left-continuous at $s$. In both cases, $B$ is continuous at $s$. Since $s \geq 0$ was arbitrary, $B$ is a continuous process. [guided] The left-continuity argument is the most delicate part of the proof. The time change $s \mapsto T_s$ is only right-continuous, so there can be jumps: $T_s > T_{s-}$. When such a jump occurs, $B_s = M_{T_s}$ could a priori differ from $\lim_{r \nearrow s} B_r = M_{T_{s-}}$. The key insight is that a jump in $T$ corresponds to a "flat stretch" of $\langle M \rangle$: the quadratic variation $\langle M \rangle$ is constant on $[T_{s-}, T_s]$. During such a flat stretch, $M$ itself must be constant (this is the content of the claim). Intuitively, $\langle M \rangle$ measures the "accumulated randomness" of $M$; if no randomness accumulates on an interval, the martingale cannot move. Why can we not have $\langle M \rangle$ constant on an interval while $M$ oscillates? Because $\langle M \rangle$ is the quadratic variation — the limit of sums $\sum (M_{t_{i+1}} - M_{t_i})^2$. If $M$ oscillated on $[a,b]$, these sums would be positive, contradicting $\langle M \rangle_b - \langle M \rangle_a = 0$. The rigorous version of this argument is the [Quadratic Variation Vanishes Iff the Martingale Vanishes](/theorems/2084) theorem: $\langle N \rangle = 0$ implies $N = 0$ for a continuous local martingale $N$ with $N_0 = 0$. [/guided] [/step] [step:Prove that $B$ is a $(\mathcal{G}_s)$-martingale with $\langle B \rangle_s = s$] We show that $B_s = M_{T_s}$ is $L^2$ and that $B^2_s - s$ is a $(\mathcal{G}_s)$-martingale, which simultaneously gives the martingale property and identifies the quadratic variation. **$B_s$ is in $L^2$.** Since $\langle M \rangle_{T_s} = s$, the stopped process $M^{T_s}$ satisfies $\langle M^{T_s} \rangle_\infty = \langle M \rangle_{T_s} = s < \infty$. By the [Quadratic Variation Norm Formula](/theorems/2085), $\mathbb{E}[(M^{T_s}_\infty)^2] = \mathbb{E}[\langle M^{T_s} \rangle_\infty] = s$. Therefore $B_s = M_{T_s} = M^{T_s}_\infty \in L^2$ with $\mathbb{E}[B_s^2] = s$. **$B$ is a martingale.** The process $M$ is a continuous local martingale, and $M^2_t - \langle M \rangle_t$ is also a continuous local martingale (by definition of the quadratic variation). Since $M^{T_s}$ has bounded quadratic variation ($\langle M^{T_s} \rangle_\infty = s$), the process $M^{T_s}$ is a true $L^2$ martingale, and $(M^{T_s})^2 - \langle M^{T_s} \rangle$ is a uniformly integrable martingale (since $|(M^{T_s}_t)^2 - \langle M^{T_s} \rangle_t| \leq (M^{T_s}_t)^2 + s$ and $\sup_t (M^{T_s}_t)^2$ is integrable by [Doob's $L^2$ Maximal Inequality](/theorems/2112)). Fix $r \leq s$. Since $T_r \leq T_s$, both are stopping times for $(\mathcal{F}_t)$, and $T_s$ is bounded by any deterministic time that makes $\langle M \rangle > s$ (but $T_s$ need not be bounded in general). We apply the [Optional Stopping Theorem](/theorems/2109). The process $M^{T_s}$ is a uniformly integrable martingale (as shown above), so: \begin{align*} \mathbb{E}[M_{T_s} \mid \mathcal{F}_{T_r}] = M_{T_r}, \end{align*} i.e., $\mathbb{E}[B_s \mid \mathcal{G}_r] = B_r$. Similarly, $(M^{T_s})^2 - \langle M^{T_s} \rangle$ is a uniformly integrable martingale, so by the [Optional Stopping Theorem](/theorems/2109): \begin{align*} \mathbb{E}[(M_{T_s})^2 - \langle M \rangle_{T_s} \mid \mathcal{F}_{T_r}] = (M_{T_r})^2 - \langle M \rangle_{T_r}. \end{align*} Substituting $\langle M \rangle_{T_s} = s$ and $\langle M \rangle_{T_r} = r$: \begin{align*} \mathbb{E}[B_s^2 - s \mid \mathcal{G}_r] = B_r^2 - r. \end{align*} Therefore $B^2_s - s$ is a $(\mathcal{G}_s)$-martingale, which means $\langle B \rangle_s = s$. [guided] The key observation is that stopping $M$ at $T_s$ produces a uniformly integrable $L^2$ martingale, because the quadratic variation $\langle M^{T_s} \rangle_\infty = s$ is deterministically bounded. This is where the condition $\langle M \rangle_{T_s} = s$ is consumed: it converts the possibly wild local martingale $M$ into a well-behaved stopped martingale $M^{T_s}$ with deterministic $L^2$ norm. Why do we need $M^{T_s}$ to be uniformly integrable, rather than just a martingale? Because the [Optional Stopping Theorem](/theorems/2109) for non-bounded stopping times requires uniform integrability. The stopping times $T_r$ and $T_s$ are not bounded in general (even though they are finite), so the bounded stopping time version does not apply. The identity $\mathbb{E}[B_s^2 - s \mid \mathcal{G}_r] = B_r^2 - r$ shows that $B^2_s - s$ is a $(\mathcal{G}_s)$-martingale. This is exactly the characterizing property of the quadratic variation $\langle B \rangle_s = s$: the process $B^2 - \langle B \rangle$ must be a local martingale, and the only continuous non-decreasing process $A$ with $A_0 = 0$ such that $B^2 - A$ is a local martingale is $A_s = s$, the identity function. This is the content of the [Existence and Uniqueness of Quadratic Variation](/theorems/2113) theorem. [/guided] [/step] [step:Apply Levy's Characterization to identify $B$ as a Brownian motion and recover $M_t = B_{\langle M \rangle_t}$] The process $B$ is a continuous $(\mathcal{G}_s)$-local martingale with $B_0 = M_{T_0} = M_0 = 0$ and $\langle B \rangle_s = s$. By [Levy's Characterization of Brownian Motion](/theorems/2100) (applied with $d = 1$), $B$ is a standard one-dimensional $(\mathcal{G}_s)$-Brownian motion. It remains to show $M_t = B_{\langle M \rangle_t}$ for all $t \geq 0$. By the identity $\langle M \rangle_{T_s} = s$, the map $s \mapsto T_s$ is a right inverse of $t \mapsto \langle M \rangle_t$: composing $T$ with $\langle M \rangle$ gives \begin{align*} T_{\langle M \rangle_t} = \inf\{u \geq 0 : \langle M \rangle_u > \langle M \rangle_t\}. \end{align*} If $\langle M \rangle$ is strictly increasing at $t$ (i.e., $t$ does not lie in a flat stretch), then $T_{\langle M \rangle_t} = t$, and \begin{align*} B_{\langle M \rangle_t} = M_{T_{\langle M \rangle_t}} = M_t. \end{align*} If $\langle M \rangle$ is constant on some interval $[a, b]$ containing $t$, then $T_{\langle M \rangle_t} = b$ (the right endpoint of the flat stretch). But $M$ is constant on $[a, b]$ by the claim established above, so $M_t = M_b = M_{T_{\langle M \rangle_t}} = B_{\langle M \rangle_t}$. In both cases, $M_t = B_{\langle M \rangle_t}$ for all $t \geq 0$ almost surely. This completes the proof. [guided] The representation $M_t = B_{\langle M \rangle_t}$ is the central message of the Dubins-Schwarz theorem: every continuous local martingale with divergent quadratic variation is a time-changed Brownian motion. The time change $t \mapsto \langle M \rangle_t$ runs the Brownian motion $B$ at a speed determined by the accumulated randomness of $M$. When $M$ is volatile (large $d\langle M \rangle / dt$), the Brownian motion $B$ runs fast; when $M$ is quiescent (small $d\langle M \rangle / dt$), $B$ runs slowly. The subtlety in the reconstruction $M_t = B_{\langle M \rangle_t}$ is that $T_{\langle M \rangle_t}$ may not equal $t$ when $\langle M \rangle$ has flat stretches. However, on such stretches $M$ is forced to be constant (by the zero quadratic variation argument), and the Brownian motion $B$ is evaluated at a constant level $\langle M \rangle_t = \langle M \rangle_a$ throughout the stretch. So both sides of $M_t = B_{\langle M \rangle_t}$ are constant on the flat stretch, and the identity holds. It is worth noting why the hypothesis $\langle M \rangle_\infty = \infty$ is essential. If $\langle M \rangle_\infty < \infty$, then $T_s = \infty$ for $s > \langle M \rangle_\infty$, and $B_s = M_\infty$ would be constant for large $s$ — not a Brownian motion. The divergence of $\langle M \rangle$ ensures that the time change covers all of $[0, \infty)$. [/guided] [/step]