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Let $N$ be the zero-starting continuous local martingale, and let $V_t$ be the total variation of its path on $[0,t]$. Stop when $|N|$ or $V$ reaches $n$, and also at time $n$. The stopped process is bounded, hence a true square-integrable martingale by the bounded local martingale criterion, and its total variation is bounded by $n$.

For a deterministic partition $0=t_0<\cdots <t_m=t$, martingale increments are orthogonal in $L^2$, soThe sum inside the expectation is at most the largest absolute increment times the stopped total variation. As the mesh tends to zero, path continuity makes the largest increment tend to zero. The product is bounded by $2n^2$, so the dominated convergence theorem makes the displayed expectation zero. Thus the stopped process vanishes at every fixed time. Use rational times, continuity and ${\tau }_n\uparrow \infty$ to concludeThis proves that a continuous finite-variation local martingale is constant.

Write a normalized semimartingale decomposition as $X=X_0+N+A$, with both $N_0=A_0=0$, $N$ a continuous local martingale and $A$ of finite variation. The difference of two such martingale parts is also the difference of two finite-variation parts, and hence vanishes by the result just proved. The other parts coincide too. The normalized continuous semimartingale decomposition is unique. Fixing the initial constants is necessary: without normalization a constant can be transferred between the two parts.

On the finite horizon $[0,T]$, strict positivity gives equivalence of the restricted probability measures: for $A\in {\mathcal{F}}_T$, $\mathbb{Q}(A)={\mathbb{E}}_{\mathbb{P}}(Z_T{1}_A)$ vanishes if and only if $\mathbb{P}(A)$ vanishes. The standard change-of-measure theorem gives semimartingale invariance under equivalent measures: a process is a semimartingale under $\mathbb{P}$ on this horizon if and only if it is one under $\mathbb{Q}$. The sample-path quadratic variation is unchanged, but the local-martingale and finite-variation parts generally change.

More explicitly, set $N=\int Z^{-1}dZ$, the zero-start stochastic logarithm of the density. For every continuous $\mathbb{P}$-local martingale $L$, the Girsanov theorem states thatis a continuous $\mathbb{Q}$-local martingale. Consequently, if the continuous semimartingale decomposition under $\mathbb{P}$ is $X=X_0+L+V$, its decomposition under $\mathbb{Q}$ isThe covariation correction is a continuous finite-variation process, locally well-defined because $Z$ stays strictly positive. In the Brownian case, if $d{Z}_t=Z_t{\theta }_{t}d{B}_t$, thenis a Brownian motion under $\mathbb{Q}$. The inverse finite-horizon density is $1/Z$, giving the reverse change of measure. These assertions apply on each horizon on which the density is strictly positive; they do not assert equivalence on a larger terminal sigma-algebra where the full density $D$ may vanish.

Write the semimartingale decomposition as $U=M+A$, where $M$ is a continuous local martingale and $A$ has finite variation. Applying Itô formula to $\log Z_t(z)$ shows that its finite-variation part isIt vanishes for every $z$. Multiplying by $Z_t(z{)}^2$ givesSubtract this identity for two points with distinct $Z_t$ to obtain $d{A}_t=0$; then $d⟨M{⟩}_t=4dt$. Since the curve starts at zero, $U_0=0$. The Lévy characterization of Brownian motion now gives $U_t=2B_t$. Hence the Loewner chain is

Each $A^{ϵ}$ is continuous, adapted, and increasing. From ucp convergence one can choose a subsequence that converges uniformly almost surely on every compact interval. Its limit $A$ is therefore also continuous and increasing, hence a finite-variation process. The stochastic integral $M_t={\int }_0^t\mathrm{sgn}(B_s)d{B}_s$ is a continuous local martingale, and part e gives the semimartingale decompositionConsequently $|B|$ is a semimartingale. In fact, comparison with the Tanaka formula identifies $A_t$ as the local time of a semimartingale $L_t^0(B)$.

Apply Itô formula to $Z_t^{(A)}=\exp (A{X}_t-A^{2}t/2)$. Its semimartingale decomposition isThe second term is a continuous finite-variation process. Since $Z^{(A)}$ is assumed to be a local martingale, uniqueness of the semimartingale decomposition makes this term identically zero. Both $A$ and $Z^{(A)}$ are nonzero, so the quadratic variation of $X$ is $[X{]}_t=t$.

The Lévy characterization of Brownian motion now says that $X_t-X_0$ is a Brownian motion. Consequentlyis a constant multiple of an exponential Brownian martingale. It is therefore a true martingale for every $a\in \mathbb{R}$.

On a fixed finite horizon, equivalent probability measures have the same semimartingales. For a strictly positive continuous Radon-Nikodym density martingale $Z$ and a continuous local martingale $L$ under the original measure, $L-\int Z^{-1}d[L,Z]$ is a local martingale under the new one, by the Girsanov theorem. Thus the finite-variation part of a continuous semimartingale decomposition changes by the opposite covariation correction.