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The occupation time of a measurable set $A$ through time $t$ is ${\int }_0^t{1}_{\{B_s\in A\}}ds$. The occupation-times formula expresses such integrals through local time of a semimartingale. Recurrence and the Strong Markov property imply that one-dimensional Brownian motion spends an unbounded total time in every nonempty open interval.

If $M$ is a continuous local martingale and $f\in C^2(\mathbb{R})$, then finite variation of $f(M)$ forces $f(M_t)=f(M_0)$. The Itô formula and the quadratic variation of a stochastic integral give ${\int }_0^t{f}^{\prime }(M_s{)}^{2}d[M{]}_s=0$. The occupation-times formula implies that $L_t^a(M)=0$ almost everywhere where $f^{\prime }(a)\ne 0$. On the zero set of $f^{\prime }$, its derivative $f^{\prime \prime }$ vanishes almost everywhere: zeros at which $f^{\prime \prime }\ne 0$ are isolated and hence countable. Thus ${\int }_0^t|f^{\prime \prime }(M_s)|d[M{]}_s=\int |f^{\prime \prime }(a)|L_t^a(M)da=0$. Both terms in the Itô formula vanish.

The Itô formula decomposes the smooth transform into a continuous local martingale and a finite-variation process:If $f(M)$ has finite variation, then $N$ also has finite variation. A continuous finite-variation local martingale is constant, so $N=0$ andIt remains to show that the second-order term vanishes; the preceding observation alone does not establish this.

Let $Z=\{a:f^{\prime }(a)=0\}$. By the occupation-times formula for the local time of a semimartingale,Therefore $L_t^a(M)=0$ for almost every $a\notin Z$. Also $f^{\prime \prime }(a)=0$ for almost every $a\in Z$: if $f^{\prime }(a)=0$ and $f^{\prime \prime }(a)\ne 0$, continuity of $f^{\prime \prime }$ makes $f^{\prime }$ strictly monotone in a neighborhood of $a$, so that zero is isolated. Such isolated zeros form a countable set. Localizing to bounded paths and finite quadratic variation, then using the occupation-times formula once more, givesBoth terms in the Itô formula now vanish. Apply this on all rational times and use continuity to obtain the simultaneous conclusionThis is the finite-variation smooth transform of a continuous local martingale property.

The existence assertion is false under the hypotheses printed in the source. No measurability of $\theta$ is specified. Even requiring $\theta$ to be a Borel measurable function does not suffice: a coefficient can be positive at every point while approaching zero too rapidly near the starting point. This is the obstruction described by pointwise positivity does not ensure weak existence.

Here is a counterexample in dimension one, with $\Theta =1$:Suppose that a solution to the martingale problem with domain $C_c^{\infty }(\mathbb{R})$ existed. The cutoff argument from part (a)(ii), now with $\mathcal{L}=\theta \Delta$, makes $X$ a continuous local martingale withThe criterion for a continuous local martingale with bounded quadratic variation makes it a square-integrable martingale, and the Itô isometry gives $\mathbb{E}{X}_t^2=\mathbb{E}[X{]}_t$.

Apply the occupation-times formula with the indicator of the singleton $\{0\}$. Since that singleton has zero Lebesgue measure,Thus $X$ spends zero ordinary time at zero. Away from zero, $\theta (x)\le x^2$, soThe Gronwall inequality forces $\mathbb{E}{X}_t^2=0$ for every $t$. Continuity then forces $X\equiv 0$ almost surely, contradicting either the zero occupation time just proved or $[X{]}_t=2t$ for that constant path. Consequently

For completeness, the intended construction works if $\theta$ is a Borel measurable function and bounded away from zero on every compact set. In particular, it works for continuous strictly positive $\theta$. Let $B$ be standard $d$-dimensional Brownian motion, and define a clock and its inverse byThe Brownian path on each finite interval has compact range, so the added hypothesis makes $A_u$ finite. The clock is continuous and strictly increasing, and $A_u\ge u/(2\Theta )$, so $A_u\to \infty$. Its inverse is continuous and satisfies ${\tau }_t\le 2\Theta t$. Each ${\tau }_t$ is a stopping time for the Brownian filtration.

For $f\in C_c^{\infty }({\mathbb{R}}^d)$, the optional sampling theorem applied at these bounded random times shows thatis a true martingale in the filtration ${\mathcal{G}}_t={\mathcal{F}}_{{\tau }_t}^B$. The inverse-clock substitution $d{\tau }_t=2\theta (X_t)dt$, valid almost everywhere, rewrites it asThis proves the Brownian time change for a variable diffusion speed:The factor $2$ in the clock is necessary because standard Brownian motion has infinitesimal generator $\Delta /2$.

There need not be a weak solution of a stochastic differential equation $d{X}_t=\sqrt{2\theta (X_t)}d{B}_t$, $X_0=0$, even when $\theta$ is a bounded Borel measurable function strictly positive everywhere. For the displayed example, any solution would be a continuous local martingale with $[X{]}_t=2{\int }_0^t\theta (X_s)ds\le 2t$, hence a square-integrable martingale. The occupation-times formula applied to $1_{\{0\}}$ gives $0={\int }_0^t{1}_{\{X_s=0\}}d[X{]}_s=2{\int }_0^t{1}_{\{X_s=0\}}ds$. Since $\theta (x)\le x^2$ away from zero, $\mathbb{E}{X}_t^2\le 2{\int }_0^t\mathbb{E}{X}_s^{2}ds$. The Gronwall inequality makes $X\equiv 0$, a contradiction. Thus a diffusion amplitude positive at a single problematic point cannot repair arbitrarily slow neighbouring motion. A local positive lower bound, as in the Brownian time change for a variable diffusion speed, rules out this example.