Stopped occupation absolute continuity for uniformly elliptic Ito processes

Fix $f\ge 0$ smooth compactly supported, and take $v$ from [F1]. Convolve in spacetime with ${\rho }_{ϵ}(z)={ϵ}^{-d-1}\rho (z\mathrm{/}ϵ)$, where $\rho \ge 0$ is a smooth compactly supported unit-mass kernel. Write $v_{ϵ}=v\ast {\rho }_{ϵ}$ and $f_{ϵ}=f\ast {\rho }_{ϵ}$. For each constant $A\ge 0$, convolution of the distributional inequality gives $(v_{ϵ}{)}_t+A:D_x^2{v}_{ϵ}-(1+\mathrm{tr}A)v_{ϵ}+(\det A{)}^{1\mathrm{/}(d+1)}{f}_{ϵ}\le 0$ at every spacetime point. These simultaneous inequalities follow first for a countable dense set of matrices and then all matrices by continuity. Also $v_{ϵ}\ge 0$, $\mathrm{\parallel }{v}_{ϵ}{\mathrm{\parallel }}_{\mathrm{\infty }},\mathrm{\parallel }{D}_x{v}_{ϵ}{\mathrm{\parallel }}_{\mathrm{\infty }}\le C_d\mathrm{\parallel }f{\mathrm{\parallel }}_{d+1}$, and $f_{ϵ}\to f$ uniformly. Thus evaluation at the random matrix $a_s$ uses smooth pointwise inequalities, with no evaluation of unsmoothed weak derivatives on an unknown occupation measure.

Put $Q_t=\exp (-t-{\int }_0^{t\wedge \gamma }\mathrm{tr}{a}_{s}ds)$ and use the Itô formula on $[0,\gamma ]$ for $Q_t{v}_{ϵ}(t,Z_t)$. The ordinary product rule with this absolutely continuous scalar factor and step 1.1 give $d(Q{v}_{ϵ})\le Q{D}_x{v}_{ϵ}\mathrm{\Sigma }dW+Q{D}_x{v}_{ϵ}\cdot \beta dt-Q(\det a{)}^{1\mathrm{/}(d+1)}{f}_{ϵ}(t,Z_t)dt\mathrm{.}$ The stopped stochastic integral has expected square energy at most $2C_d^2\mathrm{\parallel }f{\mathrm{\parallel }}_{d+1}^{2}E{\int }_0^{\gamma }\mathrm{tr}{a}_{s}ds$, hence expectation zero by [F2]. The drift term has integrable absolute value at most $C_d\mathrm{\parallel }f{\mathrm{\parallel }}_{d+1}E{\int }_0^{\gamma }\mathrm{\mid }{\beta }_s\mathrm{\mid }ds$. The initial and terminal values are bounded nonnegative variables irrespective of $Z_0$. Taking expectations and discarding the terminal value gives the estimate with $f_{ϵ}$. The weight obeys $Q_s(\det a_s{)}^{1\mathrm{/}(d+1)}\le C_d(1+\mathrm{tr}{a}_s)$ by eigenvalue AM–GM, and its stopped integral has finite expectation. Since $\mathrm{\parallel }{f}_{ϵ}{\mathrm{\parallel }}_{\mathrm{\infty }}\le \mathrm{\parallel }f{\mathrm{\parallel }}_{\mathrm{\infty }}$ and $f_{ϵ}\to f$ uniformly, [F5] gives dominated convergence for the weighted integrals. Thus $E{\int }_0^{\gamma }{Q}_s(\det a_s{)}^{1\mathrm{/}(d+1)}f(s,Z_s)ds\le C_d(1+E{\int }_0^{\gamma }\mathrm{\mid }{\beta }_s\mathrm{\mid }ds)\mathrm{\parallel }f{\mathrm{\parallel }}_{d+1}\mathrm{.}$ Localization in [F2] can be removed by its square-energy bound and by [F5] for the integrable finite-variation terms.

Define $\nu (E)=E{\int }_0^{\gamma }{Q}_s(\det a_s{)}^{1\mathrm{/}(d+1)}{1}_E(s,Z_s)ds$. It is a finite Borel measure: $Q\le 1$ and the eigenvalue AM–GM bound gives $(\det a{)}^{1\mathrm{/}(d+1)}\le C_d(1+\mathrm{tr}a)$, whose stopped expectation is finite. For every open set $O$ of finite Lebesgue measure, exhaust $O$ by compact sets $K_j\uparrow O$. By [F4] choose $0\le {\chi }_j\le 1$ smooth compactly supported in $O$, equal to one on $K_j$. The estimate in step 2.1 and $\mathrm{\parallel }{\chi }_j{\mathrm{\parallel }}_{d+1}\le \mathrm{\mid }O{\mathrm{\mid }}^{1\mathrm{/}(d+1)}$ give $\nu (K_j)\le C_d(1+E{\int }_0^{\gamma }\mathrm{\mid }{\beta }_s\mathrm{\mid }ds)\mathrm{\mid }O{\mathrm{\mid }}^{1\mathrm{/}(d+1)}$. Continuity from below gives the same bound for $\nu (O)$. Every Lebesgue-null Borel set $N$ has open supersets of arbitrarily small finite measure by [F4], so monotonicity gives $\nu (N)=0$. This argument needs no unproved inner-regularity assertion for $\nu$.

The weight $Q_s(\det a_s{)}^{1\mathrm{/}(d+1)}$ is strictly positive for $dt\otimes P$-almost every $s<\gamma$: the stopped trace integral is finite almost surely by its finite expectation, and $a_s\ge \kappa I$. For a null Borel set $N$, step 3.1 and Tonelli give a nonnegative random integral of that weight times $1_N(s,Z_s)$ equal to zero almost surely. Positivity of the weight then gives ${\int }_0^{\gamma }{1}_N(s,Z_s)ds=0$ almost surely, and taking expectations proves ${\mu }_{\gamma }(N)=0$. Finally take ${\gamma }_n=T\wedge \inf \{t:{\int }_0^t(\mathrm{\mid }{\beta }_s\mathrm{\mid }+\mathrm{tr}{a}_s)ds\ge n\}$, with infimum of the empty set interpreted as $T$. The integral is continuous adapted and finite on $[0,T]$, so these are stopping times, their stopped integrals are at most $n$, and for almost every path they equal $T$ for all sufficiently large $n$. Thus ${\gamma }_n\uparrow T$ and all the asserted stopped occupation measures are absolutely continuous.