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Replace the square-root coefficient near zero by smooth globally Lipschitz cutoffs that vanish below half the cutoff level. With a common Brownian driver, local pathwise uniqueness makes the solutions agree until the larger cutoff level. The exit times increase as the levels decrease. The compatible integrands $1_{\{s\le T_n\}}\sqrt{Z_s}$ have increasing squares and expected time-integrals at most $zt$ by the nonnegative local martingale bound. Their limit is square-integrable on every finite horizon. The Itô isometry gives a continuous global limiting stochastic integral; at a finite limiting exit time its value is zero, and it remains absorbed afterward.

Suppose a finite-horizon market has a strictly positive numéraire $B$ and an equivalent measure under which all prices $S^i/B$ are local martingales. The discounted wealth of a self-financing strategy is a stochastic integral against those prices. If it is bounded below by a deterministic constant, that wealth is a supermartingale: add the lower-bound constant to obtain a nonnegative local martingale, then apply conditional Fatou to a localizing sequence. A zero-cost admissible portfolio therefore has expected discounted terminal wealth at most zero. Nonnegative terminal wealth with positive probability of a strict gain would have strictly positive expectation under the equivalent measure, a contradiction. Thus there is no admissible arbitrage. True martingales are not needed for this implication.

For the Gaussian Brownian drift filter with positive definite $V_0$, define $Z_t=\det (I+t{V}_0{)}^{1/2}\exp [-m_t^T{V}_t^{-1}{m}_t/2+m_0^T{V}_0^{-1}{m}_0/2]$. Because $V_t^{\prime }=-V_t^2$ and $d{m}_t=V_{t}d{\hat{W}}_t$, the Itô formula gives $d\log Z_t=-m_t^{T}d{\hat{W}}_t-|m_t{|}^{2}dt/2$. Thus $Z$ is a positive stochastic exponential, hence a nonnegative local martingale and a supermartingale. For a singular prior, the inverse-matrix expression is undefined; the same process can instead be defined by $Z_t=({\mathbb{E}}_{\lambda \sim N(m_0,V_0)}\exp [{\lambda }^T{Y}_t-t|\lambda {|}^2/2]{)}^{-1}$, with $Y_t$ treated as fixed inside this Gaussian integral.

For a right-continuous nonnegative supermartingale $M$ and $R>0$, the first level-hit time $\sigma$ and the optional stopping theorem give $R\mathbb{P}(\sigma \le T)\le \mathbb{E}[M_{\sigma \wedge T}]\le \mathbb{E}[M_0]$. Let $T\to \infty$ and first use levels $R-\epsilon$, then $\epsilon \downarrow 0$, to include a supremum not attained. For a continuous nonnegative local martingale on a stochastic interval, apply the argument on increasing localized compact subintervals, then pass to their limit. Letting $R\to \infty$ shows that its running supremum is finite almost surely.

Use ${\theta }_t$ for the dollar amount invested in the stock; the number of shares is ${\theta }_t/S_t$. The bank account contains $w_t-{\theta }_t$ dollars. With no intermediate consumption, the self-financing portfolio equation isIf instead $\theta$ denotes share holdings, replace each dollar ${\theta }_t$ here by $S_t{\theta }_t$. The equation says that trading only reallocates existing portfolio wealth; gains come from cash interest and stock returns.

For the standard unconstrained complete-diffusion formulation, assume the completed natural Brownian filtration, initial wealth $w_0>0$, and nonnegative admissible wealth. Define the market price of risk ${\kappa }_t=({\mu }_t-r_t)/{\sigma }_t$. The boundedness assumptions make $\kappa$ bounded. The state-price density normalized at zero isIts money-market-adjusted process $Z_t=e^{{\int }_0^t{r}_{s}ds}{\zeta }_t$ is a true martingale by the Novikov condition. Under $d\mathbb{Q}=Z_{T}d\mathbb{P}$, the Girsanov theorem gives $W_t^{\mathbb{Q}}=W_t+{\int }_0^t{\kappa }_{s}ds$ as a Brownian motion, and the discounted stock has zero drift.

The Itô product rule applied to the wealth equation givesFor nonnegative admissible wealth this is a nonnegative local martingale, hence a supermartingale. Therefore every affordable terminal payoff $X=w_T$ satisfies the state-price budget constraint $\mathbb{E}[{\zeta }_{T}X]\le w_0$. Nonzero volatility and the Brownian martingale representation theorem allow replication of nonnegative claims with finite state-price cost.

For a differentiable strictly concave utility function with the usual conditions ensuring an interior integrable optimizer, put $I=(U^{\prime }{)}^{-1}$, the inverse marginal utility. Pointwise maximization of $U(x)-y{\zeta }_{T}x$ givesThe second equation determines the positive constant $y$. To prove optimality, use the concave supporting-tangent inequality:Taking expectations and using the budget bound proves that the candidate dominates every admissible affordable payoff, whenever the utility expectations are well defined. This is the complete-market terminal utility optimizer.

The actual wealth and dollar portfolio are as explicit as general adapted coefficients permit. LetThe Brownian martingale representation theorem, with localization if only $L^1$ integrability is initially known, gives $h$. The dollar portfolio from a deflated wealth martingale isThe bank holds $w_t^{\ast }-{\theta }_t^{\ast }$ dollars. Equivalently, with $B_t=e^{{\int }_0^t{r}_{s}ds}$, $w_t^{\ast }=B_t{\mathbb{E}}^{\mathbb{Q}}[w_T^{\ast }/B_T\mid {\mathcal{F}}_t]$. These formulas need no unjustified differentiability of a value function with random coefficients.

The printed assumption that $U$ is only increasing and concave does not guarantee that the interior formula exists, or even that the optimal value is finite. In general replace the inverse formula by an affordable selectionwhere $\mathcal{D}$ is the wealth domain; boundary choices or supporting slopes handle nondifferentiability. Existence, finite cost and replication still have to hold. A concrete obstruction is the unbounded linear terminal-wealth utility: take $U(x)=x$, $r=0$, $\mu =\sigma =1$, and the Brownian filtration. Then ${\zeta }_T=e^{-W_T-T/2}$. For $A_n=\{{\zeta }_T<1/n\}$, the bounded nonnegative claimis replicable, costs $w_0$, and has $\mathbb{E}{X}_n>n{w}_0$. Thus there is no finite optimal expected utility under the literal utility hypotheses alone. Also, bounded coefficients do not ensure market completeness in an arbitrarily enlarged filtration containing untraded randomness; the full replication formula uses the Brownian market information assumption stated above.

Fix $t>0$. For $0\le s\le t$, the residual observed path isThis Brownian bridge is independent of $(\lambda ,Y_t)$. Indeed it is independent of $\lambda$, while its covariance with $W_t$ is $sI-(s/t)tI=0$; zero covariance implies independence in the joint Gaussian law. The same argument for every finite collection of bridge times, followed by continuity and generation from rational times, gives independence of the entire bridge. Since the full observed path is determined by its endpoint and this independent bridge, Brownian endpoint sufficiency for a constant drift gives

First suppose the prior covariance matrix $V_0$ is positive definite. The Bayes' theorem multiplies the prior Gaussian density by the conditional observation density, whose dependence on $\lambda$ is $\exp ({\lambda }^T{Y}_t-t|\lambda {|}^2/2)$. Thus the posterior exponent, up to terms not depending on $\lambda$, isCompleting the square gives the Gaussian Brownian drift filter:For invertible $V_0$ these are equivalently $V_t=(V_0^{-1}+tI{)}^{-1}$ and ${\lambda }_t=V_t(V_0^{-1}{\lambda }_0+Y_t)$. At $t=0$ they reduce to the prior parameters.

The boxed formulas remain valid for a singular positive semidefinite matrix $V_0$. To see this directly without inverse prior matrices, the joint Gaussian random vector $(\lambda ,Y_t)$ hasThe latter matrix is invertible for $t>0$. The conditional multivariate normal distribution therefore has meanand covariance matrixAlternatively, subtract this mean from $\lambda$; the resulting Gaussian random vector has zero covariance with $Y_t$ and is therefore independent of it, which proves these conditional parameters.

Write the proposed innovation process in the observable formIt is adapted to the observation filtration and continuous. To prove the martingale property, take $0\le u<v$. Future increments of $W$ are independent of the joint information generated by $\lambda$ and the past of $W$, soAlso the tower property of conditional expectation gives $\mathbb{E}[{\lambda }_s\mid {\mathcal{G}}_u]={\lambda }_u$ for $s\ge u$. Conditional integration is legitimate because $\mathbb{E}|{\lambda }_s{|}^2\le \mathbb{E}|\lambda {|}^2<\infty$. Subtracting the integral proves $\mathbb{E}[{\hat{W}}_v-{\hat{W}}_u\mid {\mathcal{G}}_u]=0$. Removing a continuous finite variation process does not change quadratic covariation, henceThe Lévy characterization of multidimensional Brownian motion now applies: a continuous zero-start martingale with this bracket matrix is a standard $d$-dimensional Brownian motion relative to its given filtration. The observable innovation is a Wiener process in the observation filtration.

Differentiating the posterior formulas yields $V_t^{\prime }=-V_t^2$ andFor the final expression, the printed inverse matrices require the extra assumption that $V_0$ is positive definite. Under that assumption put $A_t=V_t^{-1}=V_0^{-1}+tI$. The Itô formula givessince $A_t{V}_t=I$ and $\mathrm{tr}(A_t{V}_t^2)=\mathrm{tr}{V}_t$. On the other hand,The trace terms cancel in the innovation exponential for Gaussian drift filtering, leavingThus $Z_0=1$ and $Z$ is a positive stochastic exponential, hence a nonnegative local martingale. To obtain the requested supermartingale conclusion explicitly, choose localizing stopping times ${\tau }_n\uparrow \infty$ for which $Z^{{\tau }_n}$ is a true martingale. The Conditional Fatou lemma givesThe same argument at $u=0$ gives $\mathbb{E}{Z}_v\le 1$, supplying integrability. The process is a supermartingale in the observation filtration. No unproved exponential-integrability condition on the random drift is required.

For a singular prior the literal inverse-matrix expression in the PDF is undefined, but there is a natural extension with the same conclusion. Define, for fixed $y$,For positive definite $V_0$, a Gaussian integral shows this reciprocal is exactly the printed expression. For any positive semidefinite $V_0$, the integral is positive and finite, $L_t=-{\Delta }_{y}L/2$, and ${\nabla }_y\log L(t,Y_t)={\lambda }_t$, by the same posterior weighting. The Itô formula consequently gives $d\log L(t,Y_t)={\lambda }_t^{T}d{Y}_t-|{\lambda }_t{|}^{2}dt/2={\lambda }_t^{T}d{\hat{W}}_t+|{\lambda }_t{|}^{2}dt/2$. Its reciprocal satisfies the boxed stochastic differential equation and the same supermartingale proof. This completes the singular-covariance extension while explicitly separating it from the undefined printed formula.

Use the usual stochastic-integrability assumptions implicit in the price stochastic differential equation: the coefficients have progressively measurable versions, the volatility is locally square integrable, and initial prices are finite deterministic positive numbers. A no-consumption admissible trading strategy consists of predictable risky share holdings $\pi$ for which the price stochastic integral exists and the portfolio wealth of the self-financing portfolio isThe remainder $X_t-{\pi }_t^T{S}_t$ is held in the unit cash asset. Admissibility requires $X_t\ge -a$ throughout the horizon for some deterministic finite $a$; the more restrictive nonnegative-wealth convention also works. A finite-horizon arbitrage is such a strategy with $x=0$, $X_T\ge 0$ almost surely and $P(X_T>0)>0$. The lower bound is essential to exclude doubling strategies.

Interpret the stated uniform ellipticity as ${\sigma }_t{\sigma }_t^T\ge \epsilon I$ for a deterministic $\epsilon >0$. If $|{\mu }_t|\le K$, the market price of risk is bounded:The authoritative PDF puts both integrals inside the exponent. With that interpretation,On every finite horizon $T$, boundedness of $\lambda$ gives the Novikov condition:Hence the positive stochastic exponential $Z$ is a true martingale with $\mathbb{E}{Z}_T=1$. Define the equivalent probability measure $Q_T$ by $d{Q}_T=Z_{T}dP$. The Girsanov theorem says that $W_t^Q=W_t+{\int }_0^t{\lambda }_{s}ds$ is a Brownian motion under $Q_T$, up to time $T$. Consequentlyso the prices are local martingales under $Q_T$.

Every admissible zero-initial-capital self-financing portfolio has portfolio wealth which is therefore a $Q_T$-local martingale bounded below by $-a$. The shifted process $X+a$ is a nonnegative local martingale, hence a supermartingale by localization and the Conditional Fatou lemma. In particular ${\mathbb{E}}_{Q_T}{X}_T\le 0$. If the strategy were an arbitrage, nonnegativity of $X_T$ would force $X_T=0$ $Q_T$-almost surely. Equivalence of $P$ and $Q_T$ would then force $P(X_T>0)=0$, a contradiction. The market has no admissible arbitrage on any finite horizon. This proves the exclusion argument itself, rather than just asserting the fundamental theorem of asset pricing.

For the deflated price calculation, let ${\sigma }_{t,i}$ denote the column vector whose transpose is the $i$th row of ${\sigma }_t$. The Itô product rule includes the quadratic covariation termSince ${\sigma }_{t,i}^T{\lambda }_t={\mu }_t^i$, the full product equation isLocalization makes the last stochastic integral a true martingale on each stopped interval, proving each component of the deflated price vector is a local martingale under the original measure. This is the bounded-coefficient asset deflator identity; bounded volatility has not yet been used.

If $\sigma$ is bounded, then ${\sigma }_{t,i}-{\lambda }_t$ is bounded as well. Solving the last linear stochastic differential equation givesThe Novikov condition now holds for this stochastic exponential on every finite horizon, because its bracket is bounded by a deterministic constant times the horizon. ThereforeThe entire deflated price vector is a true martingale when volatility is bounded. This finite-horizon argument does not assert uniform integrability as time tends to infinity, which is not required. Integrable random initial prices can also be handled by the conditional martingale argument; without any integrability assumption on initial prices, a true-martingale assertion would not be meaningful.

For fixed $\epsilon >0$, the derivative of ${\sigma }_{\epsilon }$ is zero on $(-\infty ,\epsilon /2]$, bounded on the compact transition interval $[\epsilon /2,\epsilon ]$, and equals $1/(2\sqrt{x})\le 1/(2\sqrt{\epsilon })$ for $x\ge \epsilon$. Thus ${\sigma }_{\epsilon }$ is globally Lipschitz, with at most linear growth. The global existence theorem for stochastic differential equations with Lipschitz coefficients supplies a global strong stochastic solution and pathwise uniqueness for the modified equation.

The solution stays nonnegative. More precisely, if $z>\epsilon /2$, it cannot cross $\epsilon /2$: on reaching this level the constant continuation solves the equation, and pathwise uniqueness forces that continuation. If $z\le \epsilon /2$, it is constant from the outset. In either case $Z^{\epsilon }$ is a nonnegative local martingale, so $\mathbb{E}{Z}_s^{\epsilon }\le z$.

For ${\epsilon }^{\prime }<\epsilon <z$, both coefficients agree with $\sqrt{x}$ on $[\epsilon ,\infty )$. Stop the two solutions when either first reaches $\epsilon$, and also at a common upper bound. On the resulting compact state interval, the coefficient is Lipschitz. The Itô isometry applied to their difference, followed by the Gronwall inequality, gives zero expected squared difference. Remove the upper bound using the nonexplosion of each modified solution. The two solutions agree until the first of these lower exits, and by continuity both have value $\epsilon$ there, so their exits at that level coincide. In particular,up to indistinguishability. If $\epsilon \ge z$, the first exit time is zero and the same assertion is immediate from their initial values.

Now take a decreasing sequence $0<{\epsilon }_n<z$ with ${\epsilon }_n\downarrow 0$. Write $Z^n=Z^{{\epsilon }_n}$ and ${\tau }_n=T_{{\epsilon }_n}$. Compatibility gives ${\tau }_n\le {\tau }_{n+1}$; indeed, a continuous path cannot reach the smaller level before reaching the larger one. Let $\tau ={\lim }_n{\tau }_n$, a stopping time. On $[0,\tau )$ the compatible solutions define an adapted continuous process $Z$ by setting $Z_t=Z_t^n$ whenever $t\le {\tau }_n$. We still need to justify the finite limiting endpoint and the integral equation there, rather than assume they exist.

For this purpose use the compatible predictable integrandsTheir squares increase pointwise with $n$: on the smaller interval compatibility makes the values equal, and outside it the old integrand is zero. Their limit $H$ is predictable and equals $\sqrt{Z_s}$ before $\tau$, and zero after $\tau$, apart from irrelevant single-time endpoints. For every finite $t$,The monotone convergence theorem gives $\mathbb{E}{\int }_0^t{H}_s^{2}ds\le zt$. Moreover, compatibility makes $(H-H^n{)}^2=H^2-(H^n{)}^2$ away from endpoints, so its expected time-integral tends to zero. The Itô isometry, and the Doob L2 maximal inequality if uniform convergence on a finite time interval is desired, therefore give a continuous stochastic-integral limitBefore ${\tau }_n$, the integrand agrees with $\sqrt{Z^n}$, which is the original square-root coefficient there. Locality of the stochastic integral yields $L_{t\wedge {\tau }_n}=Z_{t\wedge {\tau }_n}^n$. A countable intersection in $n$ and path continuity give this equality simultaneously for all times.

On $\{\tau <\infty \}$, every ${\tau }_n$ is finite and $L_{{\tau }_n}={\epsilon }_n$. Continuity of $L$ thus gives $L_{\tau }=0$. The limiting integrand is zero after $\tau$, so $L$ remains zero afterward. On $\{\tau =\infty \}$, every finite time is covered by one of the compatible positive solutions. Consequently $L$ is nonnegative everywhere and $H_s=\sqrt{L_s}$ for almost every $s$, including after absorption. The cutoff construction of an absorbed square-root diffusion has producedAll modified solutions, stopping times and integrands were constructed from the same prescribed Brownian motion; hence this is a strong stochastic solution. The integral is square-integrable on every finite horizon by the bound $zt$, which also rules out a hidden finite-time explosion in the patching argument.

For any $C^2$ function $h$, the Itô formula and the ratio equation giveThe bracket is zero for the functions just found. On compact localized intervals the stochastic integrand is bounded, so the stochastic integral is a martingale. Removing localization proves that $h(Z_t)$, on $0\le t<T(x)$, is a local martingale. It is nonnegative because $Z_t>1$ and $C>0$.

In particular it is a nonnegative local martingale, hence a supermartingale after the usual stopping and extension by its endpoint limit. This is the justification for using its maximal inequality; it is not automatically an unstopped uniformly integrable martingale.

Let $h_t$ and $k_t$ be predictable holdings in the stock and account, with the stochastic integrability needed for a self-financing portfolio. Then $V=hS+kB$ and $dV=hdS+kdB$. The Itô product rule and self-financing identity giveSince $V=hS+kB$ and $\sigma \lambda =\mu -r$, the finite-variation term is zero. Thereforeand the deflated wealth is a local martingale.

For zero-capital nonnegative wealth under a local deflator, a nonnegative local martingale is a supermartingale, by localization and the Conditional Fatou lemma. Under the required nonnegative-wealth condition, $ZV\ge 0$ and starts at zero, so $0\le \mathbb{E}(Z_t{V}_t)\le 0$. Thus $Z_t{V}_t=0$ almost surely at every fixed $t$. Strict positivity of $Z$ gives $V_t=0$ almost surely. Taking a countable intersection over rational times and then using continuous wealth paths strengthens this toThis argument only needs the local deflator, so it remains valid without promoting $ZB$ to a true martingale.

Write $V_t=⟨X{⟩}_t$ for the quadratic variation and setThe Hölder factorization of stochastic exponentials follows by adding exponents:ThusThe subtraction inside the numerator is $\sqrt{pq}-1$, outside the square root.

The stochastic exponential $\mathcal{E}(\sqrt{pq}X)$ starts at one and is a nonnegative local martingale, hence a supermartingale. The optional sampling theorem for a supermartingale gives expectation at most one at bounded stopping times. For a finite, possibly unbounded, stopping time $T$, apply this to $T\wedge n$, then use continuity and the Fatou lemma:The Holder inequality with exponents $q$ and $q/(q-1)$ now givesThe inequality also holds when the right side is infinite.

Use deterministic times tending to infinity and the Fatou lemma:Apply the terminal scaling inequality for stochastic exponentials to $X=aM$ and $r=1/a$. Part (c) gives $\mathbb{E}[\mathcal{E}(aM{)}_{\infty }]=1$, soLetting $a\uparrow 1$ shows that the terminal expectation is at least one. The nonnegative local martingale $\mathcal{E}(M)$ starts at one and is a supermartingale, so the Fatou lemma gives the opposite inequality. Consequently

The terminal expectation criterion for a nonnegative local martingale now closes the argument. Conditional Fatou lemma applied to the supermartingale at times tending to infinity givesThe left side has expectation one and the right side at most one, so equality holds almost surely. Thusand uniform integrability of conditional expectations proves that $\mathcal{E}(M)$ is a uniformly integrable martingale. This establishes the required stopped-moment form of the Kazamaki criterion.

Use the Chordal Loewner equation ${\partial }_t{g}_t=2/(g_t-U_t)$ with $U_t=\sqrt{\kappa }{B}_t$. For a fixed $z\in \mathbb{H}$, let $T_z$ be its Loewner swallowing time, and write $Z_t=X_t+i{Y}_t=g_t(z)-U_t$ and $J_t=|g_t^{\prime }(z)|$ before that time. The Loewner conformal radius is ${{\rm Y}}_t=Y_t/J_t$, half of the conformal radius of $D_t=\mathbb{H}\setminus K_t$ at $z$. The Koebe quarter theorem bounds the conformal radius above by four times the distance to the boundary. For the reverse comparison, if $\varphi :\mathbb{D}\to D_t$ maps zero to $z$, apply the Schwarz lemma to ${\varphi }^{-1}(z+dw)$ on the unit disc, where $d=\mathrm{dist}(z,\partial D_t)$; it gives $d\le |{\varphi }^{\prime }(0)|$. ThusWe use the basic trace theorems that, for $\kappa >8$, the Loewner trace is continuous and Transience of chordal SLE gives $|\gamma (t)|\to \infty$. These facts make its image relatively closed in $\mathbb{H}$; they do not assume the space-filling conclusion. The continuity and transience statements are available in Rohde and Schramm's basic trace theorems.

Take $\rho =\kappa -8>0$ in the supplied SLE interior-point martingale. The derivative exponent vanishes, leavingA nonnegative local martingale is a supermartingale. Applying the optional stopping theorem after localization at its first hit of a level $R$ gives the maximal inequality for a nonnegative supermartingaleThus $C={\sup }_{t<T_z}{M}_t<\infty$ almost surely.

Suppose the Loewner trace avoids some open ball $B(z,r)$ whose closure is inside $\mathbb{H}$. Before $T_z$, the whole open ball is in $D_t$: a connected open ball disjoint from the trace cannot be partly in the unbounded connected component. Hence ${{\rm Y}}_t\ge r/2$. The bound on $M$ forces $S_t\ge c>0$ throughout this interval, for a positive random constant $c$.

The Chordal Loewner equation and its derivative yieldThe first identity forces $T_z<\infty$, since otherwise $Y_t^2\le Y_0^2-4c^{2}t$ becomes negative. At a finite maximal lifetime, $Y_t\to 0$; otherwise the continuous Loewner driving function and the ordinary differential equation continue past that time. Integrating the last identity, using $S_t\ge c$, givescontradicting ${{\rm Y}}_t\ge r/2$. Thus every fixed rational ball inside $\mathbb{H}$ is hit almost surely. A countable intersection makes the trace a dense subset almost surely, and its relatively closed image, established by continuity and transience, then contains all of $\mathbb{H}$:This proves space-filling SLE above parameter eight. It rules out unvisited open regions, rather than inferring visits merely from membership in the filled compact H-hulls. The choice $\rho =\kappa -8$ does not address the critical value eight.

Let ${\tau }_n$ localize the nonnegative local martingale $M$. For $s\le t$,Conditional Fatou lemma and nonnegativity giveThus $M$ is a supermartingale, recovering the general fact about a nonnegative local martingale.

For a continuous local martingale $M$ with $M_0=0$, its stochastic exponential isThe Itô formula gives $d\mathcal{E}(M{)}_t=\mathcal{E}(M{)}_{t}d{M}_t$, so $\mathcal{E}(M)$ is a positive continuous local martingale. Every nonnegative local martingale is a supermartingale, because localization and the Conditional Fatou lemma turn the localized martingale equality into the supermartingale inequality.

Suppose an equivalent measure $Q$ made $S$ a local martingale. Stop at the positive time furnished by part (i), chosen so that $S_{t\wedge \tau }\ge 0$. A nonnegative local martingale is a supermartingale, and this one starts from $S_0=0$. Hence $S_{t\wedge \tau }=0$ $Q$-almost surely for every deterministic $t$.

But $Q$ and $P$ have the same null events, while $\tau >0$ almost surely. Some positive rational $q$ therefore satisfies $Q(q<\tau )>0$, and on that event part (i) gives $S_q>0$, a contradiction. No such equivalent measure exists.

A nonnegative local martingale is a supermartingale. If its integrable terminal limit has the same expectation as its initial value, conditional Fatou lemma gives $\mathbb{E}[Z_{\infty }\mid {\mathcal{F}}_t]\le Z_t$, and equality of expectations makes this an equality. It is therefore a uniformly integrable martingale.