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Choose the common dominating probability measure $\nu =(\mu +{\mu }^{\prime })/2$ and let $p=d\mu /d\nu$, $q=d{\mu }^{\prime }/d\nu$ be the Radon-Nikodym derivatives. The assumed second moments imply first-moment integrability by the Cauchy-Schwarz inequality. Since $Y$ is a separable Banach space, the measurable $f$ is a strongly measurable function, and both expected values exist as Bochner integrals.

The difference of these Bochner integrals satisfiesThe first inequality is the norm bound for a Bochner integral, and the second is the scalar Cauchy-Schwarz inequality. Since $(\sqrt{p}+\sqrt{q}{)}^2\le 2(p+q)$, the first factor is at mostWith the Hellinger distance convention of the original PDF, the second factor is $\sqrt{2}{d}_{\mathrm{Hell}}(\mu ,{\mu }^{\prime })$. Multiplication proves the Hellinger bound for differences of expectations:The proof uses only the two laws and their common dominating measure; no absolute continuity of measures assumption between $\mu$ and ${\mu }^{\prime }$ is needed.

TakeThis is a finite measure, hence a sigma-finite measure. If $\nu (B)=0$, nonnegativity gives $\mu (B)={\mu }_0(B)=0$, so both $\mu$ and ${\mu }_0$ are absolutely continuous with respect to $\nu$.

Let ${\mathcal{T}}_n$ be the first exit from $[1/n,1-1/n]$. On that compact interval the drift $v/2$ is bounded. The Girsanov theorem therefore gives, through every fixed time $t$, a probability measure equivalent to the original one under which the stopped process $X_{\cdot \wedge {\mathcal{T}}_n}$ has Brownian increments before ${\mathcal{T}}_n$.

If $A\subset (0,1)$ has Lebesgue measure zero, the normal distribution of Brownian motion givesSince $\{t<\mathcal{T}\}={\bigcup }_n\{t<{\mathcal{T}}_n\}$, countable subadditivity gives $\mathbb{P}(X_t\in A,t<\mathcal{T})=0$. Thus the killed law at time $t$ is absolutely continuous with respect to Lebesgue measure on $(0,1)$.

For a measurable set $E\subseteq U$, definewhere $1_E$ is its indicator function. This is finite because $U$ has finite measure and hence $1_E\in L^p(U)$. If the sets $E_k$ are pairwise disjoint and $E={\bigcup }_{k\ge 1}{E}_k$, thenThe continuity of the positive linear functional $\Lambda$ therefore givesso $\nu$ is a measure by countable additivity. Moreover, a set of zero Lebesgue measure has zero indicator in $L^p$, so $\nu$ is absolutely continuous with respect to Lebesgue measure.

The Radon-Nikodym theorem now supplies a measurable function $\omega \ge 0$ such thatConsequently, first for nonnegative simple functions and then, by approximation by nonnegative simple functions, for every nonnegative $f\in L^p(U)$,Applying the stated characterization of the Lp norm to $\omega$ shows that $\omega \in L^q(U)$ and $\Vert \omega {\Vert }_q\le \Vert \Lambda \Vert$. The Holder inequality makes $f\mapsto {\int }_{U}f\omega$ continuous on $L^p(U)$, while bounded simple functions form a dense subset; henceThis is the positive functional representation on Lp.

The relation $P\ll \tilde{P}$ is an instance of absolute continuity of measures: it means that every $\tilde{P}$-null event is also $P$-null,By the Radon-Nikodym theorem, this is equivalent to the existence of a nonnegative Radon-Nikodym derivative $dP/d\tilde{P}$ whose $\tilde{P}$-expectation is one.

If $Q\ll P$, uniform convergence on compacts in probability under $P$ implies the same convergence under $Q$. Indeed, let $Z=dQ/dP$ be the Radon-Nikodym derivative. For every event $E$,Integrability makes the second term arbitrarily small for large $K$. Thus $P(E_n)\to 0$ implies $Q(E_n)\to 0$; apply this to each event where the supremum on $[0,T]$ exceeds a fixed tolerance. Only absolute continuity of measures is needed, not equivalence.

The usual conditions are completeness and right continuity of a filtration: ${\mathcal{F}}_0$ contains all subsets of null events in the ambient probability space, and ${\mathcal{F}}_t={\bigcap }_{s>t}{\mathcal{F}}_s$. They allow the standard continuous-time martingale, stopping time and stochastic integration theorems to be used without repeated augmentation qualifications. An absolute continuity of measures change preserves old null sets but can introduce new ones; the usual completion under the new measure may therefore be understood when needed.