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Let $\mu_0$ be the standard Wiener measure on $C[0,T]$. Let $\mu_n$ be a sequence of measures with $\mu_n\ll \mu_0$ for all $n$ and so that the weak$^\ast$ limit of $\mu_n$ exists, call it $\mu$. Is it true that $\mu \ll \mu_0$?

I know for general measures this is not true. For example we can have a Gaussian with variance $\varepsilon$ and send $\varepsilon \to 0$.

But what about for Girsanov measures?

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  • $\begingroup$ What's your definition of a Girsanov measure? Just that it should be absolutely continuous to Wiener measure (definitely not sufficient for your conclusion), or that it should be the shift of $\mu_0$ by a drift, or what? $\endgroup$
    – Nate Eldredge
    Commented Jul 4, 2020 at 3:38
  • $\begingroup$ @NateEldredge My definition is that $\mu\ll \mu_0$, which by Girsanov means that there is a progressively measurable $F(t)$ with a.s. $W^{1,2}$ paths so that $B(t)-F(t)$ is a Brownian motion under $\mu$. $\endgroup$
    – user160547
    Commented Jul 4, 2020 at 3:41
  • $\begingroup$ What about taking processes with stronger and stronger drift toward zero? E.g. $\mu_n$ is the law of $X_t$ which solves $dX_t = -n X_t \,dt + dB_t$? Shouldn't that converge weakly to a point mass at 0? $\endgroup$
    – Nate Eldredge
    Commented Jul 4, 2020 at 3:54
  • $\begingroup$ @NateEldredge Is the law of $X_t$ absolutely continuous wrt $\mu_0$ for finite $n$? It doesn't seem obvious to me. $\endgroup$
    – user160547
    Commented Jul 4, 2020 at 4:06
  • $\begingroup$ math.stackexchange.com/questions/133691/… :-) $\endgroup$
    – Nate Eldredge
    Commented Jul 4, 2020 at 4:08

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This goes as wrong as can be and has nothing to do with any property of Wiener measure.

Theorem: Let $X$ be a separable metric space and $\nu$ a measure on $X$. The set of measures absolutely continuous with respect to $\nu$ is dense in the space of measures supported on the support of $\nu$ in the topology of weak convergence of measures.

Proof: Assume without loss of generality that the support of $\mu$ is all of $X$. It is known that the set of measures with finite support is dense, so it suffices to approximate such measures. So let $\mu$ be a Borel measure on $X$ with finite support $S$. By the definition of the topology of weak convergence, it suffices to show that there exists for every finite family $\mathcal{G}$ of bounded continuous functions on $X$ and every $\epsilon>0$ some measure $\tau$ on $X$ absolutely continuous with respect to $\nu$ such that $$\bigg|\int g~\mathrm d\mu-\int g~\mathrm d\tau\bigg|<\epsilon$$ for all $g\in\mathcal{G}$. For each $s\in S$, let $V_s$ be an open neighborhood on which each $g\in\mathcal{G}$ varies by less than $\epsilon/\# S$. We can and do choose the $V_s$ to be disjoint. Since $\nu$ has full support, we have $\nu(V_s)>0$ for all $s\in S$. Define a measurable function $h:X\to\mathbb{R}$ by letting $h(x)=\mu(s)/\nu(V_s)$ for $x\in V_s$ and $h(x)=0$ if $x$ is in no $V_s$. Let $\tau$ be the measure that has Radon-Nikodym derivative $h$ with respect to $\nu$. Take any $g\in\mathcal{G}$. Then, $$\bigg|\int g~\mathrm d\mu-\int g~\mathrm d\tau\bigg|=\bigg|\int g~\mathrm d\mu-\int hg~\mathrm d\nu\bigg|$$ $$=\bigg| \sum_{s\in S}\bigg(g(s)\mu(s)-\int_{V_s}hg~\mathrm d\nu\bigg)\bigg|$$ $$\leq \sum_{s\in S}\bigg|\bigg(g(s)\mu(s)-\int_{V_s}hg~\mathrm d\nu\bigg)\bigg|$$ $$<\sum_{s\in S} \mu(s)\epsilon/\# S=\epsilon.$$

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