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Consider the following family of measures on $(\mathbb R,\mathcal B(\mathbb R))$: $$ K_x(A) = \begin{cases} \int\limits_A \frac{1}{|x|\sqrt{2\pi}}\mathrm e^{-y^2/2x^2}\,dy&,\text{ if }x\neq 0, \\ I_{A}(0)&,\text{ if }x = 0. \end{cases} $$ where $I_A(t)$ is an indicator (characteristic) function of the set $A$. I wonder if it is possible to find a measure $\mu$ such that $\mu>>K_x$ for any $x$ (i.e. if $\mu(A) = 0$ then $K_x(A) = 0$) and $$ \xi(x,y):=\frac{\mathrm dK_x}{\mathrm d\mu} $$ is a continuous function of $(x,y)$.

Of course, we cannot take $\mu$ to be Lebesgue measure since $K_0(\{0\}) = 1$ and the only problem is in this point. Still I am not sure that there are no such measures at all, but I didn't fund an example or prove that there are no an example.

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A continuous density does not exist.

From the relation $\int_A \xi(0,y)\, \mu(dy)=\delta_0(A)=0$ for any Borel set $A$ that doesn't include the origin, we find that $\xi(0,y)=0$ $\mu$-almost everywhere on ${\mathbb R}\setminus\{0\}$.

Since $K_1\ll\mu$, this implies $\xi(0,y)=0$ $K_1$-almost everywhere on ${\mathbb R}\setminus\{0\}$. But $K_1$ has full support on ${\mathbb R}$ so, by continuity, we get $\xi(0,y)=0$ for all $y\in {\mathbb R}$.

But this contradicts $1=\int_{\mathbb R} \xi(0,y)\, \mu(dy)$.

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  • $\begingroup$ thank you, that's a nice proof. $\endgroup$ – Ilya Aug 10 '11 at 15:49
  • $\begingroup$ Glad to help out. $\endgroup$ – user940 Aug 10 '11 at 18:13

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