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Let $\lambda$ be the Gaussian measure on $\mathbb{R}$ defined by $$\lambda(E):=\int_{E}\frac{1}{\sqrt{2\pi}}e^{-x^2/2}dx.$$ Let $\operatorname{m}$ be Lebesgue measure on $\mathbb{R}$. What is the Radon-Nikodym derivative of $\operatorname{m}$ with respect to $\lambda$?

I've already show that $\operatorname{m}$ is absolutely continuous with respect to $\lambda$, written $\operatorname{m}\ll\lambda$. As $f:=1/\sqrt{2\pi}e^{-x^2/2}$ is non-negative and both $\operatorname{m}$ and $\lambda$ are $\sigma$-finite measures on $\mathbb{R}$, I take it the Radon-Nikodym derivative of $\lambda$ with respect to $\operatorname{m}$ is

$$\frac{d\lambda}{d\operatorname{m}} = \frac{1}{\sqrt{2\pi}}e^{-x^2/2}\tag{1}.$$

Is this correct, and if so how do I then get the Radon-Nikodym derivative of $\operatorname{m}$ with respect to $\lambda$?

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The notation for the Radon-Nikodym derivative is well chosen. That is,

$$\frac{dm}{d\lambda} = \frac{1}{\frac{d\lambda}{dm}}$$

whenever we have both $\lambda \ll m$ and $m \ll \lambda$. Of course, nothing is special about lebesgue measure $m$ in this theorem. You can find this as Corollary 3.10 in Folland.

In your case, if $\frac{d\lambda}{dm} = \frac{1}{\sqrt{2\pi}} e^{-x^2/2}$, then we would see

$$\frac{dm}{d\lambda} = \sqrt{2\pi}e^{x^2/2}$$


I hope this helps ^_^

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  • $\begingroup$ Ah, okay thanks! So, I suppose now I need to show that $\lambda\ll \operatorname{m}$. $\endgroup$ – D.Math Mar 23 at 4:23
  • $\begingroup$ Yup! But this shouldn't be hard to do, since $\lambda$ is given as integrating some function against $m$. $\endgroup$ – HallaSurvivor Mar 23 at 4:27
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The key is to show that the equality extends to $$\int_Eg\,d\lambda=\int_Egf\,dm.$$ Then if $g=dm/d\lambda$, you get $$\int_E(1-gf)\,dm=0$$ for all measurable $E$. Conclude that $fg=1$ a.e.

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