# Finding Radon-Nikodym derivative given integral

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 $$\lambda$$?

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 ^_^

• Ah, okay thanks! So, I suppose now I need to show that $\lambda\ll \operatorname{m}$. Mar 23, 2021 at 4:23
• Yup! But this shouldn't be hard to do, since $\lambda$ is given as integrating some function against $m$. Mar 23, 2021 at 4:27

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.