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 $\lambda$?
