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When we assume that a probability measure $G$ has density $g$ with respect to the Lebesgue measure $ \lambda$. Do we refer to Radon-Nikodym theorem, where we have that

$\frac{dG}{d\lambda}=g$ and $G$ is absolutely continues to $\lambda$ ??

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  • $\begingroup$ Yes, that is exactly what a density function is. $\endgroup$ Commented Oct 1, 2020 at 11:41

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For a measure $\mu$ and a non-negative function $g$ we can always define a measure $\nu$ as \begin{equation} \nu(A) = \int_A g(x) \mu(dx) \tag{1}\end{equation} and it is immediately obvious that $\mu(A)= 0 \Rightarrow \nu(A)=0$, so $\nu$ is absolute continuous with respect to $\mu$.

The Radon-Nikodym theorem however states the reverse implikation, namely that if $\mu,\nu$ are $\sigma$-finite measures with $\nu << \mu$, then there exist a non-negative function $g$ such that $(1)$ holds. The fact that such a density is unique $\mu$-almost everywhere allows us to speak of the density of $\nu$ with respect to $\mu$, which we call the Radon-Nikodym derivative and we often use the notation $$g(x) = \frac{d\nu}{d\mu} (x)$$ to motivate intuitive formulas, such as $$\int_A d\nu = \int_A \frac{d\nu}{d\mu} d\mu.$$ So just to recap. You do not need to use the Radon-Nikodym theorem to assume that $G$ has a density with respect to $\mu$, you can simply assume that $G$ is given as in $(1)$ (with $G=\nu$). However the Radon-Nikodym theorem states that it is equivalent to assume absolute continuity with respect to $\mu$ (when we are dealing with $\sigma$-finite measures).

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