# Measure has density with respect to Lebesgue measure

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 $$\lambda$$ ??

• Yes, that is exactly what a density function is. Oct 1 '20 at 11:41

For a measure $$\mu$$ and a non-negative function $$g$$ we can always define a measure $$\nu$$ as $$$$\nu(A) = \int_A g(x) \mu(dx) \tag{1}$$$$ and it is immediately obvious that $$\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).