Muhammad Haris Rao
Definition: Let $\mu, \lambda$ be measures on a measurable space $(X, \mathcal{A})$. We say that $f : X \longrightarrow \mathbb{R}_{\ge 0}$ is a Radon-Nikodym derivative or density of $\mu$ with respect to $\lambda$ if for every $A \in \mathcal{A}$, $$ \mu(A) = \int_A f \, d \lambda $$
The density of a measure can be used for changing the measure over which one integrates in evaluating Lebesgue integrals as follows:
Theorem(Change of Measure): If $\mu, \lambda$ are measures on a measurable space $(X, \mathcal{A})$ with $\mu$ having density $\frac{d\mu}{d \lambda} : X \longrightarrow \mathbb{R}_{\ge 0}$, then for any measurable $f : X \longrightarrow \mathbb{R}$ integrable with respect to $\mu$, it holds that $$ \int_X f \, d \mu = \int_X f \cdot \frac{d \mu}{d \lambda} \, d \lambda $$
Proof. As is usual for these kinds of results, we will start by proving the result for indicator functions and work our way up. For any $A \in \mathcal{A}$, we have $$ \int_X \mathbf{1}_A \, d \mu = \mu(A) = \int_A \frac{d \mu}{d \lambda} \, d \lambda = \int_X \mathbf{1}_A \frac{d\mu}{d \lambda} \, d \lambda $$ By linearity, the desired result holds for all simple functions. Now suppose $f : X \longrightarrow \mathbb{R}_{\ge 0}$, and let $f_n : X \longrightarrow \mathbb{R}_{\ge 0}$ be a sequence of simple functions converging upwards to $f$ pointwise. Then $$ \int_X f \, d\mu = \lim_{n \to \infty} \int_X f_n \, d \mu = \lim_{n \to \infty} \int_X f_n \frac{d \mu}{d \lambda} \, d \mu $$ Since the Radon-Nikodym derivative is non-negative, the sequence $f_n \frac{d \mu}{d \lambda}$ converges upwards monotonically to $f \frac{d \mu}{d \lambda}$. So we can apply the monotone convergence theroem to obtain $$ \int_X f \, d\mu = \lim_{n \to \infty} \int_X f_n \frac{d \mu}{d \lambda} \, d \mu = \int_X f \, \frac{d \mu}{d \lambda} \, d \mu $$ This proves the desired result for all measurable $f \ge 0$. If $f : X \longrightarrow \mathbb{R}$ is integrable with respect to $\mu$, then $$ \int_X f \, d \mu = \int_X f^+ \, d \mu - \int_X f^- \, d \mu = \int_X f^+ \frac{d \mu}{d \lambda} \, d \lambda - \int_X f^- \frac{d \mu}{d \lambda} \, d \lambda = \int_X (f^+ - f^-) \frac{d \mu}{d \lambda} \, d \lambda = \int_X f \, \frac{d \mu}{d \lambda} \, d\lambda $$ This completes the proof. $\blacksquare$
We have the following result guaranteeing the existence of densities:
Theorem(Radon-Nikodym): Let $(X, \mathcal{A})$ be a measurable space, and $\mu, \nu : \mathcal{A} \longrightarrow [0, \infty]$ two $\sigma$-finite measures. If $\mu$ is absolutely continuous with respect to $\nu$, that is, whenever $A \in \mathcal{A}$ is such that $\nu(A) = 0$ then also $\mu(A) = 0$, then the Radon-Nikodym derivative of $\mu$ with respect to $\nu$ exists. That is, there is $\frac{d \mu}{d \nu} : X \longrightarrow \mathbb{R}_{\ge 0}$ such that if $A \in \mathcal{A}$ then $$ \mu \left( A \right) = \int_A \frac{d \mu}{d \nu} \, d \nu $$