Change of Measures Formula for Integrals

Theorem
Let $\struct {X, \Sigma}$ be a measurable space.

Let $\mu$ and $\nu$ be $\sigma$-finite measures on $\struct {X, \Sigma}$ such that:


 * $\nu$ is absolutely continuous with respect to $\mu$.

Let $g$ be a Radon-Nikodym derivative of $\nu$ with respect to $\mu$.

Let $f : X \to \overline \R$ be a positive $\Sigma$-measurable function.

Then:


 * $\ds \int f \rd \nu = \int \paren {f \cdot g} \rd \mu$

where:


 * $f \cdot g$ is the pointwise product of $f$ and $g$
 * $\ds \int \cdot \rd \nu$ denotes the integral of a positive $\Sigma$-measurable function with respect to $\nu$.

Proof
First consider $f$ a positive simple function.

From Simple Function has Standard Representation, there exists:


 * a finite sequence $a_0, \ldots, a_n$ of non-negative real numbers
 * a partition $E_0, E_1, \ldots, E_n$ of $X$ into $\Sigma$-measurable sets

such that:


 * $\ds f = \sum_{i \mathop = 0}^n a_i \chi_{E_i}$

Then:

Now suppose that $f$ is a general positive $\Sigma$-measurable function.

From Measurable Function is Pointwise Limit of Simple Functions, there exists an increasing sequence $\sequence {f_n}_{n \mathop \in \N}$ of positive simple functions such that:


 * $\ds \map f x = \lim_{n \mathop \to \infty} \map {f_n} x$

for each $x \in X$.

For each $n \in \N$, we have:


 * $\ds \int f_n \rd \nu = \int \paren {f_n \cdot g} \rd \mu$

From the Monotone Convergence Theorem (Measure Theory), we have:


 * $\ds \lim_{n \mathop \to \infty} \int f_n \rd \nu = \int f \rd \nu$

Note that $\sequence {f_n \cdot g}_{n \mathop \in \N}$ is an increasing sequence with $f_n \cdot g \to f \cdot g$.

So, from the Monotone Convergence Theorem (Measure Theory), we have:


 * $\ds \lim_{n \mathop \to \infty} \int \paren {f_n \cdot g} \rd \mu = \int \paren {f \cdot g} \rd \mu$

So, we have:


 * $\ds \int f \rd \nu = \int \paren {f \cdot g} \rd \mu$