User:Caliburn/s/mt/Change of Measure Formula for Integrals

Theorem

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

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

$\mu$ is absolutely continuous with respect to $\lambda$.

Let $g : X \to \overline \R$ be a $\mu$-integrable function.

Let:

$\ds f \in \frac {\d \mu} {\d \lambda}$

where $\dfrac {\d \mu} {\d \lambda}$ is the Radon-Nikodym derivative of $\mu$ with respect to $\lambda$.

Then:

$\ds \int g \rd \mu = \int g f \rd \lambda$

As an abuse of notation, this can be written more pleasingly as:

$\ds \int g \rd \mu = \int g \frac {\d \mu} {\d \lambda} \rd \lambda$

Proof

Since:

$\ds f \in \frac {\d \mu} {\d \lambda}$

we have:

$\ds \map \mu A = \int_A f \rd \lambda$

for each $A \in \Sigma$.

We first show the demand for $g$ a positive simple function.

Write the standard representation of $f$ as:

$g = \ds \sum_{j \mathop = 0}^n a_j \chi_{E_j}$

where:

$a_0 = 0$

$a_1, \ldots, a_n$ are non-negative real numbers

$E_0, E_1, \ldots, E_n$ is a partition of $X$ into $\Sigma$-measurable sets

Then we have:

\(\ds \int g f \rd \lambda\)

\(=\)

\(\ds \int f \paren {\sum_{j \mathop = 0}^n a_j \chi_{E_j} } \rd \lambda\)

\(\ds \)

\(=\)

\(\ds \sum_{j \mathop = 0}^n a_j \int f \chi_{E_j} \rd \lambda\)

\(\ds \)

\(=\)

\(\ds \sum_{j \mathop = 0}^n a_j \int_{E_j} f \rd \lambda\)

\(\ds \)

\(=\)

\(\ds \sum_{j \mathop = 0}^n a_j \map \mu {E_j}\)

\(\ds \)

\(=\)

\(\ds \int g \rd \mu\)