Change of Measures Formula for Integrals/Corollary

Corollary

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 $\nu$-integrable function.

Then $f \cdot g$ is $\mu$-integrable with:

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

where:

$f \cdot g$ is the pointwise product of $f$ and $g$

$\ds \int \cdot \rd \nu$ denotes the integral of a $\nu$-integrable function with respect to $\nu$.

Proof

From Pointwise Product of Measurable Functions is Measurable, we have:

$f \cdot g$ is $\Sigma$-measurable.

We show that:

$\ds \int \paren {f \cdot g}^+ \rd \mu < \infty$

and:

$\ds \int \paren {f \cdot g}^- \rd \mu < \infty$

where $\paren {f \cdot g}^+$ and $\paren {f \cdot g}^-$ denote the positive part and negative part of $f \cdot g$ respectively.

We have:

\(\ds \int \paren {f \cdot g}^+ \rd \mu\)

\(=\)

\(\ds \int \paren {f^+ \cdot g} \rd \mu\)

Positive Part of Pointwise Product of Functions, noting that $g \ge 0$

\(\ds \)

\(=\)

\(\ds \int f^+ \rd \nu\)

Change of Measures Formula for Integrals

\(\ds \)

\(<\)

\(\ds \infty\)

Definition of Integrable Function

and:

\(\ds \int \paren {f \cdot g}^- \rd \mu\)

\(=\)

\(\ds \int \paren {f^- \cdot g} \rd \mu\)

Negative Part of Pointwise Product of Functions, noting that $g \ge 0$

\(\ds \)

\(=\)

\(\ds \int f^- \rd \nu\)

Change of Measures Formula for Integrals

\(\ds \)

\(<\)

\(\ds \infty\)

Definition of Integrable Function

So $f \cdot g$ is $\mu$-integrable.

Then:

\(\ds \int \paren {f \cdot g} \rd \mu\)

\(=\)

\(\ds \int \paren {f \cdot g}^+ \rd \mu - \int \paren {f \cdot g}^- \rd \mu\)

Definition of Integral of Integrable Function

\(\ds \)

\(=\)

\(\ds \int f^+ \rd \nu - \int f^- \rd \nu\)

\(\ds \)

\(=\)

\(\ds \int f \rd \nu\)

Definition of Integral of Integrable Function

$\blacksquare$