Measure with Density is Measure

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

Let $f: X \to \overline \R_{\ge 0}$ be a positive $\Sigma$-measurable function.

Then the $f \mu$, the measure with density $f$ with respect to $\mu$ is a measure.

Proof
Note that for each $A \in \Sigma$, we have:


 * $\ds \map {\paren {f \mu} } A = \int_A f \rd \mu$

We verify each of the three conditions for a measure.

Proof of $(1)$
We have:


 * $\map {\paren {f \mu} } A \ge 0$

for each $A \in \Sigma$ from the definition of the $\mu$-integral of a positive measurable function.

Proof of $(2)$
Let $\sequence {D_n}_{n \mathop \in \N}$ be a sequence of pairwise disjoint sets.

Let:


 * $\ds D = \bigcup_{n \mathop = 1}^\infty D_n$

We then have:

which shows $(2)$.

Proof of $(3')$
We have:

verifying $(3')$.

Since all three conditions have been verified, we have that $\mu f$ is a measure.