# 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.

$\Box$

### 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:

 $\ds \map {\paren {f \mu} } D$ $=$ $\ds \int_D f \rd \mu$ $\ds$ $=$ $\ds \sum_{n \mathop = 1}^\infty \int_{D_n} f \rd \mu$ Integral of Positive Measurable Function over Disjoint Union $\ds$ $=$ $\ds \sum_{n \mathop = 1}^\infty \map {\paren {f \mu} } {D_i}$

which shows $(2)$.

$\Box$

### Proof of $(3')$

We have:

 $\ds \map {\paren {f \mu} } \O$ $=$ $\ds \int_\O f \rd \mu$ $\ds$ $=$ $\ds 0$ Empty Set is Null Set, Integral of Positive Measurable Function over Null Set

verifying $(3')$.

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

$\blacksquare$