Measure with Density is Measure

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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.

$\blacksquare$


Sources

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