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
- 2005: René L. Schilling: Measures, Integrals and Martingales ... (previous) ... (next): $\S 9$: Problem $5$
- 2005: René L. Schilling: Measures, Integrals and Martingales ... (previous) ... (next): $10.8$