Integral with respect to Discrete Measure
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Theorem
Let $\struct {X, \Sigma}$ be a measurable space.
Let $\sequence {x_n}_{n \mathop \in \N}$ be a sequence in $X$.
Let $\ds \mu = \sum_{n \mathop \in \N} \lambda_n \delta_{x_n}$ be a discrete measure on $\struct {X, \Sigma}$.
Let $f \in \MM_{\overline \R}^+, f: X \to \overline \R$ be a positive measurable function.
Then:
- $\ds \int f \rd \mu = \sum_{n \mathop \in \N} \lambda_n \map f {x_n}$
where the integral sign denotes $\mu$-integration.
Proof
We have:
\(\ds \int f \rd \mu\) | \(=\) | \(\ds \sum_{n \mathop \in \N} \lambda_n \int f \rd \delta_{x_n}\) | Integral with respect to Series of Measures | |||||||||||
\(\ds \) | \(=\) | \(\ds \sum_{n \mathop \in \N} \lambda_n \map f {x_n}\) | Integral with respect to Dirac Measure |
$\blacksquare$
Sources
- 2005: René L. Schilling: Measures, Integrals and Martingales ... (previous) ... (next): $9.10 \ \text{(ii)}$