Integral with respect to Pushforward Measure
Theorem
Let $\struct {X, \Sigma, \mu}$ be a measure space.
Let $\struct {X', \Sigma'}$ be a measurable space.
Let $T: X \to X'$ be a $\Sigma \, / \, \Sigma'$-measurable mapping.
Let $f: X' \to \overline \R$ be a positive $\Sigma'$-measurable function.
Let $\map T \mu$ be the pushforward measure of $\mu$ under $T$.
Then $f \circ T: X \to \overline \R$ is positive and $\Sigma$-measurable with:
- $\ds \int_{X'} f \rd \map T \mu = \int_X f \circ T \rd \mu$
Corollary
Then $f \circ T : X \to \overline \R$ is $\mu$-integrable if and only if $f : X' \to \overline \R$ is $\map T \mu$-integrable.
In this case, we have:
- $\ds \int_{X'} f \rd \map T \mu = \int_X f \circ T \rd \mu$
Proof
From Composition of Measurable Mappings is Measurable:
- $f \circ T$ is $\Sigma$-measurable.
Clearly $f \circ T \ge 0$, so $f \circ T$ is a positive $\Sigma$-measurable function.
We first show the proposition for characteristic functions $f$.
Suppose $f = \chi_A$ for $A \in \Sigma'$.
Then, we have:
\(\ds \int_{X'} \chi_A \rd \map T \mu\) | \(=\) | \(\ds \map T {\mu} \sqbrk A\) | Integral of Characteristic Function: Corollary | |||||||||||
\(\ds \) | \(=\) | \(\ds \map \mu {T^{-1} \sqbrk A}\) | Definition of Pushforward Measure | |||||||||||
\(\ds \) | \(=\) | \(\ds \int_X \chi_{T^{-1} \sqbrk A} \rd \mu\) | Integral of Characteristic Function: Corollary | |||||||||||
\(\ds \) | \(=\) | \(\ds \int_X \chi_A \circ T \rd \mu\) | Characteristic Function of Preimage |
So, we have:
- $\ds \int_{X'} \chi_A \rd \map T \mu = \int_X \chi_A \circ T \rd \mu$
whenever $A \in \Sigma'$.
Now suppose that $f$ is a positive simple function.
From Simple Function has Standard Representation, there exists:
- a finite sequence $a_0, \ldots, a_n$ of non-negative real numbers
- a partition $E_0, E_1, \ldots, E_n$ of $X$ into $\Sigma'$-measurable sets
such that:
- $\ds f = \sum_{i \mathop = 0}^n a_i \chi_{E_i}$
Then:
\(\ds \int_{X'} f \rd \map T \mu\) | \(=\) | \(\ds \int_{X'} \paren {\sum_{i \mathop = 0}^n a_i \chi_{E_i} } \rd \map T \mu\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \sum_{i \mathop = 0}^n \int_{X'} a_i \chi_{E_i} \rd \map T \mu\) | Integral of Positive Measurable Function is Additive | |||||||||||
\(\ds \) | \(=\) | \(\ds \sum_{i \mathop = 0}^n a_i \int_{X'} \chi_{E_i} \rd \map T \mu\) | Integral of Positive Measurable Function is Positive Homogeneous | |||||||||||
\(\ds \) | \(=\) | \(\ds \sum_{i \mathop = 0}^n a_i \int_X \chi_{E_i} \circ T \rd \mu\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \sum_{i \mathop = 0}^n \int_X \paren {a_i \chi_{E_i} } \circ T \rd \mu\) | Integral of Positive Measurable Function is Positive Homogeneous | |||||||||||
\(\ds \) | \(=\) | \(\ds \int_X \paren {\sum_{i \mathop = 0}^n a_i \chi_{E_i} } \circ T \rd \mu\) | Integral of Positive Measurable Function is Additive | |||||||||||
\(\ds \) | \(=\) | \(\ds \int_X f \circ T \rd \mu\) |
Now suppose that $f$ is a general positive $\Sigma'$-measurable function.
From Measurable Function is Pointwise Limit of Simple Functions, there exists an increasing sequence $\sequence {f_n}_{n \mathop \in \N}$ of positive simple functions such that:
- $\ds \map f x = \lim_{n \mathop \to \infty} \map {f_n} x$
for each $x \in X$.
For each $n \in \N$, we have:
- $\ds \int_{X'} f_n \rd \map T \mu = \int_X f_n \circ T \rd \mu$
Then:
\(\ds \int_{X'} f \rd \map T \mu\) | \(=\) | \(\ds \lim_{n \mathop \to \infty} \int_{X'} f_n \rd \map T \mu\) | Monotone Convergence Theorem (Measure Theory) | |||||||||||
\(\ds \) | \(=\) | \(\ds \lim_{n \mathop \to \infty} \int_X f_n \circ T \rd \mu\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \int_X f \circ T \rd \mu\) | Monotone Convergence Theorem (Measure Theory) |
$\blacksquare$
Sources
- 2005: René L. Schilling: Measures, Integrals and Martingales ... (previous) ... (next): $14.1$