Pushforward Measure

Muhammad Haris Rao

Definition: Let $(X, \mathcal{A}, \mu)$ be a measure space, and $(Y, \mathcal{B})$ a measurable space. Given a measurable function $f : X \longrightarrow Y$, the pushforward measure $\mu^* : \mathcal{B} \longrightarrow [0, \infty]$ is defined at $A \in \mathcal{B}$ by $$ \mu^* (A) = \mu \left( f^{-1} (A) \right) $$

Because $f$ is measurable, we have $f^{-1}(A) \in \mathcal{A}$ for every $A \in \mathcal{B}$ so that $\mu^*$ is well-defined. $\mu^*$ is a measure because $$ \mu^*(\emptyset) = \mu(f^{-1}(\emptyset)) = \mu(\emptyset) = 0 $$ and if $A_1, A_2, \cdots \in \mathcal{B}$ are disjoint, then $f^{-1}(A_1), f^{-1}(A_2), \cdots \in \mathcal{A}$ are also disjoint so that $$ \mu^* \left( \bigcup_{n \ge 1} A_n \right) = \mu \left( f^{-1} \left( \bigcup_{n \ge 1} A_n \right) \right) = \mu \left( \bigcup_{n \ge 1} f^{-1}(A_n) \right) = \sum_{n \ge 1} \mu \left( f^{-1} (A_n) \right) = \sum_{n \ge 1} \mu^* (A_n) $$ Integrals over the pushforward measure can be evaluated as follows:

Theorem: Let $(X, \mathcal{A}, \mu)$ be a measure space, $(Y, \mathcal{B})$ a measurable space, $f : X \longrightarrow Y$ measurable, and $\mu^* : \mathcal{B} \longrightarrow [0, \infty]$ the pushforward measure. Then for any $\phi : Y \longrightarrow \mathbb{R}$ integrable, $$ \int_Y \phi \, d \mu^* = \int_X \phi \circ f \, d \mu $$ Moreover, $\phi$ is integrable with respect to $\mu^*$ if and only if $\phi \circ f$ is integrable with respect to $\mu$.

Proof. First, we deal with indicator functions. For any $B \in \mathcal{B}$, we have $$ \int_Y \mathbf{1}_B \, d \mu^* = \mu^*(B) = \mu \left( f^{-1} (B) \right) = \int_X \mathbf{1}_{f^{-1}(B)} \, d \mu = \int_Y \mathbf{1}_B \circ f \, d \mu $$ Now to extend to simple functions. Notice that if $B_1, B_2, \cdots, B_n \in \mathcal{B}$ are disjoint and $a_1, a_2, \cdots, a_n \in \mathbb{R}$ then \begin{align*} \sum_{i = 1}^n a_i \mathbf{1}_{B_i} \circ f = \sum_{i = 1}^n a_i \mathbf{1}_{f^{-1}(B_i)} = \left( \sum_{i = 1}^n a_i \mathbf{1}_{B_i} \right) \circ f \end{align*} Hence, we have $$ \int_Y \sum_{i = 1}^n a_i \mathbf{1}_{B_i} \, d \, \mu^* = \sum_{i = 1}^n a_i \int_Y \mathbf{1}_{B_i} \, d \mu^* = \sum_{i = 1}^n a_i \int_Y \mathbf{1}_{B_i} \circ f \, d \mu = \int_Y \sum_{i = 1}^n a_i \mathbf{1}_{B_i} \circ f \, d \mu = \int_Y \left( \sum_{i = 1}^n a_i \mathbf{1}_{B_i} \right) \circ f \, d \mu $$ which proves the result for simple functions. If $\phi : Y \longrightarrow \mathbb{R}_{\ge 0}$ is positive, let $\phi_n : Y \longrightarrow \mathbb{R}_{\ge 0}$ be a sequence of simple functions converging upwards to $\phi$. Then we will have $\phi_n \circ f$ converging upwards to $\phi \circ f$ as will, so $$ \int_Y \phi \, d \mu^* = \lim_{n \to \infty} \int_Y \phi_n \, d \mu^* = \lim_{n \to \infty} \int_Y \phi_n \circ f \, d \mu = \int_Y \phi \circ f \, d \mu $$ Finally, if $\phi : Y \longrightarrow \mathbb{R}$ is integrable with respect to $\mu^*$, then $(\phi \circ f)^\pm = \phi^\pm \circ f$ so that $$ \int_Y \phi \, d \mu^* = \int_Y \phi^+ \, d \mu^* - \int_Y \phi^- \, d \mu^* = \int_Y \phi^+ \circ f \, d \mu - \int_Y \phi^- \circ f \, d \mu = \int_Y \left( \phi \circ f \right)^+ \, d \mu - \int_Y \left( \phi \circ f \right)^- \, d \mu = \int_Y \phi \circ f \, d \mu $$ This proves the desired result for all integrable functions.

The second claim that $\phi$ is integrable if and only if $\phi \circ f$ is follows because the result holds unconditinoally for positive measurable functions and $$ \int_Y |\phi| \, d \mu^* = \int_Y |\phi| \circ f \, d \mu = \int_Y \left| \phi \circ f \right| \, d \mu^* $$ This completes the proof.$\blacksquare$