Show $\mu^*$ is a measure, and $\int_Xg\circ f d\mu = \int_Y g d\mu^*$

Question

Suppose $(X,\mathcal{M})$ and $(Y,\mathcal{N})$ are two measure spaces and $f:X\rightarrow Y$ is such that $f^{-1}(B)\in\mathcal{M}$ for all $B\in\mathcal{N}$. Suppose $\mu$ is a positive measure on $(X,\mathcal{M})$ and let $\mu^*$ be defined by the formula: $$\mu^*(B)=\mu(f^{-1}(B)), \ \ \ \ \forall B\in\mathcal{N}$$

Show that $\mu^*$ is a measure on $(Y,\mathcal{N})$ and if $g\in L^1(\mu^*)$ then $g\circ f$ is in $L^1(\mu)$ and $$\int_Xg\circ f\ d\mu = \int_Y g\ d\mu^*$$

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My solution:

$(\star)$ Because $\mu^*(\varnothing)=\mu(f^{-1}(\varnothing))$, then $\mu^*(\varnothing)=0$

Take $B\subseteq \bigcup\limits_{k=1}^\infty B_k$, then we w.t.s. that $\mu^*(B)\le \sum_{k=1}^\infty \mu^*(B_k)$

But $\mu*(B)=\mu(f^{-1}(B))$, and $\mu^*(B_k)=\mu(f^{-1}(B_k))$

And since $\mu$ is a measure, then we have: $f^{-1}(B)\subseteq f^{-1}(B_k)$, and:

\begin{align*}\mu(f^{-1}(B)) & \le\sum_{k=1}^\infty \mu(f^{-1}(B_k))\\ \mu^*(B)&\le\sum_{k=1}^\infty \mu^*(B_k) \end{align*}

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I feel like I have to justify $(\star)$ better, and then don't know how to show the equality of the integrals.

Answer 1

It suffices to show that th integral formula holds for simple functions (why?). I'll leave you to verify that $g \circ f \in L^1(\mu)$. Let $\phi:Y \to \mathbb{R}$ be a simple function, i.e., a function with finite range. We can then write $\phi(x) = \sum_{i=1}^n c_i \chi_{E_k}$ where the $c_k$ are the distinct elements of the range and $E_k = \phi^{-1}(\{c_k\})$. Note that the $E_k \in \cal{N}$ and the $E_k$'s are pairwise disjoint. Then $$(\phi \circ f)(x) = \sum_{i=1}^n c_i \chi_{f^{-1}(\phi^{-1}(x))} = \sum_{i=1}^n c_i \chi_{f^{-1}(E_k)}$$ and we have

\begin{align*}\int_Y \phi \, d\mu^* & = \int_Y\sum_{i=1}^n c_i \chi_{E_k} \, d\mu^*\\ & = \sum_{i=1}^n c_i\int_Y \chi_{E_k} \, d\mu^*\\ & = \sum_{i=1}^n c_i\mu^*(E_k)\\ & = \sum_{i=1}^n c_i\mu(f^{-1}(E_k))\\ & = \sum_{i=1}^n c_i\int_X \chi_{f^{-1}(E_k))} \, d\mu\\ & = \int_X \sum_{i=1}^n c_i \chi_{f^{-1}(E_k))} \, d\mu\\ & = \int_X \phi \circ f \, d\mu. \end{align*}