Question about the exercise 2.5.50 of Folland's textbook
Suppose $(X, \mathcal{M}, \mu)$ is a $\sigma$-finite measure space and $f \in L^{+}(X)$ where $f: X\to [0,\infty]$ are all measurable functions. Let $$G_f = \{ (x,y) \in X \times [0, \infty]: y \le f(x) \}.$$
The book gives the hint that the map $(x,y) \to f(x)-y$ is the composition of $(x,y) \to (f(x),y)$ and $(z,y) \to z -y$.
(1)My question is how to prove the map $h: (x,y)\to (f(x),y)$ is measurable by the coordinate function is measurable?
(2)And why $(z,y)=z-y$ is measurable by the continuous function?
(3)Also, what does the "$\sigma$-finite " work?
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$\begingroup$Theorem (2.2): If $X$ and $Y$ are measurable topological spaces and $f : X \to Y$ is continuous, the $f$ is measurable. Here $s := (z,y) \mapsto z - y : \mathbb{R}^2 \to \mathbb{R}$ is continuous.
The map $p:= (x,y) \mapsto (f(x), y) : \mathbb{R}^2 \to \mathbb{R}^2$ is the tensor product of the measurable maps $f = x \mapsto f(x) : \mathbb{R} \to \mathbb{R}$ and $\mathrm{id} = x \mapsto x : \mathbb{R} \to \mathbb{R}$ and therefore measurable. (Proposition (2.4))
A composition of measurable functions is measurable. (Text at beginning of section 2.1). Thus $h = s \circ p : \mathbb{R}^2 \to \mathbb{R}$ is measurable.
I'm not sure, but the $\sigma$-finiteness is probably needed to prove the $\mu \times m(G_f) = \int f \, d\mu$ part of the exercise, since it's a premise of (2.37) The Fubini-Tonelli Theorem.
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