# Prove that region under graph of function is measurable

In the measure theory book that I am studying, we consider the 'area' under (i.e. the product measure of) the graph of a function as an example of an application of Fubini's Theorem for integrals (with respect to measures).

The setting: $$(X,\mathcal{A}, \mu)$$ is a $$\sigma$$-finite measure space, $$\lambda$$ is Lebesgue measure on $$(\mathbb{R},\mathcal{B}(\mathbb{R}))$$ (Borel $$\sigma$$-algebra), $$f:X \to [0,+\infty]$$ is $$\mathcal{A}$$-measurable, and we are considering the region under the graph of $$f$$,

$$E=\{(x,y)\in X \times \mathbb{R}|0\leq y < f(x)\}$$.

I need to prove $$E \in \mathcal{A} \times \mathcal{B}(\mathbb{R})$$. I thought to write $$E=g^{-1}((0,+\infty])\cap(X \times [0,+\infty])$$ where $$g(x,y)=f(x)-y$$ but I can't see why $$g$$ must be $$\mathcal{A} \times \mathcal{B}(\mathbb{R})$$-measurable. Any help would be appreciated.

$$g=k\circ h$$ where $$h(x,y)=(f(x),y)$$ and $$k(a,b)=a-b$$. [ Here $$h:X\times \mathbb R \to \mathbb R^{2}$$ and $$k:\mathbb R^{2} \to \mathbb R$$]. $$k:\mathbb R^{2} \to \mathbb R$$ is Borel measurable because it is continuous. To show that $$h$$ is measurable it is enough to show that $$h^{-1} (A \times B) \in \mathcal A \times B(\mathbb R)$$ for $$A,B \in \mathcal B(\mathbb R)$$. This is clear because $$h^{-1} (A \times B)=f^{-1}(A) \times B$$.
I have assumed that $$f$$ takes only finite values. To handle the general case let $$g(x)=f(x)$$ if $$f(x) <\infty$$ and $$0$$ if $$f(x)=\infty$$. Let $$F=\{(x,y):0\leq y . Then $$E=(f^{-1}\{\infty\}\times [0,\infty)) \cup [(f^{-1}\{\mathbb R\}\times \mathbb R) \cap F]$$.
• I suppose you mean $h^{-1}(A \times B) \in \mathcal{A} \times \mathcal{B}(\mathbb{R})$. This proof makes sense to me, but it doesn't take infinite values of $f$ into account. I would imagine it still passes through with the correct domains/codamins? But it seems a bit more awkward in that case. – AlephNull Dec 7 '18 at 0:36