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 $F=\{(x,y):0\leq y <g(x)\}$. Then $E=(f^{-1}\{\infty\}\times [0,\infty)) \cup [(f^{-1}\{\mathbb R\}\times \mathbb R) \cap F]$.

  • $\begingroup$ 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. $\endgroup$ – AlephNull Dec 7 '18 at 0:36
  • $\begingroup$ @AlephNull I have edited my answer. $\endgroup$ – Kavi Rama Murthy Dec 7 '18 at 5:19
  • $\begingroup$ Ah, that's a clear way of doing it, instead of trying to handle the finite and infinite cases in one. $\endgroup$ – AlephNull Dec 7 '18 at 11:25

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