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I have seen a lot of questions trying to show that graphs are measurable. However, I'm asking for the other direction. This is the question:

Suppose $(X, \mathcal{S})$ is a measurable space and $f : X \to [ 0, \infty ] $ is a function. Let $\mathcal{B}$ denote the σ-algebra of Borel subsets of $( 0, \infty )$. Prove that $U_f \in S \otimes \mathcal{B}$ if and only if $f$ is an $\mathcal{S}$-measurable function.

Definition of $U_f$:

Suppose $X$ is a set and $f : X \to [ 0, ∞ ]$ is a function. Then the region under the graph of $f$ , denoted $U_f$ , is defined by $U_f = \{( x, t ) \in X \times ( 0, \infty ) : 0 < t < f ( x )\} $.

I'm only asking how to show the forward direction: if $U_f \in S \otimes \mathcal{B}$ then $f$ is an $\mathcal{S}$-measurable function.

Here $\mathcal{S} \otimes \mathcal{B}$ is defined as the smallest $\sigma$-algebra that contains $\{A \times B: A \in \mathcal{S}, B \in \mathcal{B} \}$.

I'm trying to show that $f^{-1}((a, \infty)) \in \mathcal{S}$, but I'm completely stuck. Can I get some help?

user1691278
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1 Answers1

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The mapping $$X \ni x \mapsto h(x):=(x,a) \in X \times (0,\infty)$$ is measurable for any fixed $a>0$. Thus, $h^{-1}(U_f) \in \mathcal{S}$. Noting that $$h^{-1}(U_f) = \{x \in X; a<f(x)\}=f^{-1}((a,\infty)),$$ it follows that $f$ is measurable.

saz
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