# Show that $E=\{(x,\alpha)\mid 0\leq \alpha<|f(x)|\}$ is measurable if $f$ is measurable.

I have to show that $E=\{(x,\alpha)\mid 0\leq \alpha<|f(x)|\}$ is measurable if $f$ is measurable.

My attempt :

If $f=\boldsymbol 1_F$ where $f$ is measurable, then $$E=\big((\mathbb R\cap F)\times [0,1[\big)\cup \big((\mathbb R\cap F^c)\times \{0\}\big)$$ which is measurable since it's intersection, union and product of measurable set. If $f$ is simple (remark that $f$ simple $\iff$ $|f|$ simple) i.e. $$|f|=\sum_{i=1}^n a_i\boldsymbol 1_{F_i}$$ where $a_i>0$ and $F_i$ measurable, we obtain in the same way $$E=\bigcup_{i=1}^n\big((\mathbb R\cap F_i)\times [0,a_i[\big)\cup \bigcap_{i=1}^n \big((\mathbb R \cap F_i^c)\times \{0\}\big)$$ which is also measurable since it's finite union, intersection and product of measurable set. Finally, if $|f|\geq 0$ we have a sequence of simple function $\{\varphi_k\}_{k\in\mathbb N}$ s.t. $$\varphi_k(x)\nearrow |f(x)|\quad\text{ for all }x.$$

Let $$E_n=\{(x,\alpha)\mid 0\leq \alpha\leq \varphi_n(x)\}.$$ Therefore $E_n\subset E_{n+1}$ and thus $E_n\nearrow E$ i.e. $$E=\bigcup_{n=1}^\infty E_n,$$ and since $E_n$ is measurable, the set $E$ is measurable.

Is it correct ?

$E=\bigcup_{i=1}^n\big((\mathbb R\cap F_i)\times [0,a_i[\big)\cup \bigcap_{i=1}^n \big((\mathbb R \cap F_i)\times \{0\}\big)$

$E=\bigcup_{i=1}^n\big((\mathbb R\cap F_i)\times [0,a_i[\big)\cup \bigcap_{i=1}^n \big((\mathbb R \cap F_i^{\color{red}{c}})\times \{0\}\big).$

(Edit: The OP has fixed it.)

Here is an idea for an alternative proof: Consider the measurable space $(\mathbb{R}^2,\mathcal{B}(\mathbb{R}^2))$. Since $f$ is measurable, it is not difficult to see that

$$(\mathbb{R}^2,\mathcal{B}(\mathbb{R}^2)) \ni (x,\alpha) \mapsto |f(x)| \in (\mathbb{R},\mathcal{B}(\mathbb{R}))$$

is measurable. Moreover, also the mapping

$$(\mathbb{R}^2,\mathcal{B}(\mathbb{R}^2)) \ni (x,\alpha) \mapsto \alpha \in (\mathbb{R},\mathcal{B}(\mathbb{R}))$$

is measurable. Hence,

$$(\mathbb{R}^2,\mathcal{B}(\mathbb{R}^2)) \ni (x,\alpha) \mapsto g(\alpha,x) := |f(x)| -\alpha \in (\mathbb{R},\mathcal{B}(\mathbb{R}))$$

is measurable. This implies that

$$E = \{(x,\alpha); g(x,\alpha)>0\} \cap \{(x,\alpha); \alpha \geq 0\} = g^{-1}((0,\infty)) \cap (\mathbb{R} \cap [0,\infty))$$

is measurable.

• thank you. Just a question, is the reciprocal of a measurable function is measurable ? – idm Oct 25 '15 at 15:05
• @idm If it's well-defined (i.e. we do not divide by $0$): Yes. – saz Oct 25 '15 at 15:11