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How can I prove that the graph of a measurable function is measurable. I recall that the graph of $f:\mathbb R\longrightarrow \mathbb R$ $$\Gamma(f)=\{(x,f(x))\mid x\in \mathbb R\}.$$

Attempt

1) if $f=1_{[a,b]}$ then $\Gamma(f)=[a,b]\times \{1\}\cup[a,b]^c\times \{0\}$, and thus $\Gamma(f)$ is measurable.

2) If $f$ is a step function, same thing.

3) If $f\geq 0$, then there are step function $f_n$ s.t. $f_n\nearrow f$. Now, I would like to have $\Gamma(f)=\bigcup_{n\in\mathbb N}\Gamma(f_n)$, but unfortunately it doesn't look to be the case. Any idea ?

4) Same for $f$ measurable. I have that $f=f^+-f^-$, but I don't think that $\Gamma(f)=\Gamma(f^+)\cup \Gamma(f^-)$. Any idea ?

user330587
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2 Answers2

26

Hint:

  1. Show that the mapping $$(x,y) \mapsto T(x,y) := f(x)-y$$ is measurable.
  2. Conclude that $\Gamma(f) = T^{-1}(\{0\})$ is measurable.
saz
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  • Just for the argument : If $g(x,y)=f(x)$ then ${(x,y)\mid g(x,y)\leq \alpha }=\mathbb R\times {x\mid f(x)\leq \alpha }$ and thus is measurable. Same, $h(x,y)=y$ is measurable. Therefore $T$ is measurable ? – user649261 Mar 02 '19 at 12:09
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    @user649261 Yes, that's correct (however it should read ${x \mid f(x) \leq \alpha} \times \mathbb{R}$ rather than $\mathbb{R} \times {x \mid f(x) \leq \alpha}$ but that's only a typo, I suppose.) – saz Mar 02 '19 at 13:01
  • Thank you. Your hint gives this exercise soooooo easy... (Surb make it very complicated next to you :)) – user649261 Mar 02 '19 at 13:02
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Hint

One can remark that $$\Gamma(f)=\bigcap_{n\in\mathbb N}\bigcup_{m\in\mathbb Z} f^{-1}\left(\left[\frac{m}{2^n},\frac{m+1}{2^{n}}\right]\right)\times \left[\frac{m}{2^n},\frac{m+1}{2^{n}}\right].$$

Surb
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  • What definition of boundary are you using? The set of irrational numbers is measurable, consists entirely of boundary points, and has infinite Lebesgue measure. – wanderingmathematician Jul 05 '18 at 15:34
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    @user334137: Great remark ! Indeed, it was a wrong hint. Thank you for pointing this. I corrected my answer. – Surb Jul 05 '18 at 16:09