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If $f,g$ are real-value measurable functions defined in $E$ and $\phi$ is a real-value continuous function defined in $\mathbb{R} \times \mathbb{R}$, show that $\phi(f,g)$ is measurable.

We know that if $h$ is real-value measurable function and $\phi$ is a continuous function, then $\phi \circ f$ is measurable.

My idea is to prove that $h = f \times g$ is measurable, but I'm not sure if this is true. If two bounded sets are measurable, then the cartesian product is also, right? But we also cannot ensure that the inverse image of an interval is bounded. Can someone help me?

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    $\begingroup$ Can you remind me what $f\times g$ means? $\endgroup$ – zhw. Sep 29 '18 at 18:23
  • $\begingroup$ Maybe is a bad notation. I mean $f \times g: E \times E \to \mathbb{R} \times \mathbb{R}$ defined by $(f \times g)(x) = (f(x),g(x))$. $\endgroup$ – Lucas Corrêa Sep 29 '18 at 18:24
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Rudin does it like this:

set $h(x)=\phi(f(x),g(x))$ and consider $G_a=\left \{ (u,v):\phi(u,v)>a \right \}$, which is an open set in $\mathbb R^2$, and so a union of basis elements of the form

$I_n =(a_n,b_n)\times (c_n,d_n)$, i.e. $G_a=\bigcup_n I_n$.

Now $\left \{ x:a_n<f(x)<b_n \right \}$ and $\left \{ x:c_n<g(x)<d_n \right \}$ are measurable, and so therefore is

$\left \{ x:(f(x),g(x))\in I_n \right \}=\left \{ x:a_n<f(x)<b_n \right \}\cap\left \{ x:c_n<g(x)<d_n \right \}.$

But then we have that

$\left \{ x:h(x)>a \right \}=\left \{ x:(f(x),g(x))\in G_a \right \}=\bigcup_n \left \{ x:(f(x),g(x))\in I_n \right \}$

is measurable, being a countable union of measurable sets.

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  • $\begingroup$ Nice and simple proof! Thank you! $\endgroup$ – Lucas Corrêa Sep 29 '18 at 19:06
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    $\begingroup$ Rudin is a master of technique! I believe this proof is from his "Principles of Mathematical Analysis", but his "Real and Complex Analysis" is a real masterpiece. The way he gets all the basic measure theory results is like magic.... $\endgroup$ – Matematleta Sep 29 '18 at 19:12
  • $\begingroup$ I still cannot read whole "Principles of Mathematical Analysis", but I agree with you! $\endgroup$ – Lucas Corrêa Sep 30 '18 at 4:34

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