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Let $C:\mathbb{R}^n\times\mathbb{R}\rightarrow \mathbb{R}^{n+1}:((x_1,\cdots,x_n),x_{n+1})\mapsto (x_1,\cdots,x_{n+1})$

Let $f:\mathbb{R}^{n+1}\rightarrow [0,\infty]$ be a Lebesgue measurable function.

Let $g:\mathbb{R}\rightarrow [0,\infty]$ be a Lebesgue measurable function.

Now define $F(x,y)=(f\circ C)(x,y)g(y)$, $\forall (x,y)\in \mathbb{R}^n\times\mathbb{R}$

How do i prove that $F\circ C^{-1}:\mathbb{R}^{n+1}\rightarrow [0,\infty]$ is Lebesgue measurable?

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The product of measurable functions is measurable.

share|cite|improve this answer
    
I'm the OP. This question is equivalent to this statement "Let $f:(X\times Y,\mathfrak{M}\otimes \Sigma) \rightarrow [0,\infty]$ and $g:(Y,\Sigma)\rightarrow [0,\infty]$ be measurable functions. Then, $H(x,y)=f(x,y)g(y)$ is $\mathfrak{M}\otimes\Sigma$-measurable. – John. p Feb 26 '14 at 7:49
    
I don't think this is trivial.. Because they share some variable and don't share some variable.. How do you assure that this is trivial? – John. p Feb 26 '14 at 7:50
    
I have proved that "If $f:(X,\mathfrak{M})\rightarrow [0,\infty]$ and $g:(Y,\Sigma)\rightarrow [0,\infty]$ are measurable, then $H(x,y)=f(x)g(y)$ is $\mathfrak{M}\otimes \Sigma$ measurable." I don't see any relation between this theorem and the statement in the above comment. How are they related? – John. p Feb 26 '14 at 7:53
    
They are both measurable on $X \times Y$ with the product $\sigma$-algebra. – Robert Israel Feb 26 '14 at 15:39

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