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Let $(X, A)$ and $(Y,B)$ be two measurable space. Let $f \geq 0$ be measurable w.r.t $A \times B$ (product $\sigma$ -algebra). Let $g(x)=\sup_{y \in Y} f(x,y)$ and suppose $g(x)< \infty$ for each $x$. Is g necessarily measurable w.r.t $A$?

I am not sure whether the answer is no or yes. But, I haven't been able to produce a proof. Thanks in advance for any help!

Gautam Shenoy
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Riju
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  • For starters, try to prove it for a simple indicator function. Let $A_1 \in A$, $B_1\in B$. Can you show the above is true for $1_{A_1\times B_1}(x,y)$? – Gautam Shenoy May 27 '17 at 14:53
  • Yeah I did it as you suggested and it is true for $A_{1} \times B_{1}$ but now I would like to take any arbitrary $E \in A\times B$ and consider $1_{E}$. But I am unable to determine $g(x)= sup_{y \in Y} 1_{E}(x,y)$. – Riju May 27 '17 at 16:53
  • https://math.stackexchange.com/questions/876544/measurability-of-supremum-of-a-class-of-functions –  May 27 '17 at 17:36
  • @d.k.o. I wanted to discuss on the answer by Nate. The answer is no. The function that has been taken is $1_{A}$ where $A={\ (y,y) | y \in Y}\ $. Here Y is a non-measurable set in $[0,1]$ w.r.t the lebesgue sigma algebra. My question is why is the above function measurable w.r.t the product $\sigma$- algebra $"m \times m"$. – Riju May 27 '17 at 18:06
  • @Riju $A$ is closed in $[0,1]\times Y$ and thus a Borel set. Also ${f≤1}=[0,1]\times Y$ and ${f≤0}=A^c$. –  May 27 '17 at 19:16

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