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I think that the answer is positive: It is enough to show that the set $( x | g(x) < c )$ is borelBorel. as you saed it is an image under $Proj_x$ of an open set $U$. devide divide $[0,1]^2$ to a union of its interior $(0,1)^2$ and the boundary. Corispondingly devide Correspondingly divide $U$ into $U_0:= (0,1)^2 \cap U$ and its compement complement $Z$. it is enouf enough to sho show the the image of each of them under $Proj_X$ is Borel. weach Which is evident. |
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I think that the answer is positive: It is enough to show that the set $( x | g(x) < c )$ is borel. as you saed it is an image under $Proj_x$ of an open set $U$. devide $[0,1]^2$ to a union of its interior $(0,1)^2$ and the boundary. Corispondingly devide $U$ into $U_0:= (0,1)^2 \cap U$ and its compement $Z$. it is enouf to sho the the image of each of them under $Proj_X$ is Borel. weach is evident |
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