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Nasir Kim

Nasir Kim

Answered question

2022-05-22

Let E R d be a (Lebesgue) measurable set and let f, g be two measurable functions defined on E. I would like to show that if Φ is a continuous function on R 2 , then the function h : x Φ ( f ( x ) , g ( x ) ) is measurable. The proof remains unknown to me, but I can address the problem if it is only one-dimensional. Specifically, if Φ is a continuous function on R , then I can show that Φ f is measurable. Indeed, since { Φ < a } is an open set G, we can conclude that { Φ f < a } = f 1 ( G ) is measurable. How about the two-dimensional problem? Does anyone have an idea? Thank you.

Answer & Explanation

hoffwnbu

hoffwnbu

Beginner2022-05-23Added 13 answers

Sketch: you need to show that { x : h ( x ) > a } is measurable for each a R . So, set U = { ( u , v ) : Φ ( u , v ) > a } .   U is open because Φ is continuous.
so it is a countable union of rectangles U = n ( a n , b n ) × ( c n , d n ) .
Now, for each integer n ,   { x : ( f ( x ) , g ( x ) ) ( a n , b n ) × ( c n , d n ) } is measurable because
it is equal to { x : a n < f ( x ) < b n } { c n < g ( x ) < d n } and f and g are measurable.
But, { x : h ( x ) > a } = { x : ( f ( x ) , g ( x ) ) U } = n { x : ( f ( x ) , g ( x ) ) ( a n , b n ) × ( c n , d n ) }
from which the claim follows because countable unions of measurable sets are measurable.

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