Proving the Measurability of a Function Composition

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SUMMARY

The discussion centers on proving that the function F : Rn * Rn -> R, defined by F(x, y) = f(x - y), is Lebesgue measurable given that f : Rn -> R is Lebesgue measurable. It is established that the mapping (x, y) -> x - y is continuous and therefore measurable. The key conclusion is that the composition of measurable functions is also measurable, which directly supports the measurability of F.

PREREQUISITES
  • Understanding of Lebesgue measurable functions
  • Familiarity with function composition in real analysis
  • Knowledge of continuous functions and their properties
  • Basic concepts of Rn and measurable spaces
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  • Study the properties of Lebesgue measurable functions in detail
  • Learn about the implications of continuous mappings in real analysis
  • Explore the concept of function composition and its measurability
  • Investigate the relationship between measurable functions and integration
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Homework Statement



If f : Rn -> R is Lebesgue measurable on Rn, prove that the function F : Rn * Rn -> R de fined by F(x, y) = f(x - y) is Lebesgue measurable on Rn * Rn.

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The Attempt at a Solution



I am confused by the expression of F(x,y), it seems x-y is equal to x of f(x), how can I prove this question?
 
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You know that [tex]\mathbb{R}^n\times\mathbb{R}^n\rightarrow \mathbb{R}^n:(x,y)\rightarrow x-y[/tex] is continuous and thus measurable right?

The only thing you need to show then is that the composition of measurable functions is measurable...
 

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