E Measurable Implies E + y Measurable

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SUMMARY

The discussion centers on the proof that if a subset E of R is measurable, then each translate E + y of E is also measurable. The key equation used is m^*(A) = m^*(A ∩ E) + m^*(A ∩ (R \setminus E)), where m* represents the outer measure. The proof is straightforward for y = 0, but for non-zero y, the author explores the implications of monotonicity of m* and the relationships between E, F, and their complements. Ultimately, the author finds a solution in another post, indicating that the proof is established.

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  • Familiarity with outer measure and its properties
  • Knowledge of monotonicity in measure theory
  • Basic concepts of set operations and complements
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Homework Statement
Show that if a subset E or R is measurable, then each translate E + y of E is also measurable.


Relevant equations
E is measurable if for all subsets A of R

m^*(A) = m^*(A \cap E) + m^*(A \cap (\textbf R \setminus E))

where m* is the outer measure.

The attempt at a solution
The result is trivial if y = 0, so supposes y is not 0. I will denote the complement of a subset of R using '. Let F = E + y. By monotonicity of m*,

m^*(A) \le m^*(A \cap F) + m^*(A \cap F')

In the other direction, I have determined the following: Since E is a subset of F' and F is a subset of E', we have that A \cap E \subseteq A \cap F' and A \cap F \subseteq A \cap E'. By monotonicity of m*, m^*(A \cap E) \le m^*(A \cap F') and m^*(A \cap F) \le m^*(A \cap E'). The former inequality is no good, but the latter one isn't. I don't know how to proceed from here. Any tips?
 
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