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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
[tex]m^*(A) = m^*(A \cap E) + m^*(A \cap (\textbf R \setminus E))[/tex]
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*,
[tex]m^*(A) \le m^*(A \cap F) + m^*(A \cap F')[/tex]
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 [itex]A \cap E \subseteq A \cap F'[/itex] and [itex]A \cap F \subseteq A \cap E'[/itex]. By monotonicity of m*, [itex]m^*(A \cap E) \le m^*(A \cap F')[/itex] and [itex]m^*(A \cap F) \le m^*(A \cap E')[/itex]. The former inequality is no good, but the latter one isn't. I don't know how to proceed from here. Any tips?
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
[tex]m^*(A) = m^*(A \cap E) + m^*(A \cap (\textbf R \setminus E))[/tex]
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*,
[tex]m^*(A) \le m^*(A \cap F) + m^*(A \cap F')[/tex]
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 [itex]A \cap E \subseteq A \cap F'[/itex] and [itex]A \cap F \subseteq A \cap E'[/itex]. By monotonicity of m*, [itex]m^*(A \cap E) \le m^*(A \cap F')[/itex] and [itex]m^*(A \cap F) \le m^*(A \cap E')[/itex]. The former inequality is no good, but the latter one isn't. I don't know how to proceed from here. Any tips?