Your approach is mostly correct, but here are a few clarifications:
- When you say "Then a+E = a union E", you should actually say "Then a+E = {a+e | e ∈ E} = {x | x = a+e for some e ∈ E}" to make it clear that you're taking the sum of every element in E with a.
- When you say "since {a} in R and E both are L. mble", you should clarify what you mean by "{a} in R". I assume you're using {a} to mean the set whose only element is a, but you should explicitly state that.
Here's a revised version of your proof with these clarifications:
Suppose E is measurable. Then a+E = {x | x = a+e for some e ∈ E} is the image of E under the translation map T(x) = x+a, which is a measure-preserving map since it's a homeomorphism of R. Therefore, a+E is also measurable.
Conversely, suppose a+E is measurable. Then E = (a+E) \ (-a) is the difference of two measurable sets, and therefore measurable.
Hence, E is measurable if and only if a+E is measurable.
Question:
For a subset E of R and a number a ∈ R, let a+E = {a+e | e ∈ E}. Show that E is measurable if and only if a+E is measurable.
My approach:
==>
Suppose E is L. measurable.
Then a+E = a union E
SInce {a} in R and E both are L. mble, its union also is L. mble.
Hence, E is L.mble ==> a+E is L.mble -->(1)
Since (1) is true for any a in R and any subset E of R, replace E and a+E in (1) by a+E and E respectively to get,
a+E is L.mble ==> E is L.mble --->(2)
So from 1 and 2, E is L.mble iff a+E is L.mble
Hence the proof.
Is this a correct approach?
Thanks in advance for your help!
1 answer