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:
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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!
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.
Your approach is mostly correct, but I would make a few clarifications and adjustments in your explanation. Here's an improved version of your approach:
Suppose E is a measurable subset of R.
To show that a+E is measurable, we need to show that for any measurable set A, the set (a+E) ∩ A is measurable.
Let B = A - a, which is also measurable since the translation of measurable sets by a is measurable.
Now, (a+E) ∩ A = (a+E) ∩ (B+a) = ((a+E) ∩ B) + a.
Since both B and E are measurable, their intersection (a+E) ∩ B is also measurable, as the intersection of measurable sets is measurable.
Therefore, (a+E) ∩ A = ((a+E) ∩ B) + a is measurable, since it is the sum of a measurable set and a single point.
Hence, we have shown that if E is measurable, then a+E is also measurable.
To show the converse, suppose a+E is measurable.
Let C = (a+E) - a, which is also measurable since the translation of measurable sets is measurable.
Now, (a+E) ∩ (C+a) = ((a+E) ∩ C) + a = (C+a) + a.
Since both C and E are measurable, their intersection (a+E) ∩ C is also measurable.
Therefore, (a+E) ∩ (C+a) = ((a+E) ∩ C) + a = (C+a) + a is measurable, since it is the sum of a measurable set and a single point.
Hence, we have shown that if a+E is measurable, then E is also measurable.
Therefore, we conclude that E is measurable if and only if a+E is measurable.
Overall, your approach is correct but it would be helpful to provide more justification for certain steps, particularly in the relationships between the different sets involved.