Question:

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!

Answer 1

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.