Chechk for Lebesgue Measurable Property when a is in R and E is a Subset of R

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

Claim:

Let E be a subset of R and let a be a real number. Define:

a + E = { a + e | e in E }

Then E is Lebesgue measurable if and only if a + E is Lebesgue measurable.

Proof:

Suppose E is Lebesgue measurable.

Lebesgue measurability is translation invariant, meaning that if E is measurable, then so is a + E.

So:

If E is measurable, then a + E is measurable. (1)

Now suppose a + E is measurable.

Then E = (-a) + (a + E), which is just a translation of a + E.

By translation invariance, E is also measurable. (2)

From (1) and (2), we conclude:

E is measurable if and only if a + E is measurable.