Assume E is measurable. Then we know that for any ε > 0, there exists a cover {I_i} of E such that the sum of the lengths of the intervals is less than ε. Now consider a+E. For any element y in a+E, we have y = a + e for some e in E. Since {I_i} covers E, we can express e as the intersection of E with some interval J_i. Thus, y can be expressed as a sum of a and J_i, and so {J_i + a} is a cover of a+E. Moreover, the sum of the lengths of these intervals is equal to the sum of the lengths of the original intervals {I_i}, which is less than ε. Hence, a+E is also measurable.
Conversely, assume a+E is measurable. By definition, this means that for any ε > 0, there exists a cover {J_i} of a+E such that the sum of the lengths of the intervals is less than ε. Consider the sets {J_i - a}. Each of these sets is a translate of a set in the cover {J_i}, and so their sum of lengths is also less than ε. Moreover, we claim that {J_i - a} is a cover of E. Indeed, if e is in E, then e+a is in a+E by definition, and hence e+a is in some interval J_i. It follows that e is in the translate of J_i that is obtained by adding -a to each of its endpoints, which is one of the sets {J_i - a}. Hence, {J_i -a} is a cover of E, and so E is measurable.
Therefore, E is measurable if and only if a+E is measurable.
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.
1 answer