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.
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.
To prove that E is measurable if and only if a+E is measurable, we need to break it down into two parts: (1) if E is measurable, then a+E is measurable, and (2) if a+E is measurable, then E is measurable.
Part (1): If E is measurable, then a+E is measurable.
Let E be a measurable set in ℝ. We want to show that a+E is measurable for any real number a.
To prove this, we will show that the collection of all sets of the form a+E, where E is measurable, forms a σ-algebra. Recall that a σ-algebra on a set X is a collection of subsets of X that includes the empty set, is closed under complements, and closed under countable unions.
i) The empty set is a+E: Since E is measurable, the empty set is a subset of E, so a+∅ = ∅, where ∅ represents the empty set. Thus, the empty set belongs to the collection of all sets of the form a+E.
ii) Complement of a+E: Let B be a set of the form a+E, where E is measurable. We want to show that the complement of B, denoted by B', is also a set of the form a+E.
Note that (a+E)' = {x | x ∉ a+E} = {x | x-a ∉ E}. Since E is measurable, its complement E' is also measurable. Therefore, for any x-a ∉ E, we have x-a ∈ E'. Thus, we can rewrite (a+E)' as {x | x-a ∈ E'}. Since E' is measurable, we can conclude that (a+E)' is measurable.
iii) Countable union of a+E: Let B₁, B₂, B₃, ... be a sequence of sets of the form a+E, where E is measurable. We want to show that the countable union of these sets, denoted by ∪ Bᵢ, is also a set of the form a+E.
Since each Bᵢ is of the form a+E, we can write Bᵢ as {a+eᵢ | eᵢ ∈ E}. Then, the countable union of these sets can be written as {a+e | e ∈ E}, where e ranges over all elements in the countable union of sets E. Hence, the countable union of sets of the form a+E is also a set of the form a+E.
Therefore, we have shown that the collection of all sets of the form a+E forms a σ-algebra, concluding that a+E is measurable if E is measurable.
Part (2): If a+E is measurable, then E is measurable.
Let a+E be a measurable set in ℝ for some real number a. We want to show that E is measurable.
To prove this, we will use the fact that the complement of a measurable set is also measurable.
Since a+E is measurable, its complement (a+E)' is also measurable. By using the properties of set complement, we can expand (a+E)' as {x | x-a ∉ E}. Notice that if x-a ∉ E, then x ∉ (a+E), which means x belongs to the complement of a+E.
Hence, we can conclude that E is measurable since its complement (a+E)' is measurable.
By proving both parts (1) and (2), we have shown that E is measurable if and only if a+E is measurable.