Lesbegue Measure of a Set Given the Union of a Set is Measurable

Question

Question:

Let A ⊆R such that m∗(A) = 0.

Show that if A ∪ B ∈ L, then B ∈ L.

My approach:

We can write B as

B = (A ∪ B) ^ (A^(B^c)), where ^ denote intersection.

We know that the intersection of two Lebesgue measurable sets is also a Lebesgue measurable set.

A ∪ B is already given as a measurable set.

A is a set with Lebesgue outer measure 0.

Now I want to know whether B or B^c is measurable, to prove A^(B^c) is measurable, which again comes to what we need to prove in this question.

Also, we can see that

A^(B^c) is a subset A.

I think to find the Lebesgue outer measure of a set, we must first have that set is Lebesgue measurable.

If A^(B^c) is Lebesgue measurable, we can write by the monotone property of m*,

m*(A^(B^c)) <= m*(A)

which gives m*(A^(B^c)) = 0, provided that A^(B^c)) is Lebesgue measurable.

But again, is that set measurable?

How do we prove the required result?

Answer 1

By definition (of the Lebesgue measurability), in order to show that B is Lebesgue measurable, it suffices to show that for any set E⊆R¹(the set of real numbers),

m*(E) = m*(E∩B) + m*(E∩B^c)

where B^c is the complement of B.

Since it is known that A∪B is Lebesgue measurable, we already have

m*(E) = m*(E∩(A∪B)) + m*(E∩(A∪B)^c).

We show that m*(E∩B) = m*(E∩(A∪B)) and m*(E∩B^c) = m*(E∩(A∪B)^c).

Step I. m*(E∩B) = m*(E∩(A∪B))

By the (countable) monotonicity of the outer measure m*, we have

m*(E∩B) ≤ m*(E∩(A∪B)) = m*((E∩A)∪(E∩B))

≤ m*(E∩A)+m*(E∩B) ≤ m*(A)+m*(E∩B) = .

Thus m*(E∩B) = m*(E∩(A∪B)).

Step II. m*(E∩B^c) = m*(E∩(A∪B)^c)

Again by the monotonicity of the outer measure, and notice that

E∩B^c = [(E∩B^c)∩(E∩A^c)]∪[(E∩B^c)∩(E∩A^c)^c]

⊆ [E∩(A∪B)^c]∪[(E ∩B^c)∩A] ⊆ [E∩(A∪B)^c]∪A,

we have

m*(E∩B^c) ≤ m*(E∩(A∪B)^c) + m*(A) = m*(E∩(A∪B)^c) ≤ m*(E∩B^c).

Therefore m*(E∩(A∪B)^c) = m*(E∩B^c). This completes the proof of the Lebesgue measurability of B.