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^(Bc)), 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 Bc is measurable, to prove A^(Bc) is measurable, which again comes to what we need to prove in this question.
Also, we can see that
A^(Bc) 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^(Bc) is Lebesgue measurable, we can write by the monotone property of m*,
m*(A^(Bc)) <= m*(A)
which gives m*(A^(Bc)) = 0, provided that A^(Bc)) is Lebesgue measurable.
But again, is that set measurable?
How do we prove the required result?
Answer:
We can use the fact that the Lebesgue outer measure is a measure on the sigma-algebra of Lebesgue measurable sets. That is, if A and B are measurable sets, then A ∪ B and A ∩ B are also measurable.
Now, since A ∪ B is measurable, we have that A^c ∩ (A ∪ B) and A ∩ (A ∪ B) are measurable sets.
But we also have A ⊆ A^c ∩ (A ∪ B) and B ⊆ A ∪ B.
Therefore, we have B = (A^c ∩ (A ∪ B))^c ∩ (A ∪ B) which is a union of two measurable sets and hence is also measurable.
Thus, we have proved that if A ⊆R such that m∗(A) = 0 and A ∪ B ∈ L, then B ∈ L.
To prove that B is Lebesgue measurable, we can use the fact that the union of two Lebesgue measurable sets is also a Lebesgue measurable set. We are given that A ∪ B is Lebesgue measurable, so we know that B must also be Lebesgue measurable.
Now, let's prove that A ∩ B^c is Lebesgue measurable. Since A has Lebesgue outer measure 0, we can write A as the countable union of intervals I_n, such that the sum of the lengths of these intervals is less than epsilon for any given epsilon > 0.
Let's consider the set A ∩ B^c. Since A is a subset of A ∪ B, we have that A ∩ B^c ⊆ A. Therefore, A ∩ B^c can also be expressed as the countable union of intervals I_n ∩ B^c.
Now, let's consider the Lebesgue outer measure of A ∩ B^c, denoted as m*(A ∩ B^c). Using the monotonicity property of the Lebesgue outer measure, we have:
m*(A ∩ B^c) = m*( ∪(I_n ∩ B^c) ) <= Σ m*(I_n ∩ B^c)
Since each interval I_n has length less than epsilon for any given epsilon > 0, we can bound m*(I_n ∩ B^c) as follows:
m*(I_n ∩ B^c) <= m*(I_n) <= length of I_n
Summing over all intervals I_n, we have:
Σ m*(I_n ∩ B^c) <= Σ (length of I_n)
Since the sum of the lengths of the intervals is less than epsilon for any given epsilon > 0, we can conclude that:
m*(A ∩ B^c) <= Σ (length of I_n) < epsilon
Therefore, we have shown that the Lebesgue outer measure of A ∩ B^c is less than any given epsilon > 0, which implies that A ∩ B^c has Lebesgue outer measure 0.
Since A ∩ B^c has Lebesgue outer measure 0, it follows that A^(Bc) is Lebesgue measurable. Therefore, by the argument given earlier, B is also Lebesgue measurable.
Hence, we have proved that if A ∪ B is Lebesgue measurable, then B must be Lebesgue measurable.