Question

Let \(N \subseteq \mathbb{R}\) be a non-measurable set with respect to the Lebesgue measure on \(\mathbb{R}\).
Consider the following statements:
P: If \(M = \{x \in N : x \text{ is irrational}\}\), then \(M\) is Lebesgue measurable.
Q: The boundary of \(N\) has positive Lebesgue outer measure.
Then

• A. both P and Q are TRUE
• B. P is FALSE and Q is TRUE
• C. P is TRUE and Q is FALSE
• D. both P and Q are FALSE

Hint:

Remember these key facts: 1. The union of two measurable sets is measurable. 2. Countable sets have measure zero. 3. A set is measurable if and only if its boundary has measure zero (for sets of finite outer measure). A non-measurable set must have a "fuzzy" boundary with positive outer measure.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
This question tests fundamental properties of Lebesgue measure, including the properties of measurable and non-measurable sets, the measure of countable sets, and the relationship between a set's measurability and the measure of its boundary.
Step 3: Detailed Explanation:
Analysis of Statement P:
Let \(\mathbb{Q}\) be the set of rational numbers and \(\mathbb{I} = \mathbb{R} \setminus \mathbb{Q}\) be the set of irrational numbers.
We are given \(M = \{x \in N : x \text{ is irrational}\} = N \cap \mathbb{I}\).
We can express the non-measurable set \(N\) as the union of its rational and irrational parts: \[ N = (N \cap \mathbb{Q}) \cup (N \cap \mathbb{I}) = (N \cap \mathbb{Q}) \cup M \] The set of rational numbers \(\mathbb{Q}\) is countable, and any countable set has Lebesgue measure zero. Therefore, \(\mathbb{Q}\) is a measurable set with \(m(\mathbb{Q})=0\).
The set \(N \cap \mathbb{Q}\) is a subset of \(\mathbb{Q}\). Since the Lebesgue measure is complete, any subset of a measure-zero set is measurable and has measure zero. Thus, \(N \cap \mathbb{Q}\) is a measurable set.
Now, assume for the sake of contradiction that \(M\) is Lebesgue measurable.
If \(M\) were measurable, then \(N\), being the union of two measurable sets (\(N \cap \mathbb{Q}\) and \(M\)), would also be measurable. This contradicts the given information that \(N\) is a non-measurable set.
Therefore, our assumption must be false. \(M\) cannot be measurable.
Thus, P is FALSE.
Analysis of Statement Q:
A fundamental theorem in Lebesgue measure theory states that a set \(A \subset \mathbb{R}\) with finite outer measure is Lebesgue measurable if and only if its boundary, \(\partial A\), has Lebesgue measure zero.
Let's restate this: \(m^*(\partial A) = 0 \iff A\) is measurable (assuming \(m^*(A)<\infty\), which is true for standard constructions of non-measurable sets like Vitali sets, which are bounded).
We are given that \(N\) is a non-measurable set. By the contrapositive of the theorem, since \(N\) is not measurable, its boundary cannot have measure zero.
Since measure (and outer measure) is non-negative, if it is not zero, it must be positive.
Therefore, the boundary of \(N\) must have a positive Lebesgue outer measure.
Thus, Q is TRUE.
Step 4: Final Answer:
Statement P is FALSE and statement Q is TRUE.
Step 5: Why This is Correct:
P is false because if it were true, the non-measurable set N would be a union of two measurable sets, making it measurable, a contradiction. Q is true as it is a direct consequence of a key theorem linking the measurability of a set to the measure of its boundary.