Show that a Given Set is Lebesgue Measurable

Question

Question:

Show that {a} is Lebesgue measurable, where a belongs to the set of real numbers.

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My approach towards the question:

We know that a subset E of the set of real numbers(R) is said to be Lebesgue measurable if

m*(A) = m*(A^E) + m*(A^E^c), where A, m* and ^ denote a subset of R(set of real numbers), Lebesgue outer measure of a given set, and intersection of sets, respectively.

Here, I considered two cases as follow:

Case 1: {a} is a subset of B^c, where B is a subset of R

Then, we can see that,

m*(B^{a}) = m*(empty set) = 0 and,

m*(B^({a}^c)) = m*(B)

which gives, m*(B) = m*(B^{a}) + m*(B^({a}^c)) that shows {a} is Lebesuge measurable when {a} is a subset of B^c

Case 2: {a} is a subset of B

In this case,

m*(B^{a}) = m*({a}) and m*(B^({a}^c)) = m*({a}^c)

But,

m*(B^{a}) + m*(B^({a}^c)) = m*({a}) + m*({a}^c)

(1) How can I show that m*({a}) + m*({a}^c) is equal to m*({a})?

(2) Also, do I need to consider a case where B = {a}?

(3) Is thisthe correct approach to solve this problem?

Answer 1

Your case 1 is correct but your case 2 has some problems. For instance m*(B^({a}^c)) = m*({a}^c) is incorrect. When a is an element of B, it should be the case that m*(B^({a}^c)) = m^*(B\a). Since you already correctly wrote that m*(B^{a}) = m*({a}) , we can add these two to get

m*(B^{a}) + m*(B^({a}^c)) = m*({a}) + m*(B\a) = m^*(B)

as desired. Here we used the fact that m*(A union B) = m*(A) + m*(B) when A and B are disjoint sets.

Another thing that might be helpful to use is that in the other question you posted on Wyzant you proved that a set with outer measure 0 is always lebesbue measurable. Since a set with a single point has outer measure 0, this question is a direct consequence of your other question.