(1) Yes, this is a correct way to prove this. Your approach specifically shows that for any subset A of R, we have m*(A^E) + m*(A^(Ec)) = m*(A), which is the key condition for E to be Lebesgue measurable.
(2) One alternative, more general approach to prove this is by using Carathéodory's criterion for Lebesgue measurability. It states that a set E is Lebesgue measurable if and only if for every subset A of R, m*(A) = m*(A^E) + m*(A^(Ec)). This criterion is actually a generalization of the property you have used in your proof.
In this case, you can use the fact that m*(E) = 0 as a starting point, and then for any subset A of R, observe that:
m*(A^E) <= m*(E) = 0
and
m*(A^(Ec)) <= m*(A)
Adding these inequalities gives
m*(A^E) + m*(A^(Ec)) <= m*(A).
Since m*(A^E) and m*(A^(Ec)) are both non-negative, it follows that:
m*(A) >= m*(A^E) + m*(A^(Ec)).
Note that we've already shown the reverse inequality in your original approach. Thus, we arrive at the equality:
m*(A) = m*(A^E) + m*(A^(Ec))
and hence, E is Lebesgue measurable by Carathéodory's criterion.
Question:
If m*(E) = 0, show that E is Lebesuge measurable, where m*(E) denote the Lebesgue Outer Measure of E.
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Notations I've used:
^ refers to intersection.
eg: A^E refers ro the intersection of the two sets A & E
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My approach towards the question:
We know that a subest E of R is Lebesuge measurable if,
m*(A) = m*(A^E) + m*(A^(Ec) --->(*) for any subset A of R.
Now consider m*(E).
m*(E) = 0 (Given) --->(1)
But,
m*(E^E) = m*(E) = 0 (given in the question)
and,
m*(E^(Ec)) =m*(null set) = 0
This gives,
m*(E^E) = m*(E) + m*(E^(Ec)) =0 --->(2)
From (1) & (2),
m*(E) = m*(E) + m*(E^(Ec)) which can be considered a special case of (*), which proves E is Lebesgue measurable when m*(E) = 0
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My concerns regarding my approach to this question are:
(1) Is this a correct way to prove this?
(2) Are there any other better ways to prove this?
1 answer