Discussion
Let L be a countable subset of \(\mathbf{R}\) and M be a Lebesgue measurable subset of \(\mathbf{R}\) with \(m(M) \gt 0\). If \(N = L \cup M\) and \(O = (L \cup M) \backslash (L \cap M)\), then :
Answer: A
The answer is \(m(M) = m(N) = m(O)\).
A countable set has Lebesgue measure 0, so L and every subset of it are null. N differs from M only by \(L \backslash M\), a null set, so \(m(N) = m(M)\).
O is the symmetric difference of L and M. It differs from M by removing \(L \cap M\) and adding \(L \backslash M\), both null, so \(m(O) = m(M)\) as well.
Adding or removing countably many points never changes Lebesgue measure, so no strict inequality is possible, which rules out the other three options.
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