Discussion
Let \((\mathbf{R}, M, \mu)\) denote the real line with Lebesgue measure \(\mu\). Define \(\nu_{1}(A) = \int_{A} f(x)\,dx\) where :
\(f(x) = e^{-x}\) for \(x \gt 1\)
\(= 0\) otherwise
and \(\nu_{2}(A) = \int_{A} g(x)\,dx\) where
\(g(x) = e^{-1}\) for \(0 \le x \le 1\)
\(= 0\) otherwise
Then :
Answer: C
\(\nu_{1}\) and \(\nu_{2}\) are mutually singular.
\(\nu_{1}\) lives on \((1, \infty)\): it gives measure 0 to every subset of \((-\infty, 1]\). \(\nu_{2}\) lives on \([0, 1]\): it gives measure 0 to every subset of \((1, \infty)\). So the disjoint sets \((1, \infty)\) and \((-\infty, 1]\) separate them, which is the definition of mutual singularity.
Absolute continuity fails both ways: \(\nu_{2}((1, 2)) = 0\) but \(\nu_{1}((1, 2)) \gt 0\), and \(\nu_{1}([0, 1]) = 0\) but \(\nu_{2}([0, 1]) = e^{-1} \gt 0\). And Lebesgue measure gives positive mass to \((-2, -1)\), where both vanish, so it is not absolutely continuous with respect to either.
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