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Consider a measurable space $(M, \mathfrak{F})$. Let the $\sigma$- finite measures $\mu$ and $\nu$ are defined on $\mathfrak{F}$ and $\mu\prec\nu\prec\mu$. Suppose $f\in L_{1}(M ,d\mu)$. Is from this fact follow that $f\in L_{1}(M ,d\nu)$? Thanks.

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What does $\prec$ mean? Absolutely continuous? –  Stefan Hansen Feb 12 '13 at 9:27
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Presumably. One uses more often $\ll$ (\ll). –  Did Feb 12 '13 at 9:37
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By the way, (homework)? –  Did Feb 12 '13 at 9:54
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1 Answer

No. Try $(M,\mathfrak F)=(\mathbb R,\mathcal B(\mathbb R))$, $\nu$ the Lebesgue measure, $\mu$ any (nondegenerate) gaussian measure, and $f=1$.

For probability measures on the Borel real line, try $\nu$ the standard double exponential measure with density proportional to $\mathrm e^{-|x|}$, $\mu$ the standard gaussian measure with density proportional to $\mathrm e^{-x^2/2}$, and any function $f$ "inbetween", say $f(x)=\exp(|x|^{3/2})$.

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