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Does there exists a finite measure $\mu$ on the $\sigma-$algebra of Lebesgue measurable subsets of $\mathbb{R}$ such that $\mu << m$ and $m <<\mu$, where $m$ is the Lebesgue measure?

My attempt is to use the Radon-Nikodym theorem to arrive at a contradiction but I am stuck. Any help will be appreciated.

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    $\begingroup$ How about $\mu(A)=\int_A\exp(-x^2)m(dx)$? $\endgroup$ – kimchi lover May 2 '19 at 21:48
  • $\begingroup$ How do we compute $\mu(E)$ if $m(E)=0$? $\endgroup$ – John Thompson May 2 '19 at 21:54
  • $\begingroup$ I realized that I think your $\mu$ will actually work. Thank you! $\endgroup$ – John Thompson May 2 '19 at 22:09
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Pick your favorite positive function $f\in L^1(\mathbb{R},\mathrm dm(x))$ and define $$ \mu(A)=\int_A f(x)\;\mathrm d m(x) $$ for $A$ a Lebesgue measurable set.

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  • $\begingroup$ Hmmm, should $f$ be necessarily integrable on $\mathbb{R}$? $\endgroup$ – John Thompson May 3 '19 at 3:11
  • $\begingroup$ @JohnThompson I missed the important caveat that the measure be finite! Of course yes take $f$ to be $L^1$ $\endgroup$ – qbert May 3 '19 at 3:17
  • $\begingroup$ I agree that any such integrable $f$ will work. Thanks! $\endgroup$ – John Thompson May 3 '19 at 3:19
  • $\begingroup$ @JohnThompson sure thing. Good luck $\endgroup$ – qbert May 3 '19 at 3:24
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How about defining the function $$ \phi(x) = {\pi \over 2} + \arctan(x) $$ and defining, for every interval $A = ]a, b[$ in $\mathbb{R}$, even if $a = -\infty$ or $b = +\infty$ or both, $$ \mu(A) = \phi(b) - \phi(a), $$ taking the necessary limits if the interval is not bounded?

The idea of $g(x)$ is to map unbounded intervals to bounded ones, guaranteeing the constructed measure will be finite.

I suppose $g(x) = \phi'(x)$ can be seen as a Radon-Nykodym derivative that will do the job.

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Take $\mu(A) =\sum \frac {m(A \cap (-n,n))} {2^{n}}$.

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