# Existence of absolutely continuous measure

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.

• How about $\mu(A)=\int_A\exp(-x^2)m(dx)$? – kimchi lover May 2 '19 at 21:48
• How do we compute $\mu(E)$ if $m(E)=0$? – John Thompson May 2 '19 at 21:54
• I realized that I think your $\mu$ will actually work. Thank you! – John Thompson May 2 '19 at 22:09

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.

• Hmmm, should $f$ be necessarily integrable on $\mathbb{R}$? – John Thompson May 3 '19 at 3:11
• @JohnThompson I missed the important caveat that the measure be finite! Of course yes take $f$ to be $L^1$ – qbert May 3 '19 at 3:17
• I agree that any such integrable $f$ will work. Thanks! – John Thompson May 3 '19 at 3:19
• @JohnThompson sure thing. Good luck – qbert May 3 '19 at 3:24

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.

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