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Let $\mu$ be a $\sigma$- finite measure on $(X,M)$. Prove that there exists a finite measure $\lambda$ on $M$ such that $\lambda\ll\mu$ and $\mu\ll\lambda$.

Can anyone give me a hint on how to start on this problem?

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2 Answers 2

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Some hints:

  1. What would happen if you have a measure $\lambda = f\,\mathrm d\mu$? Say, if $f > 0$ everywhere? Does integral of positive function w.r.t. $\lambda$ has to be infinite if $\lambda$ is infinite?

  2. To find such $f$, think of the fact that there exists an increasing sequence of sets on which $\lambda$ is finite, and union covers the whole space. How would you define $f$ on each of those sets?

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  • $\begingroup$ Thanks. I completed my solution using your hint! Was much easier than I expected. $\endgroup$
    – dezdichado
    Nov 4, 2015 at 10:10
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Hint: For each $n$, $\mu(E_n) <\infty$ by $\sigma$-finiteness, and define a new measure $$\mu_n = \frac{2^{-n}}{\mu(E_n)}\mu_{E_n}.$$

Rest hidden in spoilers:

Then $\|\mu_n \| = 2^{-n}$, so $\sum_{n=1}^\infty \|\mu_n\| < \infty$, so the series of the measures $\mu_n$ converges in the normed space of measures (I assume you defined this in class). Then, define $\lambda = \sum_{n=1}^\infty \mu_n$ and obtain $\mu \ll \nu \ll \mu$.

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  • $\begingroup$ Thanks for the answer though we have not defined normed space of measures. But I think I see where this is going. $\endgroup$
    – dezdichado
    Nov 4, 2015 at 9:52

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