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In the lecture we proved the following version of the Radon–Nikodym Theorem:

Let $\mu, \nu$ be two finite measure on the measurable space $(\Omega,\mathcal{A})$. Then the following are equivalent:

  1. $\nu$ is absolutely continuous with respect to $\mu$, i.e. $\nu << \mu$.
  2. There exists a measurable function $f:\Omega \longrightarrow [0,\infty]$ such that $$\nu(A)=\int_A f d\mu$$ In that case $f$ is $\mu$ almost everywhere unique.

The definiton of absolute continuity we used is:

  • Let $\nu$ be a measure or signed measure and $\mu$ a measure on $(\Omega, \mathcal{A})$. Then we call $\mu$ absouletly continuous with respect o $\mu$ if for $A \in \mathcal{A}$ one always has $$\mu(A)=0 \implies \nu(A)=0$$

Our prof said one could extend this even to the case where $\mu$ and $\nu$ are measures (not necesserily finite) and $\mu$ is $\sigma$-finite on $\Omega$, with a relatively short argument.

How do I reduce then the general Theorem to the finite case we proved in the lecture?

Thanks in advance!

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If $\nu$ is finite and $\mu$ is $\sigma$-finite, then $\Omega = \bigcup_{n = 1}^{\infty} A_{n}$, where $\{A_{1},A_{2},\dots,\} \subseteq \mathcal{A}$ is disjoint and $\mu(A_{j}) < \infty$ for each $j$. For each $j$, consider the restriction of $\nu$ to $A_{j}$, i.e. the measure $\nu_{j}$ on $(A_{j},\mathcal{A} \cap \mathcal{P}(A_{j}))$ given by $\nu_{j}(E) = \nu(E)$. Define $\mu_{j}$ analogously. Since $\nu \ll \mu$, it follows that $\nu_{j} \ll \mu_{j}$. Since $\mu_{j}(A_{j}) < \infty$, we can apply the finite measure version of the Radon-Nikodym Theorem to find a non-negative function $f_{j} \in L^{1}(A_{j},\mu_{j})$ such that $$\forall E \in \mathcal{A} \cap \mathcal{P}(A_{j}) \quad \nu_{j}(E) = \int_{E} f_{j}(x) \, \mu_{j}(dx) = \int_{E} f_{j}(x) \, \mu(dx).$$

Finally, let $f = \sum_{j = 1}^{\infty} f_{j} \chi_{A_{j}}$. If $E \in \mathcal{A}$, then $E = \bigcup_{n = 1}^{\infty} (E \cap A_{n})$ so $$\int_{E} f(x) \, \mu(dx) = \sum_{n = 1}^{\infty} \int_{E \cap A_{n}} f_{j}(x) \, \mu_{j}(dx) = \sum_{n = 1}^{\infty} \nu(E \cap A_{n}) = \nu(E).$$ This proves $d \nu = f \, d \mu$, which yields the Radon-Nikodym Theorem in the $\sigma$-finite case.

Notice that the above argument generalizes also to the case when $\nu$ is $\sigma$-finite as well. Can you see why?

Edit: When $\nu$ is $\sigma$-finite but not finite, then $\Omega = \bigcup_{n = 1}^{\infty} B_{n}$, where $\{B_{1},B_{2},\dots\} \subseteq \mathcal{A}$ is disjoint and $\nu(B_{j}) < \infty$. Repeat the construction above, this time working with the restriction of $\nu$ to $B_{j}$ and then "gluing them together" analogously to what we did above.

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  • $\begingroup$ No I actually do not see how it generalizes to the $\nu$ $\sigma$-finite case. But even then, $\nu$ can be assumed to be a measure which is not necessarily $\sigma$-finite and the Theorem should still be true. How would I generalize it then for an arbitrary $\nu$ measure? $\endgroup$ – vaoy Jul 27 '17 at 19:09
  • $\begingroup$ I edited my answer. I don't think the theorem is true if $\nu$ is not $\sigma$-finite, but I'm still working on finding a counterexample. $\endgroup$ – fourierwho Jul 28 '17 at 1:43
  • $\begingroup$ how can you say that $\nu$ is finite on $B_j$, if $\nu$ is not $\sigma$-finite? $\endgroup$ – vaoy Jul 28 '17 at 5:16
  • $\begingroup$ It's a typo. Corrected. $\endgroup$ – fourierwho Jul 28 '17 at 18:04

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