$Prove that $\frac{d \lambda}{d \mu} = \frac{d \lambda}{d \nu} \frac{d \nu}{d \mu}$ for $\sigma$-finite measures $\mu,\nu, \lambda$.$

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$Prove that $\frac{d \lambda}{d \mu} = \frac{d \lambda}{d \nu} \frac{d \nu}{d \mu}$ for $\sigma$-finite measures $\mu,\nu, \lambda$.$

Suppose that $\mu,\nu, \lambda$ are $\sigma$-finite measures on a measurable space $(X,\mathcal{M})$ and that \[ \lambda \ll \nu \textrm{ and } \nu \ll \mu .\] Prove that \[ \frac{d \lambda}{d \mu} = \frac{d \lambda}{d \nu} \frac{d \nu}{d \mu} \;\; \mu\textrm{-a.e.}. \]

Real Analysis Measure Theory

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Since $\lambda \ll \nu$ and $\nu \ll \mu$, then $\lambda \ll \mu$. Indeed if
\[ \mu(E)=0 \Rightarrow \nu(E)=0 \Rightarrow \lambda(E)=0. \]
Also, by Lebesgue-Radon-Nikodym Theorem there exist a $\nu$-integrable function such that
\[ (1) d\lambda=f d\nu \ \ i.e. \ \ \frac{d \lambda}{d \nu}=f. \]
Moreover, if $d \lambda=f' d \nu$, then $f'=f$ $\nu$-a.e. Similarly,
\[(2) d \nu= g d \mu, \mu-a.e. \ \ i.e. \frac{d \nu}{d \mu}=g. \]
Hence it follows from (1) and (2) that
\[ d \lambda= f d\nu=fg d \mu. \] Thus, \[ \frac{d \lambda}{d \mu}=fg=\frac{d \lambda}{d \nu}.\frac{d \nu}{d \mu}, \mu-a.e. . \]
The proof is complete.

Erdos

$Prove that $\frac{d \lambda}{d \mu} = \frac{d \lambda}{d \nu} \frac{d \nu}{d \mu}$ for $\sigma$-finite measures $\mu,\nu, \lambda$.$

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