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Let $m$ be Lebesgue measure on $\mathbb R_+=(0,\infty)$ and $\mathcal A = \sigma\left(( \frac 1{n+1} , \frac 1n ]:n=1,2,...\right)$. Define a new measure $\lambda$ on $\mathcal A$, for each $E \in \mathcal A$, by $\lambda(E)= \int_E fdm $ . Find the Radon-Nikodym derivative $\frac{d\lambda}{dm}$.

It is clear that $\lambda $ is absolutely continuous with respect to $m$ by definition, and Lebesgue measure is $\sigma$-finite in $(\mathbb R_+ , \mathfrak M_+)$, where $\mathfrak M_+$ is the collection of all Lebesgue measurable subsets of $\mathbb R_+$, so $m$ is $\sigma$-finite in $(\mathbb R_+, \mathcal A)$ since every element of $\mathcal A$ is also Lebesgue measurable.

However, to apply Radon-Nikodym theorem, $\lambda$ must be a finite measure, but $\lambda(E)= \int_E fdm = \infty$ if $E=\mathbb R_+ - ( \frac 12,1]$, so we cannot apply that theorem. Is there any other approach to this problem?

Let $m$ be Lebesgue measure on $\mathbb R_+=(0,\infty)$ and $\mathcal A = \sigma\left(( \frac 1{n+1} , \frac 1n ]:n=1,2,...\right)$. Define a new measure $\lambda$ on $\mathcal A$, for each $E \in \mathcal A$, by $\lambda(E)= \int_E fdm $, where $f(x)=2x^2$ . Find the Radon-Nikodym derivative $\frac{d\lambda}{dm}$.

It is clear that $\lambda $ is absolutely continuous with respect to $m$ by definition, and Lebesgue measure is $\sigma$-finite in $(\mathbb R_+ , \mathfrak M_+)$, where $\mathfrak M_+$ is the collection of all Lebesgue measurable subsets of $\mathbb R_+$, so $m$ is $\sigma$-finite in $(\mathbb R_+, \mathcal A)$ since every element of $\mathcal A$ is also Lebesgue measurable.

However, to apply Radon-Nikodym theorem, $\lambda$ must be a finite measure, but $\lambda(E)= \int_E fdm = \infty$ if $E=\mathbb R_+ - ( \frac 12,1]$, so we cannot apply that theorem. Is there any other approach to this problem?