# measure $\lambda(E)=0$ or $\lambda(E)=+\infty$

Let $\mu$ be a finite measure, let $\lambda<<\mu$ ($\lambda$ is absolutely continuous wrt. $\mu$)

let $P_n$,$N_n$ be a Hahn decomposition for $\lambda-n\mu$. Let $P=\cap P_n$ and $N=\cup N_n$.

How to prove: $N$ is $\sigma-$finite for $\lambda$ and

if $E\subset P$ f,$E\in \mathbf{X}$, then measure $\lambda(E)=0$ or $\lambda(E)=+\infty$

we thought that from $\lambda<<\mu$, then $\lambda$ should be finite too. but the question says $\lambda(E)=+\infty$. we got stuck. could you please help

Assuming that $P_n$ is positive set in the decomposition for $\lambda-n\mu$, note that for each $n$, $$(\lambda-n\mu)(N_n)\leq 0$$ Since $0\leq \mu(N_n)<\infty$, this is equivalent to $\lambda(N_n)\leq n\mu(N_n)<\infty$, that is, each $N_n$ has finite measure with respect to $\lambda$. It follows that $N$ is $\sigma$-finite.
Next, let $E$ be a measurable subset of $P$. Since $E\subset P_n$ for each $n$ and $P_n$ is a positive set with respect to $\lambda-n\mu$ then $$0\leq (\lambda-n\mu)(E)$$ Again, since $0\leq \mu(E)<\infty$, this is equivalent to $n\mu(E)\leq \lambda(E)$ and this is valid for every positive integer $n$. If $\mu(E)=0$, then $\lambda(E)=0$ because $\lambda\ll\mu$. If $0<\mu(E)$ then $\lambda(E)=\infty$ because in this case, we can made $n\mu(E)$ as big as we want and the inequality $n\mu(E)\leq \lambda(E)$ will always hold.