Past exam of the mathematics course of the University of Cambridge / 2025 / ii / Paper 1 / 23H (Analysis of Functions)

Since $\nu \ll \mu$ and both measures are finite, the Radon--Nikodym theorem supplies a measurable $w:E\to [0,\infty ]$ such thatfor every measurable $A$. Moreover,so $w$ is finite $\mu$-almost everywhere and is $\mu$-integrable. The reverse absolute continuity implies $w>0$ $\mu$-almost everywhere: if $w=0$ on $A$, then $\nu (A)=0$, hence $\mu (A)=0$.

For $0<p\le 1$, the elementary bound $t^p\le 1+t$ givesNo exponent $p>1$ is forced. On $E=(0,1)$ with Lebesgue measure, take $w(x)=x^{-1/p}$. Then $w$ is positive and integrable, so $d\nu =w,d\mu$ defines a finite measure mutually absolutely continuous with $\mu$, butThus the required range is exactly $0<p\le 1$.

Yes. Define the probability measureCountable additivity follows by Tonelli's theorem and $\mu (E)={\sum }_n{2}^{-n}=1$. If $\mu (A)=0$, every nonnegative term $2^{-n}{\nu }_n(A)$ is zero, so ${\nu }_n(A)=0$ for every $n$. Hence every ${\nu }_n\ll \mu$.