Question 57, Chapter 18 of Royden Real Analysis Let $(\mu_{n})$ a sequence of measures on a measurable space $(X,\mathcal{M})$, for which there is a constant $c>0$ such that $\mu_{n}(X)\leq c$ for all $n$. Define $\mu:\mathcal{M}\rightarrow [0,\infty]$ by
$$\mu:=\sum\limits^{\infty}_{n=1}\frac{\mu_{n}}{2^{n}}$$
Show that $\mu$ is a measure on $\mathcal{M}$ and that each $\mu_{n}$ is absolutely continuous with respect to $\mu$.
Proof: First, note that since $\mu_{n}(\emptyset)=0$ for all $n$, we have $\left(\sum\limits^{m}_{n=1}\frac{\mu_{n}}{2^{n}}\right)(\emptyset)=\sum\limits^{m}_{n=1}\frac{\mu_{n}(\emptyset)}{2^{n}}=0$, for all $m$. Then,
$$\mu(\emptyset)=\lim\limits_{m\rightarrow \infty}\left(\sum\limits^{m}_{n=1}\frac{\mu_{n}}{2^{n}}\right)(\emptyset)=0.$$
Let $(E_{k})$ a sequence of measurable disjoint sets. Then,  since $\mu_{n}$ is countably additive and finite, for all $n$, we have
$$\left(\sum\limits^{m}_{n=1}\frac{\mu_{n}}{2^{n}}\right)\left(\bigcup^{\infty}_{k=1} E_{k}\right)=\sum\limits^{m}_{n=1}\frac{1}{2^{n}}\mu_{n}\left(\bigcup\limits^{\infty}_{k=1} E_{k}\right)=\sum\limits^{m}_{n=1}\frac{1}{2^{n}}\sum\limits^{\infty}_{k=1}\mu_{n}\left(E_{k}\right)=\sum\limits^{\infty}_{k=1}\sum\limits^{m}_{n=1}\frac{1}{2^{n}}\mu_{n}\left(E_{k}\right)$$
On the other hand, $\mu(E_{k})=\lim\limits_{m\rightarrow \infty}\sum\limits^{m}_{n=1}\frac{1}{2^{n}}\mu_{n}\left(E_{k}\right)$, then
$$\mu\left(\bigcup^{\infty}_{k=1} E_{k}\right)=\lim\limits_{m\rightarrow \infty}\left(\sum\limits^{m}_{n=1}\frac{\mu_{n}}{2^{n}}\right)\left(\bigcup^{\infty}_{k=1} E_{k}\right)$$
$$=\lim\limits_{m\rightarrow \infty}\sum\limits^{\infty}_{k=1}\sum\limits^{m}_{n=1}\frac{1}{2^{n}}\mu_{n}\left(E_{k}\right)=\sum\limits^{\infty}_{k=1}\lim\limits_{m\rightarrow \infty}\sum\limits^{m}_{n=1}\frac{1}{2^{n}}\mu_{n}\left(E_{k}\right)=\sum\limits^{\infty}_{k=1}\mu(E_{k})$$
This means that $\mu$ is countably additive, and thus, a measure. 
To show that $\mu_{n}<<\mu$, for all $n$, let $E\in \mathcal{M}$ such that $\mu(E)=0$. This implies that
$$\left(\sum\limits^{m}_{n=1}\frac{\mu_{n}}{2^{n}}\right)(E)\leq \mu(E)=0, \hspace{0.3cm} \forall m\geq 1$$
and since each $0\leq \mu_{n}(E)<\infty$, we have $\mu_{n}(E)=0$ for all $n$. Therefore, $\mu_{n}<<\mu$ for all $n$.
My question is: How use this to calculate the Raddon-Nickodym derivative of $\delta_{1/n}$ with respect to $\mu$? Thank you.
 A: We have, for each $X\in \mathcal{M}$,
$$\delta_{1/n}(X)=\int_{X}\frac{d\delta_{1/n}}{d\mu}d\mu$$
So, 
$$\mu(X)=\sum\limits^{\infty}_{n=1}\frac{\delta_{1/n}(X)}{2^{n}}=\sum\limits^{\infty}_{n=1}\frac{1}{2^{n}}\int_{X}\frac{d\delta_{1/n}}{d\mu}d\mu=\int_{X}\left(\sum\limits^{\infty}_{n=1}\frac{1}{2^{n}}\frac{d\delta_{1/n}}{d\mu}\right)d\mu $$
and, by uniqueness of the Radon-Nickodym derivative, we have
$$\sum\limits^{\infty}_{n=1}\frac{1}{2^{n}}\frac{d\delta_{1/n}}{d\mu}=1\hspace{0.4cm}\mu-\textrm{almost everywhere} $$
A: Taking a look at my copy of Royden's book, I see that you have stated the question verbatim as well as a solution to it (which looks correct to me, by the way). Then at the very end you have asked an unrelated question, namely to find Radon-Nikodym derivative of an unrelated Dirac measure.
Therefore, I interpret that you meant to ask for the Radon-Nikodym derivative of $\mu_n$ with respect to $\mu$. Now there are two cases. (1) If $\mu_n,\mu_m$ are mutually singular measures for all $m\not=n$, then the Radon-Nikodym derivative is very simply $2^{-n}$. Otherwise, (2) there is no way to simplify the general answer: it is simply the sum of the Radon-Nikodym derivatives of the measures adding up to $\mu$.
To prove my claims in the last paragraph, simply use the linearity of the Radon-Nikdoym derivative.
