Here is an alternative proof base on J.L. Doob's notes on measure theory.
Define $\mu:\mathcal{A}\rightarrow[0,\infty)$ as
$\mu(A)=\lim_n\mu_n(A)$. Clearly $\mu(\emptyset)=0$, $\mu(A)\geq0$ and $\mu$ is additive, i.e., $\mu(A_1\cup A_2)=\mu(A_1)+\mu(A_2)$ for all $A_j\in\mathcal{A}$ with $A_1\cap A_2=\emptyset$. It remains to show that $\mu$ is continuous at $\emptyset$. That is, for any monotone nonincreasing sequence $A_m\in\mathcal{A}$ with $\bigcap_mA_m=\emptyset$, $\lim_m\mu(A_m)=0$.
Let $(A_m:m\in\mathbb{N})$ be such a sequence. Then $\alpha=\lim_m\mu(A_m)$ exists. Suppose $\alpha>0$. Notice that for each $n\in\mathbb{N}$
$$\lim_m\mu_n(A_m)=0\tag{0}\label{zero}$$
for each $\mu_n$ is a finite measure.
Set $n_1=m_1=1$ and fixed a number $0<\lambda<1$ to be determine later. Procedding by induction, once $m_\ell, n_\ell$, $1\leq \ell\leq k$ have been constructed, by set wise convergence, there is $n_{k+1}>n_k$ such that
$$|\mu_n(A_{m_k})-\mu(A_{m_k})|<\lambda\alpha\qquad\forall n\geq n_{k+1}$$
It follows that
$$\mu_{n_{k+1}}(A_{m_k})> \mu(A_{n_k})-\lambda\alpha\geq(1-\lambda)\alpha\tag{1}\label{one}$$
By \eqref{zero}, there is $m_{k+1}>m_k$ such that
$$\mu_{n_{k+1}}(A_{m_{k+1}})<\lambda\alpha\tag{2}\label{two}$$
For each $k\in\mathbb{N}$ define $B_k=A_{m_k}\setminus A_{m_{k+1}}$. Then
$$\mu_{n_{k+1}}(B_k)=\mu_{n_{k+1}}(A_{m_k})-\mu_{n_{k+1}}(A_{m_{k+1}})>(1-2\lambda)\alpha\tag{3}\label{three}
$$
We will require that $1-2\lambda>0$.
The sets $B_k$ form a pairwise disjoint sequence of measurable sets, and for each $k\in\mathbb{N}$,
$$A_{n_k}=\bigcup^\infty_{\ell=k}B_\ell=\bigcup\{B_\ell:\ell\,\text{odd}, \,\ell\geq k\}\cup\bigcup\{B_\ell:\ell\,\text{even}, \,\ell\geq k\}$$
If $2j+1>k$, then by \eqref{three}
$$\mu_{n_{2j+1}}\left( \bigcup\{B_\ell:\ell\,\text{even}, \,\ell\geq k\}\right)\geq \mu_{n_{2j+1}}(B_{2j})>(1-2\lambda)\alpha$$
and so,
$$\mu\left( \bigcup\{B_\ell:\ell\,\text{even}, \,\ell\geq k\}\right)\geq (1-2\lambda)\alpha\tag{4}\label{four}
$$
If $2j>k$, then by \eqref{three}
$$\mu_{n_{2j}}\left( \bigcup\{B_\ell:\ell\,\text{odd}, \,\ell\geq k\}\right)\geq \mu_{n_{2j}}(B_{2j-1})>(1-2\lambda)\alpha$$
and so,
$$\mu\left( \bigcup\{B_\ell:\ell\,\text{odd}, \,\ell\geq k\}\right)\geq (1-2\lambda)\alpha\tag{5}\label{five}$$
From \eqref{four} and \eqref{five} we have that
$\mu(A_{n_k})\geq 2(1-2\lambda)\alpha$ for all $k$ and so,
$$\alpha=\lim_n(A_n)=\lim_k\mu(A_{n_k})\geq2(1-2\lambda)\alpha$$
If we further choose $\lambda$ so that $2(1-2\lambda)>1$, that is $0<\lambda<\frac14$, we reach a contradiction. Therefore, $\alpha=0$.
This concludes the proof of Nikodym's convergence theorem.
Comments:
- This result is known as Nikodym convergence theorem.
- There is another result by Vitali-Hahn-Saks (see Dunford, N. and Schwartz, J. T., Linear operators: Part I. New York, Interscience, 1957, pp. 158–159 or Yosida, K., Functional Analysis, 6th edition, Springer-Verlag, Berlin, Heidelberg, New York, 1980, pp. 70-71) that is when restricted to finite measures (nonnegative measures) is equivalent to Nikodym's result above.
- The simple proof presented above follows the ideas presented by Doob, J. L., Measure Theory, Springer Verlag, GTM 143, New York, 1993, pp. 31-32). Doob published his treatment on Measure Theory originally as an excuse to buy a computer.
