TITLE: Convergence in some measure implies convergence in some other measure absolutely continuous to the first
QUESTION [6 upvotes]: Suppose that $f_n$ converges to $f$ in measure $\mu$ and $\nu$ is a different measure absolutely continuous with respect to $\mu$ (which of course means that both $\mu$ and $\nu$ are defined on the same measure space and  $\mu(E)=0\Longrightarrow \nu(E)=0$ for all sets $E$ in the sigma field). Then I'm trying to prove that $f_n$ converges to $f$ in $\nu$ also. 
I think a proof using the Radon-Nikodym theorem may exist, but I tried doing this from the first principles. So I argued that since $f_n$ converges to $f$ in $\mu$, it has a subsequence $f_{n_k}$ that converges to $f$ almost everywhere w.r.t $\mu$ and hence also w.r.t $\nu$ as $\nu$ is absolutely continuous with respect to $\mu$. So if I assume that all measures are finite, then convergence of $f_{n_k}$ almost everywhere w.r.t $\nu$ implies convergence of $f_{n_k}$ to $f$ in measure $\nu$. I can't finish the argument after this. Can a proof on these lines be completed?
Also in hindsight, I assumed finiteness of all involved measures, so a general question is whether it works for $\sigma-$finite measures. I think not, but would a appreciate a counterexample.

REPLY [2 votes]: I happened to come across this awhile ago when I was reading a textbook on semimartingales. The author mentioned and used this as a minor fact without proof. This is how I think about this, hope it helps.
Suppose $\nu$ and $\mu$ are two finite measures on a measurable space $ ( \Omega, \mathscr{F} ) $ such that $\nu << \mu$. Then by the Radon-Nikodym theorem, there exists a measurable function $ \xi $ such that for every $ A \in \mathscr{F} $  $$ \nu(A) = \int_A \xi d\mu. $$
Note that since $\nu$ is finite, $ \int \xi d\mu \leq M $ for some $M > 0$.
Now, given $ \varepsilon > 0 $, let $A_n$ be the set $ \{ \omega \in \Omega : \left|f_n(\omega) - f(\omega)\right| > \varepsilon \} $. Then, choose a $c > 0$, \begin{align} \nu(A_n) &= \int_{A_n} \xi d\mu \\ 
&= \int_{A_n} \xi \mathbb{1}_{\{\xi \leq c\}} d\mu + \int_{A_n} \xi \mathbb{1}_{\{\xi > c\}} d\mu \\ &\leq c \mu(A_n) + \int_{A_n} \xi \mathbb{1}_{\{\xi \leq c\}} d\mu.
\end{align}
By the dominated convergence and the fact that $ \mathbb{1}_{A_n} \rightarrow 0 $ $\mu$-$a.e$ as $ n \to \infty$, the second term in the last inequality vanished as $n$ tends to infinity. Thus $$ \lim_{n \to \infty} \nu(A_n) \leq c \lim_{n \to \infty}\mu(A_n) = 0, $$
proving the assertion.