Convergence in some measure implies convergence in some other measure absolutely continuous to the first 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.
 A: As of the first question, this characterisation

Let $\mu$ be a finite measure. A sequence $f_n$ converges to $f$ in measure with respect to $\mu$ if and only if any subsequence $f_{n_k}$ admits a sub-subsequence $f_{n_{k_h}}$ that converges to $f$ almost everywhere.

provides far more than a solid path. It's almost a proof, because the $f_{n_{k_h}}$ given by the convergence in the measure $\mu$ works for the measure $\nu$ too.
For the second question, recall the following result:

Let $\mu$ be a $\sigma$-finite measure on $(\Omega,E)$. Then, there exists a probability $\mathbb{P}$ on the same space that is equivalent to $\mu$, i.e. such that $\forall A\in E\ (\mu(A)=0\leftrightarrow\mathbb{P}(A)=0).$

An example of this is $\Omega:=\mathbb{R}$, $\mu:=$Lebesgue measure, $\mathbb{P}:=\frac{1}{\sqrt{2\pi}}e^{-\frac{x^2}{2}}\cdot\mu$
In this example you can show that $f_n:=1_{[n,+\infty)}$ convergese in measure for $\mathbb{P}$ but not for $\mu$
The idea behind this is that, while the notion of almost-everywhere convergence depends only on negligible sets, it implies convergence in measure for finite ones, but not for $\sigma$-finite ones.
A: 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.
