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.
Best Answer
As of the first question, this characterisation
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:
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.