Under what assumptions does setwise convergence of signed measures imply convergence in total variation? Suppose the sequence $\{\mu_n : n \in \mathbb{N} \}$ of signed measures on $(X, \mathcal{F})$ converges setwise to the signed measure $\mu$, by which we mean that $\lim_{n \to \infty}\mu_n(A) = \mu(A)$ for every $A \in \mathcal{F}$. 

Under what further assumptions does $\{\mu_n : n \in \mathbb{N} \}$ converge in total variation to $\mu$?

Recall that the total variation norm $\| \cdot \|$ is defined by $\|\mu \| = \sup \sum_i |\mu(E_i)|$, where the supremum is over countable partitions of $X$, and $\{\mu_n : n \in \mathbb{N} \}$ converges in total variation to $\mu$ if $\|\mu_n - \mu \| \to 0$ as $n \to \infty$.
I had thought this would be a standard topic, but after googling and looking through some textbooks I still haven't been able to find any results.
 A: I'm going to try to mimick the proof of Corollary 4.2 in the paper linked by @Clement C. Comments and generalizations are appreciated.
We first suppose that $\{\mu_n \}$ is a sequence of probability measures on $(X, \mathcal{F})$ that converges setwise to the probability measure $\mu$. Moreover, suppose $\mu_n(A) = \int f_n d \lambda$ and $\mu(A) = \int_A f d \lambda$, where $\lambda$ is a $\sigma$-finite measure on $(X, \mathcal{F})$ and $f_n \to f$ a.e. ($\lambda$).
Now, setwise convergence implies $\int f_n d\lambda \to \int f d\lambda$. By Scheffe's Lemma, $\int |f_n - f|d\lambda \to 0$. And this in turn implies $\|\mu_n - \mu \| \to 0$.

In order to upgrade setwise convergence to convergence in total variation, we have assumed that $\mu, \mu_1, \mu_2,...$ are probabilities (although this should work for any finite, non-negative measures) that are absolutely continuous with respect to $\lambda$, and that the densities $f_n = d\mu_n/d\lambda$ converge almost everywhere to $f = d\mu/d\lambda$.

A: I don't have an answer to your question, just a few comments. If you
want $\Vert\mu_{n}-\mu\Vert(X)\rightarrow0$ then a necessary condition is that
$\sup_{n}\Vert\mu_{n}\Vert(X)<\infty$. So let's assume that.
Now If you assume that all the measures $\mu_{n}$ and $\mu$ are absolutely
continuous with respect to a positive measure $\nu$, then $\mu_{n}(E)=\int
_{E}f_{n}\,d\nu$ and $\mu(E)=\int_{E}f\,d\nu$ and $\int_{E}f_{n}%
\,d\nu\rightarrow\int_{E}f\,d\nu$ for every $E$. Together with the fact that
$\sup_{n}\Vert\mu_{n}\Vert(X)=\sup_{n}\int_{X}|f_{n}|\,d\nu<\infty$, this
condition is equivalent to weak convergence in $L^{1}(X,\nu)$ (it's a theorem
of Dunford-Pettis). Thus $f_{n}\rightharpoonup f$ in $L^{1}(X,\nu)$. It
follows (again by a theorem of Dunford-Pettis) that for every $\varepsilon>0$
there is $\delta>0$ such that for all $n$,
$$
\int_{E}|f_{n}|\,d\nu\leq\varepsilon
$$
for every measurable set $E$ with $\nu(E)\leq\delta$. Also for every
$\varepsilon>0$ there exists $E_{\varepsilon}$ measurable, with $\nu
(E_{\varepsilon})<\infty$ such that for all $n$,
$$
\int_{X\setminus E}|f_{n}|\,d\nu\leq\varepsilon.
$$
Next by Vitali's convergence theorem a sequence $\{f_{n}\}$ converges strongly
in $L^{1}(X,\nu)$ if and only if $\{f_{n}\}$ converges to $f$ in measure and
satisfies the previous two conditions. In other words a sequence $\{f_{n}\}$ converges
strongly to $f$ in $L^{1}$ if and only if it converges to $f$ weakly in
$L^{1}$ and in measure. 
This is a very special case, very far from replying to your original question. 
Concerning your answer and the paper linked by @Clement C., for real-valued functions it becomes
$$
\int_{X}|f_{n}|\,d\nu\rightarrow\int_{X}|f|\,d\nu.
$$
So going back to the original question, the analog would be $\Vert\mu_{n}
\Vert(X)\rightarrow\Vert\mu\Vert(X)$. If you assume this, you can prove that
$\Vert\mu_{n}\Vert(E)\rightarrow\Vert\mu\Vert(E)$. 
To see this, fix $\varepsilon>0$ and find countable partition of $E$ such that
$$
\Vert\mu\Vert(E)\leq\sum_{k}|\mu(E_{k})|+\varepsilon.
$$
Since the series is convergent, we can find $l$ such that $\sum_{k=1}^{\infty
}|\mu(E_{k})|\leq\sum_{k=1}^{l}|\mu(E_{k})|+\varepsilon$. Then
\begin{align*}
\Vert\mu\Vert(E)  & \leq\sum_{k=1}^{l}|\mu(E_{k})|+2\varepsilon=\lim
_{n\rightarrow\infty}\sum_{k=1}^{l}|\mu_{n}(E_{k})|+2\varepsilon\leq
\liminf_{n\rightarrow\infty}\sum_{k=1}^{\infty}|\mu_{n}(E_{k})|+2\varepsilon
\\
& \leq\liminf_{n\rightarrow\infty}\Vert\mu_{n}\Vert(E)+2\varepsilon.
\end{align*}
Now let $\varepsilon\rightarrow0$. To prove the other inequality, use the
previous inequality for $X\setminus E$ to write
\begin{align*}
\Vert\mu\Vert(E)  & =\Vert\mu\Vert(X)-\Vert\mu\Vert(X\setminus E)\geq\Vert
\mu\Vert(X)-\liminf_{n\rightarrow\infty}\Vert\mu_{n}\Vert(X\setminus E)\\
& =\lim_{n\rightarrow\infty}\Vert\mu_{n}\Vert(X)+\limsup_{n\rightarrow\infty
}(-\Vert\mu_{n}\Vert(X\setminus E))\\
& =\limsup_{n\rightarrow\infty}(\Vert\mu_{n}\Vert(X)-\Vert\mu_{n}
\Vert(X\setminus E))=\limsup_{n\rightarrow\infty}\Vert\mu_{n}\Vert(E).
\end{align*}
So now you have $\Vert\mu_{n}\Vert(E)\rightarrow\Vert\mu\Vert(E)$ for every
$E$. This is tight convergence. Unfortunately you are still far from $\Vert
\mu_{n}-\mu\Vert(X)\rightarrow0$. Wish I could help more but this is all I can say....
