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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.

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  • $\begingroup$ what is $\mu_\infty$? $\endgroup$ – Filburt Mar 4 '17 at 14:34
  • $\begingroup$ A typo. Thanks for pointing it out. $\endgroup$ – grndl Mar 4 '17 at 16:55
  • $\begingroup$ @aduh I myself don't have the answer off the top of my head, but if you are desperate for an answer I can place a bounty on your question. $\endgroup$ – Clement C. Mar 6 '17 at 14:18
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    $\begingroup$ Well, it's not like these points have any actual meaning anyway. $\endgroup$ – Clement C. Mar 6 '17 at 15:33
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    $\begingroup$ Mildly relevant (although it's rather about when weak convergence implies setwise (strong) convergence, with one detour about when weak convergence implies TV convergence): hindawi.com/journals/ijsa/1997/614526/abs See Lemma 4.1 and Corollary 4.2. $\endgroup$ – Clement C. Mar 8 '17 at 20:33
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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 $f = d\mu/d\lambda$.

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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....

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  • $\begingroup$ Why is my application of Scheffe's Lemma wrong? Since $f_n$ is a probability density, $|f_n| = f_n$ a.s. $\endgroup$ – grndl Mar 12 '17 at 15:40
  • $\begingroup$ I was commenting to " (although this now seems inessential) ". You need probability measures. If $f_n\ge 0$ then your proof works fine $\endgroup$ – Gio67 Mar 12 '17 at 15:45
  • $\begingroup$ Ah, I see. Thanks. What I had in mind with "although this now seems inessential" was that the proof would work for any finite ($\sigma$-finite?) measure. I agree it does not work for general signed measures, however. $\endgroup$ – grndl Mar 12 '17 at 15:47
  • $\begingroup$ I edited my answer. You are right. My comment was misleading $\endgroup$ – Gio67 Mar 12 '17 at 15:49
  • $\begingroup$ I will correct my definition of total variation. Thanks for pointing that out. I had confused it with the common definition of total variation distance. $\endgroup$ – grndl Mar 12 '17 at 15:49

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