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Let $(\mathcal{X} , \mathcal{B})$ be a measurable space. Let $(\nu_n)_{n=1}^\infty$ be a sequence of signed measures on $\mathcal{B}$ which converge setwise to a set function $\nu$. Suppose there exists a finite measure $\mu$ on $\mathcal{B}$ such that $|\nu_n(B)| \leq \mu(B)$ for every $B \in \mathcal{B}$ and $n \in \mathbb{N}$. Prove that $\nu$ is a signed measure.

I've been banging my head against the wall on this for a while, and so far this is the only progress I've made:

Let $\bigcup_{i=1}^\infty A_i$ be a union of pairwise disjoint members of $\mathcal{B}$. To show $\nu$ is a signed measure, it is enough to show that $\nu(\bigcup_{i=1}^\infty A_i) = \sum_{i=1}^\infty \nu(A_i)$. Notice that each $\nu_n$ is absolutely continuous with respect to $\mu$, so we can write $d \nu_n = f_n d\mu$ for some integrable function $f_n : \mathcal{X} \to \mathbb{R}$ (this is a result from a previous exercise/Radon-Nikodym theorem). So:

$\nu(\bigcup_{i=1}^\infty A_i) = \lim_{n \to \infty} \int_{\bigcup A_i} f_n d\mu = \lim_{n \to \infty} (\sum_{i=1}^\infty \int_{A_i} f_n d\mu)$

$\sum_{i=1}^\infty \nu(A_i) = \sum_{i=1}^\infty (\lim_{n \to \infty} \int_{A_i} f_n d\mu)$

So I've managed to reduce this question to the interchange of this limit and infinite sum. The most natural thing to try is Dominated Convergence Theorem, but I can't seem to find an integral dominator for all the $f_n$'s. Could someone please point me in the right direction here?

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Note that $|\nu_{n}(B)|\leq\mu(B)$ for all $B$ in fact implies that $|\nu_{n}|(B)\leq\mu(B)$. Indeed, $\nu_{n}^{+}(A)$ is the supremum of all $\nu_{n}(B)$'s for $B\subseteq A$, thus $\nu_{n}^{+}(A)\leq\mu(A)$ for all $A$, and the same holds for $\nu_{n}^{-}$, so the same holds for $|\nu_{n}|$. Then now you can show that $|f_{n}|$ is bounded by $1$.

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  • $\begingroup$ I haven't seen this notation before - is $|\nu_n|(B) = \nu^+(B) + \nu^-(B)$ where $\nu = \nu^+ - \nu^-$ is the Jordan Decomposition of $\nu$? $\endgroup$
    – dacian98
    Dec 5, 2020 at 2:17
  • $\begingroup$ @dacian98 Yes. (I actually prefer a more natural definition of $|\nu_{n}|$ but in the end it is what you said.) What matters here is that $|f_{n}|$ is precisely the Radon-Nikodym derivative of $|\nu_{n}|$ with respect to $\mu$. $\endgroup$
    – Junekey Jeon
    Dec 5, 2020 at 2:29
  • $\begingroup$ ahh I think I finally got it! So we have that $|\nu_n|(B) = \nu_n^+ (B) + \nu_n^- (B) = \int_B f_n^+ d\mu + \int_B f_n^- d\mu = \int_B |f_n| d\mu \leq \mu(B) = \int_B d\mu$, which implies that $|f_n| \leq 1$ $\mu$-almost everywhere. So if we take the union over n of the sets where $|f_n|>1$, we see that it has measure zero, so we can say without loss of generality that the constant function 1 (which is integrable since $\mu$ is finite) dominates $f_n$ for every $n$. $\endgroup$
    – dacian98
    Dec 8, 2020 at 15:52
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    $\begingroup$ Or just modify $f_{n}$ on a set of measure zero so that $f_{n}$ is pointwise dominated by $1$. Either way is fine. LDCT is already written in terms of a.e. dominance, not pointwise dominance, AFAIK, though. $\endgroup$
    – Junekey Jeon
    Dec 8, 2020 at 23:23

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