# Prove that the setwise limit of signed meaures is a signed measure

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?

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 $$1$$.
• I haven't seen this notation before - is $|\nu_n|(B) = \nu^+(B) + \nu^-(B)$ where $\nu = \nu^+ - \nu^-$ is the Jordan Decomposition of $\nu$? Dec 5, 2020 at 2:17
• @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 $\mu$. Dec 5, 2020 at 2:29
• 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 $n$. Dec 8, 2020 at 15:52
• 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. Dec 8, 2020 at 23:23