I'm trying to tackle Junghenn's Principles of Real Analysis' exercise 5.11:
Let $\mu, \mu_1, \mu_2, \dots$ be signed measures on $(X, \mathfrak{F})$ with $\mu_n \to \mu$. Show that $\mu^{\pm}(E) \leq \liminf_{n} \mu_n^{\pm}(E)$ and $|\mu|(E) \leq \liminf_{n} |\mu_n|(E)$.
I started off trying to prove that $\mu = \liminf_{n} \mu_n^+ - \liminf_{n} \mu_n^-$ to conclude the first inequality, but this didn't seem to pay off, since I couldn't prove that $\liminf_{n} \mu_n^+$ and $\liminf_{n} \mu_n^-$ were measures. I've also tried to use the fact that $$\mu^+(E) = \sup\{\mu(A) : A \in \mathfrak{F} \cap E\}$$ along with $$\liminf_{n} \mu_n^+ = \sup\bigg\{ \inf\Big\{ \sup\{ \mu_k(A) : A \in \mathfrak{F} \cap E \} : k \geq n \Big\} : n \geq 1 \bigg\},$$ but I kind of got tangled up between all the supremums and infimums, and I couldn't make much progress. Any help/hints would be greatly appreciated.
Edit: I was finally able to do it. I'll post my solution in case someone might want to take a look at it.
We will first show that $\mu^{\pm}(E) \leq \liminf_{n} \mu_{n}^{\pm}(E)$. Let $E \in \mathfrak{F}$. We know that for each $n \in \mathbb{N}$, $$ \mu_n^{+}(E) = \sup\{ \mu_n(A) : A \in \mathfrak{F} \cap E \} $$ $$ \mu_n^{-}(E) = - \inf\{ \mu_n(A) : A \in \mathfrak{F} \cap E \}. $$ Thus, it is clear that for each $A \in \mathfrak{F} \cap E$, $$ \mu_n(A) \leq \mu_n^{+}(E) \quad \forall n \in \mathbb{N}. $$ $$ -\mu_n(A) \leq \mu_n^{-}(E) \quad \forall n \in \mathbb{N}. $$ Since $\mu_n \to \mu$, this implies that $$ \mu(A) \leq \liminf_{n} \mu_n^{+}(E) \quad \forall A \in \mathfrak{F} \cap E $$ $$ -\mu(A) \leq \liminf_{n} \mu_n^{-}(E) \quad \forall A \in \mathfrak{F} \cap E. $$ That is, $\liminf_{n} \mu_n^{+}(E)$ is an upper bound for $\{ \mu(A) : A \in \mathfrak{F} \cap E \}$ and $\liminf_{n} \mu_n^{-}(E)$ is an upper bound for $\{ -\mu(A) : A \in \mathfrak{F} \cap E \}$. Then, $$ \sup\{ \mu(A) : A \in \mathfrak{F} \cap E \} \leq \liminf_{n} \mu_n^{+}(E) $$ $$ \sup\{ -\mu(A) : A \in \mathfrak{F} \cap E \} \leq \liminf_{n} \mu_n^{-}(E). $$ Using $\sup\{ -\mu(A) : A \in \mathfrak{F} \cap E \} = -\inf\{ \mu(A) : A \in \mathfrak{F} \cap E \}$, this means that $$ \mu^{+}(E) \leq \liminf_{n} \mu_n^{+}(E) $$ $$ \mu^{-}(E) \leq \liminf_{n} \mu_n^{-}(E), $$ and so $\mu^{\pm}(E) \leq \liminf_{n} \mu_{n}^{\pm}(E)$.
To show $|\mu|(E) \leq \liminf_{n} |\mu_{n}|(E)$ let $E \in \mathfrak{F}$. Hence, for each $n \in \mathbb{N}$, $$ |\mu_n|(E) = \sup\bigg\{ \sum_{j=1}^{k} |\mu_n(E_j)| : E_1, \dots, E_k \text{ is a measurable partition of } E \bigg\}. $$ Thus, let $E_1, \dots, E_k$ be a measurable partition of $E$. It is then clear that $$ \sum_{j=1}^{k} |\mu_n(E_j)| \leq |\mu_n|(E) \quad \forall n \in \mathbb{N}. $$ Since $\mu_n \to \mu$, we get $\lim_{n\to\infty} \sum_{j=1}^{k} |\mu_n(E_j)| = \sum_{j=1}^{k} |\mu(E_j)|$, and so $$ \sum_{j=1}^{k} |\mu(E_j)| \leq \liminf_{n} |\mu_n|(E). $$ Given that $E_1, \dots, E_k$ was an arbitrary measurable partition of $E$, it follows that $\liminf_{n} |\mu_n|(E)$ is an upper bound for $\bigg\{ \sum_{j=1}^{m} |\mu(E_j)| : E_1, \dots, E_m \text{ is a measurable partition of } E \bigg\}$, and so $$ |\mu|(E) = \sup\bigg\{ \sum_{j=1}^{m} |\mu(E_j)| : E_1, \dots, E_m \text{ is a measurable partition of } E \bigg\} \leq \liminf_{n} |\mu_n|(E). $$