Several Properties of lim sup and lim inf in measure theory

Question

Question:

Let $\mathcal{E}$ be a $\sigma$-algebra on a set $X$ and $\mu: \mathcal{E} \to [0,\infty]$ a $\sigma$-additive function on $\mathcal{E}$. If I have an arbitrary sequence $\{A_n\}_{n \in \mathbb{N}} \subset \mathcal{E}$, are the following statements true?

1. For $B \subseteq \bigcup_{n=1}^{\infty} A_n$, $\mu(B) \leq \sum_{n=1}^{\infty} \mu(A_n)$
2. $\limsup_{n \to \infty} A_n^c = (\liminf_{n \to \infty} A_n)^c$
3. $\mu(\liminf_{n \to \infty} A_n) = \liminf_{n \to \infty} \mu(A_n)$
4. $\limsup_{n \to \infty} \mu(A_n) \leq \mu(\limsup_{n \to \infty} A_n)$

My thoughts:

1. here I would say that this is true, since $\sigma$-additive $\Rightarrow$ $\sigma$-subadditive
2. i tried the following: $\limsup_{n \to \infty} A_n^c = \bigcap_{n=1}^{\infty} \bigcup_{m=n}^{\infty} A_m^c = \bigcap_{n=1}^{\infty} (\bigcap_{m=n}^{\infty} A_m)^c = (\bigcup_{n=1}^{\infty} \bigcap_{m=n}^{\infty} A_m)^c = (\liminf_{n \to \infty} A_n)^c$
3. Here i started like this, but got kinda stuck: $\mu(\liminf_{n \to \infty} A_n) = \mu(\bigcup_{n=1}^{\infty} \bigcap_{m=n}^{\infty} A_m) = \sum_{n=1}^{\infty} \mu(\bigcap_{m=n}^{\infty} A_m)$
4. I know this holds, provided $\mu(X) < \infty$, but does it also hold without this condition?

I would really appreciate some help here. Thanks a lot!

Answer 1

1. As you already mentioned, $\mu$ is $\sigma$-subadditive, so $\mu(B)\leq\mu(\cup_{n=1}^{\infty} A_n))\leq\sum\limits_{n=1}^{\infty}\mu(A_n)$.

2. This seems to be correct.

3. This is wrong. Consider $X=[0,1],\mathcal{E}=\mathcal{B}[0,1]$ and $\mu=\lambda\lvert_{[0,1]}$ (Lebesgue-measure). For $n\in\mathbb{N}$ put $A_n=[0,0.5]$ if $n$ is even and $A_n=[0.5,1]$ if $n$ is odd. Then $\liminf_{n\to\infty}A_n=\{0.5\}$, so the LHS is $0$, but the RHS is $0.5$. However, as in this example, $\leq$ always holds: For all $n\in\mathbb{N}$ we have $\mu(\liminf_{n\to\infty}A_n)=\mu(\cup_{n=1}^{\infty}\cap_{m=n}^{\infty}A_m)\leq\mu(\cap_{m=n}^{\infty}A_m)\leq\mu(A_n)$.

4. This generally doesn't hold if $\mu(X)=\infty$. Consider $X=\mathbb{R},\mathcal{E}=\mathcal{B}(\mathbb{R})$ and $\mu=\lambda$. Put $a_1=0,b_1=1$ and for $n>1$ put $a_n=b_{n-1}, b_n=a_n+(b_{n-1}-a_{n-1})+1$. Now, put $A_n=[a_n,b_n]$. Then $\limsup_{n\to\infty}\mu(A_n)=\infty>0=\mu(\emptyset)=\mu(\limsup_{n\to\infty}A_n)$.