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A proof for $\mu(\lim_{n\rightarrow\infty}{A})=\lim_{n\rightarrow\infty}{\mu(A_n)}$

Definition: If $\liminf A_n = \limsup A_n = A^{\star}$ we define $\lim_n A_n = A^{\star}$.

Problem: Assume $\mu$ a finite measure and $A_n\in\mathcal{F}, \forall{n}\in\mathbb{N}$. Show $\mu(\lim_{n\rightarrow\infty}{A})=\lim_{n\rightarrow\infty}{\mu(A_n)}$

An initial idea:

My initial idea was to use the following properties of $\liminf A_n$ and $\limsup A_n$:

  1. $w\in\liminf{A_n} \iff \exist{n}$ such that $\forall{k\geq{n}}$ we have $w\in{A_k}$.

  2. $w\in\limsup{A_n} \iff$ there are infinitely many $n$ such that $w\in{A_n}$.

However these propoerties do not reveal a connection between $\mu(\limsup{A_n})=\mu(\liminf{A_n})=\mu(A^{\star})$ and $\mu(A_n)$.

Proof: (Idea thanks to GPT)

Define the following: $$ C_k=\cup_{i=k}^{\infty}A_i \\ B_k=\cap_{i=k}^{\infty}A_i $$ We note the following -

  1. $C_k$ is a decreasing collection ${C_{k+1}}\subset..\subset{C_1}$.
  2. $B_k$ is an increasing collection $B_1\subset..\subset{B_k}$.

These imply the following statements -

$$ \bigcap_{k=1}^{n}C_k = C_n \ \bigcup_{k=1}^{n}B_k = B_n $$

Now we are ready for a set of inequlities which will conclude the proof:

$$ \mu(\bigcup_{k=1}^{n}B_k) = \mu(B_k)\leq\mu(A_k) $$

And in particular, since $B_k\subset{A_k}$ for all $i\geq{k}$ we have by monotinicity -

$$ \mu(\bigcup_{k=1}^{n}B_k) = \mu(B_n)\leq\inf_{i\geq{n}}\mu(A_i)\leq\sup_{1\leq{k}\leq{n}}\inf_{i\geq{k}}\mu(A_i) $$

Using similar reasoning for $C_k$, we have - $$ \inf_{1\leq{k}\leq{n}}\sup_{i\geq{k}}\mu(A_i)\leq\sup_{i\geq{n}}\mu(A_i) \leq \mu(C_n)=\mu(\bigcap_{k=1}^{n}C_k) $$

We put everything togther to obtain -

$$ \mu(\bigcup_{k=1}^{n}B_k)\leq\sup_{1\leq{k}\leq{n}}\inf_{i\geq{k}}\mu(A_i)\leq\inf_{1\leq{k}\leq{n}}\sup_{i\geq{k}}\mu(A_i)\leq\mu(C_n)=\mu(\bigcap_{k=1}^{n}C_k) $$

We let $n\rightarrow\infty$ and obtain -

$$ \mu(\liminf{A_n})\leq\liminf\mu(A_n)\leq\limsup\mu(A_n)\leq\mu(\limsup{A_n}) $$

But since - $\liminf{A_n}=\limsup{A_n}=\lim_{n\rightarrow\infty}{A_n}$, by our initial assumption. The two ends of the above inequality are equal and we get the desired result -

$$ \mu(\lim_{n\rightarrow\infty}{A_n})=\lim_{n\rightarrow\infty}\mu(A_n) $$

$\square$.

Alterantive proofs?

I was struggling for a while to find alternative proofs before eventually abandoning the idea.. I will see if I find other alternative proofs relying on the basic properties of $\liminf$ and $\limsup$ if possible as this one was highly non trivial for me.