Find Lebesgue measure of $\limsup A_n \cap B_n$ if $m(\limsup A_n)=m(\liminf B_n)=1$,

Question

Let $m$ be the lebesgue measure on $X=[0,1]$. if $m(\limsup\limits_{n\rightarrow{\infty}} {A_n})=1$ and $m(\liminf\limits_{n\rightarrow{\infty}} {B_n})=1$, prove that $m(\limsup\limits_{n\rightarrow{\infty}} {A_n \cap B_n})=1$. I got stuck after I expanded the $\limsup\limits_{n\rightarrow{\infty}} {A_n \cap B_n}$ by definition.

Answer 1

This solution uses the hint given in Etienne's comment. Let $\limsup_{n\to\infty}A_n=A,\liminf_{n\to\infty}B_n=B,\limsup_{n\to\infty}A_n\cap B_n=E$.

$A\cap B\subseteq E$

$$\begin{align} &x\in A\cap B\implies \color{blue}{(x\in B)}\land \color{red}{(x\in A)}\\ \implies&\color{blue}{\left(x\in\bigcup_{k=1}^\infty\bigcap_{n=k}^\infty B_n\right)}\land\color{red}{\left(x\in\bigcap_{k=1}^\infty\bigcup_{n=k}^\infty A_n\right)}\\ \implies&\color{blue}{\left(\exists q: x\in\bigcap_{n=q}^\infty B_n\right)}\land\color{red}{\left(x\in\bigcup_{n=k}^\infty A_n\;\forall\;k\right)}\\ \implies&\color{blue}{(\;\exists q: x\in B_{n}\;\forall\;n\ge q\;)}\land\color{red}{(\text{ given }k\;\exists p\ge k: x\in A_{p} )}\\ \implies&\color{blue}{(\;\exists q: x\in B_{n}\;\forall\;n\ge q\;)}\land \color{red}{(\text{ given }k\;\exists p\ge \max(k,q): x\in A_{p} )}\\ \implies&\text{ given }k\;\exists p\ge k: x\in A_p\cap B_p\\ \implies& x\in\bigcup_{n=k}^\infty(A_n\cap B_n)\qquad\forall\;k\in\mathbb{N}\\ \end{align}$$

Assuming $E$ is Lebesgue-measurable we have

$m(E)\ge 1$

$$\begin{align} & A\cap B\subseteq E\\ \implies& X\setminus E\subseteq X\setminus (A\cap B)=(X\setminus A)\cup(X\setminus B)\\ \implies& m(X\setminus E)\le m((X\setminus A)\cup(X\setminus B))\le m(X\setminus A)+m(X\setminus B)\\ \implies& m(X)-m(E)\le 2m(X)-m(A)-m(B)\\ \implies& m(E)\ge m(A)+m(B)-m(X)=1+1-1=1 \end{align}$$

$m(E)\le 1$

$$E\subseteq X\implies m(E)\le m(X)=1$$