# Show that $\bigcup\limits_{n=1}^\infty A_n \in \mathcal{A}$.

Exercise:

Let $\lambda$ be the Lebesgue measure restricted to measure space $((a,b],\mathcal{B}(a,b])$, called $S$.

Let $\mathcal{F}_{(a,b]}$ be the collection of finite unions of half-open intervals in $S$.

Let $$\mathcal{A} = \{A\in\mathcal{B}((a,b]):\forall\epsilon >0\exists F\in \mathcal{F_{(a,b]}\text{ such that }\lambda(A\triangle F)<\epsilon\}}$$

Assume that $(A_n)_{n\geq 1}$ is a disjoint sequence in $\mathcal{A}$. Use the fact that if $\bigcup\limits_{n=1}^\infty A_n = A$ and $\bigcup\limits_{n=1}^\infty B_n=B$ then $A\triangle B \subseteq \bigcup\limits_{n=1}^\infty (A_n \triangle B_n)$, to show that $\bigcup\limits_{n=1}^\infty A_n \in \mathcal{A}$.

Question: How do I solve this exercise? Isn't it trivial that $\bigcup\limits_{n=1}^\infty A_n \in \mathcal{A}$, if $(A_n)_{n\geq 1}$ is a disjoint sequence in $\mathcal{A}$?

Let $\epsilon>0$.

We have $\sum_{n=1}^{\infty}\lambda(A_n)=\lambda(A)\leq \lambda((a,b])=b-a$ so for $m$ large enough $\sum_{n=m+1}^{\infty}\lambda(A_n)<\epsilon2^{-1}$.

For every $n\leq m$ let $F_n\in\mathcal F_{(a,b]}$ such that $\lambda(A_n\Delta F_n)<\epsilon2^{-n-1}$ and for every $n>m$ let $F_n=\varnothing$.

Then $F=\bigcup_{n=1}^{\infty}F_n\in\mathcal F_{(a,b]}$ and we claim that $\lambda(A\Delta F)<\epsilon$.

For this observe that $A\Delta F\subseteq\bigcup_{n=1}^{\infty}(A_n\Delta F_n)$ so that: $$\lambda(A\Delta F)\leq\sum_{n=1}^{\infty}\lambda(A_n\Delta F_n)=\sum_{n=1}^{m}\lambda(A_n\Delta F_n)+\sum_{n=m+1}^{\infty}\lambda(A_n)<\sum_{n=1}^{m}\epsilon2^{-n-1}+\epsilon2^{-1}\leq\epsilon$$

This can be done for every $\epsilon>0$ so proves that $A\in\mathcal A$.

• How do you know that $\sum_{n=1}^\infty \lambda(A_n) = \lambda(A)$? And why does $\lambda(A)\leq b-a$ imply that for $m$ large enough $\sum_{n=m+1}^\infty \lambda(A_n) < \epsilon 2^{-1}$? – titusAdam Jan 6 '18 at 10:37
• The set $A_n$ are disjoint measurable and $A=\bigcup_{n=1}^{\infty}A_n$. Then $\lambda(A)=\sum_{n=1}^{\infty}\lambda(A_n)$ because $\lambda$ is a measure hence is $\sigma$-additive. $\sum_{n=1}^{\infty}\lambda(A_n)=\lambda(A)<\infty$ means that $s_m:=\sum_{n=1}^{m}\lambda(A_n)\to\lambda(A)$. Then $\sum_{n=m+1}^{\infty}\lambda(A_n)=\lambda(A)-s_m\to0$. – drhab Jan 6 '18 at 12:10