Dtyjlui

I found a lot of questions with the similar title on the forum, but not all of them are properly answered and those, which are, differ significantly from mine.

I work in the closed interval $[0,1]$. I use the following definition of Lebesgue measurability:

$mathbf{Def. 1}$ The set E is called measurable iff $$lambda^{*}(E)+lambda^{*}([0,1]backslash E)=1,$$ where $lambda^{*}$ is the outer lebesgue measure, i.e. $lambda^{*}(E)=inf_{Usupset E}lambda(U)$, where $U$ is an open set.

I can prove that this definition is equivalent to the following:

$mathbf{Def. 2}$ The set E is called measurable iff $forall epsilon >0 exists U_{epsilon} ,F_{epsilon}$ - open and closed sets respectively, such that $F_{epsilon} subset E subset U_{epsilon} lambda^{*}(U_{epsilon})-lambda^{*}(F_{epsilon})<epsilon$.

$sigma$-additivity for $lambda$ is also proven (for a measurable set $E$: $lambda(E)=lambda^{*}(E))$ :

If ${E_n}$ - a collection of measurable sets and $E_kcap E_l = varnothing$ then $lambda left (bigcup_{n=1}^{infty}E_n right)=sum_{n=1}^infty{lambda(E_n)}$.

And here is my question:

a) If $A$ and $B$ are measurable sets (they may intersect) prove that $Acup B, Acap B, Abackslash B$ are measurable using $mathbf{ Def. 1}$ or $mathbf{ Def. 2}$.

b) Prove that countable intersection and countable union of the measurable sets is measurable.

I'd appreciate your help, guys.

Usually, a set $E$ is called (Lebesgue) measurable, if for all sets $A$ we have
begin{equation} lambda^*(Ecap A)+lambda^*(E^complementcap A)=lambda^*(A).end{equation} We can show, that Def. 1 is equivalent to this standard definition of Lebesgue measurability. Then a) and b) follow, (you can read their proofs in a book on measure theory, they are a bit long, though).

Proof: If $E$ is Lebesgue measurable (by the standard definition), then it satisfies the equality in Def. 1 (take $A=[0,1]$).

Let $E$ be not Lebesgue measurable. Then there is a set $A$ and an $varepsilon>0$ such that
begin{equation}lambda^*(Ecap A)+lambda^*(E^complementcap A)>lambda^*(A)+4varepsilon.end{equation}
By the definition of $lambda^*$ there is a countable collection of open intervals $(I_n)_n$ where $I_n=(a_n,b_n)$ which together cover $A$ and where
begin{equation}lambda^*(J)leqsum_{n=1}^infty(b_n-a_n)<lambda^*(A)+2varepsilon,end{equation}
for $J=bigcup_nI_n$. Since $Asubseteq J$ also $lambda^*(Ecap J)+lambda^*(E^complementcap J)>lambda^*(A)+2varepsilon$. Since the tail of the above sum goes to $0$, there is an index $N$ such that
begin{equation}sum_{n=N}^infty(b_n-a_n)<varepsilon.end{equation}
Let $J_N=bigcup_{n=1}^NI_n$. Then
begin{equation}lambda^*(Ecap J)leqlambda^*(Ecap J_N)+lambda[Ecap(Jsetminus J_N)]<lambda^*(Ecap J_N)+varepsilonend{equation} and the same for $E^complement$ instead of $E$. The next thing works, because $J_N$ and $J_N^complement$ are finite unions of intervals and disjoint.
begin{align}
lambda^*(E)+lambda^*(E^complement)&=lambda^*[(Ecap J_N)cup(Ecap J_N^complement)]+lambda^*[(E^complementcap J_N)cup(E^complementcap J_N^complement)]\
&=lambda^*(Ecap J_N)+lambda^*(Ecap J_N^complement)+lambda^*(E^complementcap J_N)+lambda^*(E^complementcap J_N^complement)\
&>[lambda^*(Ecap J)+lambda^*(E^complementcap J)-2varepsilon]+lambda(J_N^complement)\
&>lambda^*(A)+4varepsilon-2varepsilon+[1-lambda(J)]\
&>1,end{align}
which means, that $E$ is neither Lebesgue measurable according to Def. 1.