$\mu(A_n \Delta B_n)=0$ for all $n.$

Question

Let $(X,S,\mu)$ a measure space, and let, $(A_n), (B_n)$ two sequences of elements of S. If $\mu(A_n \Delta B_n)=0$ for all n prove, the following are $\mu-$null sets ($\mu(E)=0$ for $E\in$S):

i) $\mu( ( \bigcup^{\infty}_{n=1}A_n) \Delta (\bigcup^{\infty}_{n=1}B_n))$.

ii) $\mu(({\bigcap}_{n=1}^\infty) A_n \Delta ({\bigcap}_{n=1}^\infty B_n))$.

iii) $\mu((\overline{\lim} A_n) \Delta ((\overline{\lim} B_n))$.

iv) $\mu((\underline{\lim} A_n) \Delta ((\underline{\lim} B_n))$.

For (i) I prove that $\mu(A_n - B_n)=\mu(B_n - A_n)=0$, because $\mu(A_n \Delta B_n)=\mu((A_n-B_n)\cup(B_n-A_n))=0$ and $B_n-A_n$, $A_n-B_n$ are disjoint, then $\mu(A_n \Delta B_n)=\mu((A_n-B_n)\cup(B_n-A_n))=\mu(A_n-B_n)+\mu(B_n-A_n)=0$ for all n, but $\mu$ is no negative, then $\mu(A_n - B_n)=\mu(B_n - A_n)=0$.

For (ii) I used that $({\bigcap}_{n=1}^\infty) A_n \Delta ({\bigcap}_{n=1}^\infty B_n)\subset ({\bigcap}_{n=1}^\infty A_n \Delta B_n)$ then $\mu(({\bigcap}_{n=1}^\infty) A_n \Delta ({\bigcap}_{n=1}^\infty B_n))\leq \mu(({\bigcap}_{n=1}^\infty A_n \Delta B_n))$.

But for (iii) and (iv) I am not sure.

Answer 1

We need some general identities:

Let $K$ an index set. Then: $$ \bigcup_{k\in K} X_k \triangle \bigcup_{k\in K} Y_k \subset \bigcup_{k\in K} X_k \triangle Y_k\\ X \triangle Y = X^{c} \triangle Y^c \\ \bigcap_{k\in K} X_k \triangle \bigcap_{k\in K} Y_k \subset \bigcup_{k\in K} X_k \triangle Y_k\\ $$ The first identity: If $a\in \bigcup_{k\in K} X_k$ but $a\notin \bigcup_{k\in K} Y_k$, then $a\in X_{k_0}$ and $a\notin Y_{k_0}$, so $a\in X_{k_0}\triangle Y_{k_0}$, for some $k_0\in K$. The other case is similar.

The second: definition of set difference.

Third: apply the first and second and De-Morgan

Answering (i) and (ii) is a simple application of the id's above + $\sigma$-subadditivity of the measure.

For (iii): set $X_n=\bigcup_{k\ge n} A_k$ and $Y_n=\bigcup_{k\ge n} B_k$. Then $X_n \triangle Y_n$ is a null set: it is covered by a union of null sets: $X_n \triangle Y_n\subset \bigcup_{k\ge n} A_k \triangle B_k$, by the first identity.

Now, the relation $\bigcap_n X_n \triangle \bigcap_n Y_n\subset \bigcup_n X_n \triangle Y_n$ implies similarly that $\mu\left( \overline{\lim}A_n \triangle \overline{\lim}B_n\right)=0$.

The $\underline{\lim}$ case is almost identical.