Prove $E_2$ is measurable.

Question

If $E_1$ measurable and $m^* (E_1 \oplus E_2)=0$, then $E_2$ measurable.

Note: $E_1 \oplus E_2 = (E_1\cup E_2)-(E_1 \cap E_2)$.

$E_1$ measurable if for all $A\subset \mathbb R$ satisfy $m^*(A)=m^*(A\cap E_1)+ m^*(A\cap E_1^C)$.

Now we want to prove $E_2$ measurable, i.e. $m^*(A)=m^*(A\cap E_2)+ m^*(A\cap E_2^C)$.

\begin{align} &m^* (E_1 \oplus E_2)=0\\ \iff& m^* ((E_1\cup E_2)-(E_1 \cap E_2))=0 \end{align}

If $m^* ((E_1\cup E_2)-(E_1 \cap E_2))=0$ then $(E_1\cup E_2)-(E_1 \cap E_2)$ is measurable, i.e. $m^*(A)=m^*(A\cap ((E_1\cup E_2)-(E_1 \cap E_2)))+ m^*(A\cap ((E_1\cup E_2)-(E_1 \cap E_2))^C)$.

\begin{align} &m^*(A)=m^*(A\cap ((E_1\cup E_2)-(E_1 \cap E_2)))+ m^*(A\cap ((E_1\cup E_2)-(E_1 \cap E_2))^C)\\ \iff& m^*(A\cap E_1)+ m^*(A\cap E_1^C) = m^*(A\cap ((E_1\cup E_2)\cap (E_1 \cap E_2))^C)+ m^*(A\cap ((E_1\cup E_2)\cap(E_1 \cap E_2)^C)^C)\\ \iff& m^*(A\cap E_1)+ m^*(A\cap E_1^C) = m^*(A\cap ((E_1\cup E_2)^C\cup (E_1 \cap E_2)^C))+ m^*(A\cap ((E_1\cup E_2)^C\cup(E_1 \cap E_2)))\\ % \iff& m^*(A\cap E_1)+ m^*(A\cap E_1^C) = m^*(A\cap ((E_1^C\cap E_2^C)\cup (E_1^C \cup E_2^C)))+ m^*(A\cap ((E_1^C\cap E_2^C)\cup(E_1 \cap E_2)))\\ \end{align}

Now, I stuck here. I can't relating $m^* ((E_1\cup E_2)-(E_1 \cap E_2))=0$ to $m^*(A)=m^*(A\cap E_2)+ m^*(A\cap E_2^C)$. What the hint to prove that?

Answer 1

We can express the set $E_2$ as follows: $$ \boxed{E_2 = \left[ E_1 \cup (E_2 - E_1) \right] - [E_1 - E_2]} \tag{1} $$

Given that $m(E_1 \Delta E_2) = 0$.

By definition, $$ E_1 \Delta E_2 = (E_1 - E_2) \cup (E_2 - E_1) $$

Any subset of a set of measure zero is measurable and has measure zero.

Therefore, $E_1 - E_2$ and $E_2 - E_1$ are both measurable and $$ m(E_1 - E_2) = 0, \ \ m(E_2 - E_1) = 0. $$

The union of two measurable sets is measurable. Thus, $E_1 \cup (E_2 - E_1)$ is measurable.

Since both $[E_1 \cup (E_2 - E_1)]$ are measurable and $E_1 - E_2$ are measurable, it follows from (1) that $E_2$ is measurable.