How to show that $m^*(A\cap (\cup_{i=1}^\infty E_i))=\sum _{i=1}^\infty (m^*(A\cap E_i))$

Question

For any set $A\subset \Bbb R$ ,if $\{E_i\}$ is a sequence of disjoint measurable sets then show that $m^*(A\cap (\cup_{i=1}^\infty E_i))=\sum _{i=1}^\infty (m^*(A\cap E_i))$ where $m^*$ denotes Lebesgue Outer Measure.

I know that $m^*(A\cap (\cup _{i=1}^n E_i))=\sum_{i=1}^n m^*(A\cap E_i)$

Now $m^*(A\cap (\cup_{i=1}^\infty E_i))$

$=m^*(\cup_{i=1}^\infty (E_i\cap A))$

$=\underline{m^*(\lim _{n\to \infty } \cup_{i=1}^n(A\cap E_i))=\lim _{n\to \infty }(m^* \cup_{i=1}^n(A\cap E_i))}$

$=\lim _{n\to \infty } \sum_{i=1}^nm^*(A\cap E_i)=\sum _{i=1}^\infty (m^*(A\cap E_i))$

I don't know whether the underlined part is true or not as I have never come across such a result ,Kindly please check.

Please help.

Answer 1

First by countable subadditivity you have $$m^* \left( A \cap \bigcup_{i=1}^\infty E_i \right) = m^* \left( \bigcup_{i=1}^\infty (A \cap E_i) \right) \le \sum_{i=1}^\infty m^*(A \cap E_i).$$

Next observe that if $E$ and $F$ are disjoint and measurable, then for any set $A$ you have $$m^*(A \cap (E \cup F)) = m^*(A \cap (E \cup F) \cap E) + m^*(A \cap (E \cup F) \setminus E) = m^*(A \cap E) + m^*(A \cap F).$$ An induction argument extends this to any finite number of measurable sets. In particular $$ m^*\left(A \cap \bigcup_{i=1}^k E_i \right) =\sum_{i=i}^k m^*(A \cap E_i)$$ for every $k$ since the $E_i$ are measurable. By monotonicity you get $$\sum_{i=i}^k m^*(A \cap E_i)\le m^*\left(A \cap \bigcup_{i=1}^\infty E_i \right) $$ for all $k$. Now let $k \to \infty$ to find $$\sum_{i=i}^\infty m^*(A \cap E_i) \le m^*\left(A \cap \bigcup_{i=1}^\infty E_i \right).$$

Answer 2

First, we note that $\leq$ is simply $\sigma$-subaditivity. Then we remember the definition of $\mu^*$, as: $$ \mu^*(A)=inf\{\mu(B):B\in\mathcal{B},A\subset B\}$$ Then, given $i>0$ (natural) and $\epsilon>0$ there is $B_i\in\mathcal{B}$ such that $\mu(B_i)\leq\mu^*(A)+\epsilon/2^{î+1}$, and therefore, if we define $B=\bigcup_{i>0}(B_i\cap E_i)$ we have that $$\sum_{i>0}\mu^*(A\cap E_i) \leq \sum_{i>0}\mu(B\cap E_i) = \mu(B\cap \bigcup_{i>0}E_i) \leq \mu^*(A\cap\bigcup_{i>0}E_i)+ \epsilon$$ Last des-equality can be probed by induction. It's easy to see that if $\mu(B)\leq\mu^*(A)+\delta$, then $\mu(B\cap(E\cup F))\leq\mu^*(A\cap(E\cup F))+\delta$. By induction $$\mu(B\cap \bigcup_{i=0}^kE_i) \leq \mu^*(A\cap\bigcup_{i=0}^kE_i)+ \epsilon\leq\mu^*(A\cap\bigcup_{i>0}E_i)+ \epsilon$$ And we conclude by continuity of $\mu$. Then, if we take $\epsilon\rightarrow0^+$ we conclude $\geq$