Let $\{E_k\}^{\infty}_{k=1}$ be a countable disjoint collection of measurable sets. Prove that for any set A... 
Let $\{E_k\}^{\infty}_{k=1}$ be a countable disjoint collection of measurable sets. Prove that for any set A, $m^*(A \cap \cup^{\infty}_{k=1})$ = $\sum^{\infty}_{k=1} m^*(A\cap E_k)$

My answer:
I tried to prove this by induction:
We know that if k=1, $m^*(A \cap E) = m^*(A \cap E)$. So now if we assume that for some n, $m^*(A \cap \cup^{n}_{k=1})$ = $\sum^{n}_{k=1} m^*(A\cap E_k)$. Now we have to prove that it is true for n+1.  $m^*(A \cap \cup^{n+1}_{k=1}) \leq m^*((A \cap \cup^{n}_{k=1} E_k) + m^*(A \cap E_{n+1})$
So now I have to prove the reverse inclusion, right?...But I'm kind of stuck here...
 A: First, there's philosophical problems with using an induction argument to prove the infinite property you're trying to show. So I don't think that will work. Additionally, I found David Mitra's solution to the problem to be rather complicated compared to the one that follows.  
Here goes nothing. 
By countable subadditivity of outer measure, 
\begin{align}
m^*\left(A\cap\bigcup_{k=1}^\infty E_k\right)&=m^*\left(\bigcup_{k=1}^\infty (A\cap E_k)\right)\\
&\leq \sum_{k=1}^\infty m^*(A\cap E_k),
\end{align}
so it really suffices to prove the opposite inequality, i.e., $\sum_{k=1}^\infty(A\cap E_k)\leq m^*\left(A\cap\bigcup_{k=1}^\infty E_k\right)$. 
Since $A\cap\bigcup_{k=1}^\infty E_k$ contains $A\cap\bigcup_{k=1}^n E_k$ for each $n$, we have:
\begin{align}
m^*\left(A\cap\bigcup_{k=1}^\infty E_k\right)&\geq m^*\left(A\cap\bigcup_{k=1}^n E_k\right)\\
&=\sum_{k=1}^n m^*(A\cap E_k)\tag{1}
\end{align}
by Proposition 6 of Royden $\S 2.3$. 
This gives:
\begin{align}
\lim_{n\to\infty} m^*\left(A\cap\bigcup_{k=1}^\infty E_k\right)&\geq \lim_{n\to\infty} \sum_{k=1}^n m^*(A\cap E_k)\\
m^*\left(A\cap\bigcup_{k=1}^\infty E_k\right)&\geq\sum_{k=1}^\infty m^*(A\cap E_k),
\end{align}
since the LHS is independent of $n$. Q.E.D. 
A: I assume $m^*$ is an outer measure; and will use the fact that a set $E$ is measurable if and only if
$$m^*(E)=m^*(E\cap T)+m^*(E\cap T')$$ for all sets $T$.
Since $\cup_{k=1}^\infty E_k$ is measurable, your result is equivalent to proving that
$$\tag{1}
m^*(A)=\sum_{k=1}^\infty m^*(E_k\cap A) +m^*\Bigr(A\cap \bigl(\ \bigcup_{k=1}^\infty E_k\ \Bigr)'\Bigr).
$$
Towards that end, note, by the the countable subadditivity of $m^*$
$$\eqalign{
 m^*(A)
&\le m^*\Bigr(A\cap \Bigl(\ \bigcup_{k=1}^\infty E_k\ \Bigr) \Bigr) +m^*\Bigr(A\cap \Bigl(\ \bigcup_{k=1}^\infty E_k\ \Bigr)'\Bigr)\cr
&\le\sum_{k=1}^\infty m^*(E_k\cap A) +m^*\Bigr(A\cap \Bigl(\ \bigcup_{k=1}^\infty E_k\ \Bigr)'\Bigr).
}
$$
With this in hand, we see that the desired result is trivial if $m^*(A)=\infty$.
Now assume $m^*(A)\ne\infty$.
To prove the reverse inequality needed,
we will prove by induction that for each $p\ge1$ and any set $A$
$$\tag{2}
m^*(A)=\sum_{k=1}^p m^*(E_k\cap A) +m^*\Bigr(A\cap \Bigl(\ \bigcup_{k=1}^p E_k \Bigr)'\ \Bigr).
$$
Once this is done, since the sequence $\Bigl(m^*\Bigr(A\cap \bigl(\cup_{k=1}^p E_k \bigr)'\Bigr)\Bigr)_{p=1}^\infty$ is nonincreasing and bounded below by 
$m^*\Bigr(A\cap \bigl(\cup_{k=1}^\infty E_k \bigr)' \Bigr)$,
it will follow that
$$
 m^*(A)\ge\sum_{k=1}^\infty m^*(E_k\cap A) +m^*\Bigr(A\cap \Bigl(\ \bigcup_{k=1}^\infty E_k\ \Bigr)'\Bigr),
$$
as desired.

So, on to the proof of $(2)$.
The base case follows from the fact that $E_1$ is measurable:
$$
m^*(A)=m^*(A\cap E_1)+m^*(E\cap E_1').
$$
Assume  $(2)$ is true for $p=k$.
Since $E_{p+1}$ is measurable, 
$$m^*(A)=m^*(A\cap E_{p+1})+  m^*(A\cap E_{p+1}')$$
Now, using $(2)$ applied to the set $A\cap E_{p+1}'$,
$$
\eqalign{
m^*(A)&=m^*(A\cap E_{p+1})+  m^*(A\cap E_{p+1}')\cr 
&=m^*(A\cap E_{p+1})+\sum_{k=1}^p m^*(A\cap E_{p+1}' \cap E_k) +m^*\Bigl( A\cap E_{p+1}'\cap \Bigl( \bigcup_{k=1}^p E_k\ \Bigr)'\Bigr)\cr
&=m^*(A\cap E_{p+1})+\sum_{k=1}^p m^*(A\cap   E_k) +m^*\Bigl( A\cap E_{p+1}'\cap \Bigl(\ \bigcup_{k=1}^p E_k\ \Bigr)'\Bigr)\cr
&=\sum_{k=1}^{p+1} m^*(E_k\cap A) +m^*\Bigr(A\cap \Bigl(\ \bigcup_{k=1}^{p+1} E_k\ \Bigr)'\Bigr);
}
$$
where, in the third equality, we used the fact that ${E_k\subset E_{p+1}'}$. This is exactly what we need to finish the proof by induction. $(2)$ holds for all $p$.
A: I don't think you can use induction to prove a property for $\infty$.
Why not using that measures are completely additive and the equality 
$A\cap \bigcup_{k=1}^\infty E_k = \bigcup_{k=1}^\infty(A\cap E_k)$
