Show that $\mu(A)=\sup \left\{\mu_1(B)+\mu_2(A \setminus B) : B \subset A \right\}$ is measure. Let $(X ,\mathfrak{m})$-measurable space.
Let $\mu_1$ and $\mu_2$ will be measures. For $A \in \mathfrak{m}$ define :
$\mu(A)=\sup \left\{\mu_1(B)+\mu_2(A \setminus B) : B \subset A \right\} $
Show that $\mu$ is measure.
1.$\mu(\emptyset)=0$
This is obvious.
2.$\mu(\bigcup_{n=1}^{\infty} A_n)=\sum_{n=1}^{\infty}\mu(A_n)$
If $B=\emptyset$ (obvious). What, if $B \neq \emptyset$ ?
 A: Fix a pairwise disjoint sequence $\{A_n\}$. 
Let $B\subset \bigcup_nA_n$. Then
\begin{align}
\mu_1(B)+\mu_2\left(\bigcup_nA_n\setminus B\right)
&=\mu_1\left(\bigcup_n(A_n\cap B)\right)+\mu_2\left(\bigcup_n(A_n\setminus B)\right)\\ \ \\
&=\sum_n\mu_1(A_n\cap B)+\sum_n\mu_2(A_n\setminus B)\\ \ \\
&=\sum_n (\mu_1(A_n\cap B)+\mu_2(A_n\setminus B))\\ \ \\
&=\sum_n (\mu_1(A_n\cap B)+\mu_2(A_n\setminus (A_n\cap B)))\\ \ \\
&\leq\sum_n\mu(A_n). 
\end{align}
As $B$ was arbitrary, we obtain 
$$\tag1
\mu\left(\bigcup_nA_n\right)\leq\sum_m\mu(A_n). 
$$
For the reverse inequality, fix $\varepsilon>0$, and for each $n$ choose $B_n\subset A_n$ such that 
$$
\mu(A_n)\leq \mu_1(B_n)+\mu_2(A_n\setminus B_n)+\frac\varepsilon{2^n}. 
$$
Let $B=\bigcup_n B_n$. Note that all the $B_n$ are pairwise disjoint, since the $A_n$ are pairwise disjoint. 
Then
\begin{align}
\sum_n\mu(A_n)
&\leq \sum_n \left(\mu_1(B_n)+\mu_2(A_n\setminus B_n)+\frac\varepsilon{2^n} \right)\\ \ \\
&=\varepsilon+\sum_n (\mu_1(B_n)+\mu_2(A_n\setminus B_n))\\ \ \\
&=\varepsilon + \mu_1(B)+\sum_n (\mu_2(A_n\setminus B))\\ \ \\
&=\varepsilon + \mu_1(B)+\mu_2\left(\bigcup_n (A_n\setminus B)\right)\\ \ \\
&=\varepsilon + \mu_1(B)+\mu_2\left(\bigcup_n A_n\setminus B\right)\\ \ \\
&\leq\varepsilon + \mu\left(\bigcup_n A_n\right). 
\end{align}
As we can do this for all $\varepsilon>0$, the inequality holds without $\varepsilon$ and so 
$$\tag2
\sum_m\mu(A_n)\leq\mu\left(\bigcup_nA_n\right) 
$$
A: By the definition of $\operatorname{sup}$, for a sequence $\{\epsilon_n\}$, where $\epsilon_n > 0$, we may find a $B_n$ corresponding to $\epsilon_n$ and $A_n$ so that
$$\mu(A_n) < \mu_1(B_n) + \mu_2(A_n\setminus B_n) + \epsilon_n
< \mu(A_n) + \epsilon_n$$
Note that, given that $\{A_n\}$ are disjoint, hence so are $\{B_n\}$ and $\{A_n\setminus B_n\}$, so apply the fact that $\mu_1,\mu_2$ are measures:
\begin{align}
\sum_{n=1}^{\infty} \mu(A_n) &\leq
\sum_{n=1}^{\infty}\left(\mu_1(B_n) + \mu_2(A_n\setminus B_n) + \epsilon_n\right)\\
&= \mu_1\left(\bigcup_{n=1}^{\infty} B_n\right) + \mu_2\left(\bigcup_{n=1}^{\infty} (A_n\setminus B_n)\right)
 + \sum_{n=0}^{\infty}\epsilon_n \\ 
&= \mu_1\left(\bigcup_{n=1}^{\infty} B_n\right) + \mu_2\left(\bigcup_{n=1}^{\infty} A_n\setminus \bigcup_{n=1}^{\infty} B_n\right)
 + \sum_{n=0}^{\infty}\epsilon_n\\
&\leq \sup\left\{\,\mu_1(B) + \mu_2\left(\bigcup_{n=1}^{\infty}A_n \setminus B\right) \colon B \subset \bigcup_{n=1}^{\infty}A_n\,\right\}
 + \sum_{n=0}^{\infty}\epsilon_n\\
&= \mu\left(\bigcup_{n=0}^{\infty}A_n\right) +  \sum_{n=0}^{\infty}\epsilon_n\\
\end{align}
Now we can make the summation on the right as small as desired: Take $\epsilon > 0$ then if we initially choose $\{\epsilon_n\}$ so that $\sum_{n=0}^{\infty} \epsilon_n \leq \epsilon$, e.g., by taking $\epsilon_n=\frac{\epsilon}{2^{n}}$, then
$$\sum_{n=1}^{\infty} \mu(A_n) \leq \mu\left(\bigcup_{n=0}^{\infty}A_n\right) +  \epsilon $$
for any $\epsilon > 0$, hence
$$\sum_{n=1}^{\infty} \mu(A_n) \leq \mu\left(\bigcup_{n=0}^{\infty}A_n\right)$$
The other way around: take $A_1,A_2$ disjoint and fix $\epsilon > 0$, then there exists some $B \subset A_1\cup A_2$ so that
\begin{align}
\mu(A_1\cup A_2) &< \mu_1(B) + \mu_2((A_1\cup A_2) \setminus B) + \epsilon\\
&= \mu_1(B\cap A_1) + \mu_1(B\cap A_2) + \mu_2(A_1\setminus \underbrace{B}_{= B\cap A_1}) + \mu_2(A_2\setminus \underbrace{B}_{= B\cap A_2} ) + \epsilon\\
&\leq \mu(A_1) + \mu(A_2) + \epsilon
\end{align}
as this holds for all $\epsilon$,
$\mu(A_1\cup A_2) \leq \mu(A_1) + \mu(A_2)$ which inductively implies 
$$\mu\left(\bigcup_{n=0}^{\infty}A_n\right) \leq\sum_{n=1}^{\infty} \mu(A_n)$$
(Really we could have just used the fact that $\operatorname{sup}$ is sub-additive, but you would prove that using an $\epsilon$-argument as above anyway).
