A formula for the infimum of two measures In this question the OP claims that given measures $\mu_1,\mu_2$ on a measurable space $(E,\mathcal E)$ we can define their infimum w.r.t. to the ordering of measures by:
$$(\mu_1 \wedge \mu_2)(A) := \inf\{\mu_1(A\cap B)+\mu_2(A \cap B^c) : B\in \mathcal E\}$$
Unfornately I'm kind of weak in Analysis and I'm asking for a proof that this 
defines a measure. 
(Not the actual question: Does this have anything to do with the concept of outer measure / Caratheodory's extension theorem? Because it looks somewhat similar).
 A: The property $(\mu_1 \wedge \mu_2)(\varnothing) = 0$ follows immediately from $\mu_1(\varnothing) = \mu_2(\varnothing) = 0$.
It remains to show the $\sigma$-additivity of $\mu_1 \wedge \mu_2$ on $\mathcal{E}$. Thus let $(A_n)_{n\in\mathbb{N}}$ a disjoint sequence in $\mathcal{E}$ and
$$A = \bigcup_{n = 0}^\infty A_n.$$
For every $B\in \mathcal{E}$ we have
\begin{align}
\mu_1(A \cap B) + \mu_2(A \cap B^c) &= \sum_{n = 0}^\infty \mu_1(A_n \cap B) + \sum_{n = 0}^\infty \mu_2(A_n \cap B^c) \\
&= \sum_{n = 0}^\infty \bigl(\mu_1(A_n \cap B) + \mu_2(A \cap B^c)\bigr) \\
&\geqslant \sum_{n = 0}^\infty (\mu_1 \wedge \mu_2)(A_n),
\end{align}
and thus, taking the infimum over all $B$ on the left, we have the $\sigma$-superadditivity of $\mu_1 \wedge \mu_2$ on $\mathcal{E}$. Now if
$$\sum_{n = 0}^\infty (\mu_1 \wedge \mu_2)(A_n) = +\infty,$$
the equality
$$(\mu_1 \wedge \mu_2)(A) = \sum_{n = 0}^\infty (\mu_1 \wedge \mu_2)(A_n)$$
follows trivially from the $\sigma$-superadditivity. So we need only look at the case
$$\sum_{n = 0}^\infty (\mu_1 \wedge \mu_2)(A_n) < +\infty.$$
Then, for any given $\varepsilon > 0$, choose a sequence $(B_n)_{n\in \mathbb{N}}$ in $\mathcal{E}$ with $B_n \subset A_n$ and
$$\mu_1(A_n \cap B_n) + \mu_2(A_n \cap B_n^c) < (\mu_1 \wedge \mu_2)(A_n) + \frac{\varepsilon}{2^{n+1}}$$
for all $n\in \mathbb{N}$. Let
$$B = \bigcup_{n = 0}^\infty B_n.$$
Then
\begin{align}
\mu_1(A \cap B) + \mu_2(A \cap B^c) &= \sum_{n = 0}^\infty \mu_1(A_n \cap B) + \sum_{n = 0}^\infty \mu_2(A_n\cap B^c) \\
&= \sum_{n = 0}^\infty \mu_1(A_n \cap B_n) + \sum_{n = 0}^\infty \mu_2(A_n \cap B_n^c) \\
&= \sum_{n = 0}^\infty \bigl(\mu_1(A_n \cap B_n) + \mu_2(A_n \cap B_n^c)\bigr) \\
&< \sum_{n = 0}^\infty \biggl((\mu_1 \wedge \mu_2)(A_n) + \frac{\varepsilon}{2^{n+1}}\biggr) \\
&= \Biggl(\sum_{n = 0}^\infty (\mu_1 \wedge \mu_2)(A_n)\Biggr) + \varepsilon,
\end{align}
and hence
$$(\mu_1 \wedge \mu_2)(A) \leqslant \Biggl(\sum_{n = 0}^\infty (\mu_1 \wedge \mu_2)(A_n)\Biggr) + \varepsilon$$
for every $\varepsilon > 0$, so the $\sigma$-subadditivity
$$(\mu_1 \wedge \mu_2)(A) \leqslant \sum_{n = 0}^\infty (\mu_1 \wedge \mu_2)(A_n)$$
also follows in the non-trivial case where the sum on the right is finite. Altogether, the $\sigma$-additivity is established.
