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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).

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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.

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  • $\begingroup$ Why can you choose $B_n \subset A_n$? And how do you get from $\mu_2(A_n \cap B^c)$ to $\mu_2(A_n \cap B_n^c)$? $\endgroup$ – Dominik Nov 10 '16 at 8:30
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    $\begingroup$ @Dominik If we have an arbitrary $B \in \mathcal{E}$ and set $B_n = A_n \cap B$, then we have $A_n \cap B = A_n \cap (A_n \cap B) = A_n \cap B_n$ and $A_n \cap B^c = A_n \cap (A_n^c \cup B^c) = A_n \cap (A_n \cap B)^c = A_n \cap B_n^c$. $\endgroup$ – Daniel Fischer Nov 10 '16 at 10:08

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