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).


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

$$B = \bigcup_{n = 0}^\infty B_n.$$


\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.

  • $\begingroup$ 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
  • 1
    $\begingroup$ @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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.