# Infimum of two measures

Let $\Sigma$ be a $\sigma$-algebra on a set $X$. (Unsigned) measures $\mu,\nu:\Sigma\rightarrow[0,+\infty]$ can be ordered "elementwise" ($\mu \le \nu$ if $\mu(E) \le \nu(E)$ for each measurable set $E$). So given two such measures $\mu_1$, $\mu_2$, what would be their infimum (greatest lower bound) $\mu$ in this sense?

Given a measurable set $E\in\Sigma$, and another measurable set $A\in\Sigma$, $\mu(E)$ has to satisfy $$\mu(E) = \mu(E\cap A)+\mu(E\cap A^c) \le \mu_1(E\cap A)+\mu_2(E\cap A^c)$$ so $\mu(E)$ should not exceed the infimum of the right hand side, and it is not hard to show that this infimum defines a measure (use usual $\epsilon/2^n$ argument to show countable additivity). So the infimum of $\mu_1$ and $\mu_2$ is given by $$\mu(E) = \inf\{\mu_1(E\cap{A})+\mu_2(E\cap{A^c})\mid A\text{ measurable set}\}$$ Equivalently, one could limit the sets $A$ to be subsets of $E$.

Now my question: Can this infimum always be realized? Given $E\in\Sigma$, can we always find some measurable subset $A$ that achieves the infimum?

For example, if $\mu_1$, $\mu_2$ are given by density functions $f_1$, $f_2$ with respect to Lebesgue measure, $A$ can be chosen to be the set of points where $f_1(x) \le f_2(x)$. But I don't see how to show it is possible in general measure spaces. Is there a counterexample in the general case?

It does generalize, at least if $\mu_1$ and $\mu_2$ are $\sigma$-finite.

Let $\nu = \mu_1 + \mu_2$. Then $\nu$ is a measure and $\mu_1$ and $\mu_2$ are both absolutely continuous wrt $\nu$, so by Radon-Nikodym there are $f_1, f_2$ that are densities for $\mu_1$ and $\mu_2$ wrt $\nu$.
Then you can take $A = \{x: f_1(x) \le f_2(x)\}$.

EDIT: Yes, the infimum is attained. If $\mu(E) = \infty$, you must have $\mu_1(E) = \infty$ so you can take $A=E$. So suppose $\mu(E) = m < \infty$. For each positive integer $n$ there is some measurable $A_n \subseteq E$ such that $\mu_1(E \cap A_n) + \mu_2(E \cap A_n^c) < m + 1/n$. Let $B = \bigcap_n A_n$, $C = \bigcap_n A_n^c$, and $D = E \backslash (B \cup C)$. Then $\mu_1$ and $\mu_2$ are $\sigma$-finite on $D$, so the restrictions of $\mu_1$ and $\mu_2$ to $D$ have densities $f_1$, $f_2$ with respect to their sum, and you can take $A = B \cup \{x \in D: f_1(x) \le f_2(x)\}$.

However, one thing you can't do in general is find a set $A \in \Sigma$ that works for every set $E \in \Sigma$.

Let $\Sigma$ be a $\sigma$-algebra that contains all singletons, and $B$ a subset of $X$ that is not in $\Sigma$. Let $\mu_1$ and $\mu_2$ be counting measure on $B$ and $B^c$ respectively. Thus for any set $E \in \Sigma$, $\mu_1(E)$ and $\mu_2(E)$ are the numbers (possibly $\infty$) of points in $E \cap B$ and $E \cap B^c$ respectively. If $x$ is a single point we have $\mu_1(\{x\}) = 1$ and $\mu_2(\{x\}) = 0$ if $x \in B$, $\mu_1(\{x\}) = 0$ and $\mu_2(\{x\}) = 1$ otherwise. Thus the only possible $A$ that works for all $E$ is $B^c$, but this is not in $\Sigma$.

• Nice general argument. – PatrickR Jan 3 '15 at 0:40
• In the last sentence, I think you mean the only possible $A$ that works would be $B^c$. – PatrickR Jan 3 '15 at 0:41
• Thanks for catching that -- edited. – Robert Israel Jan 4 '15 at 2:16