# Minimum of two measures is no measure

Let $(X,\mathcal A, \mu ), (X, \mathcal A, \nu)$ be two measure spaces. Show $\lambda(A)=\min(\mu(A), \nu(A))$ is in general no measure on $(X, \mathcal A)$

It may be an easy question but I am really going crazy as I can't find a counterexample. I tried combinations of the trivial measure and counting measure but never got the desired results. Please release me from this pain.

• For whatever reason I get the impression that subadditivity may fail in some cases, but I don't have any great reason why yet. Feb 6, 2017 at 21:56
• Let your space have two points, and let $\mu$ vanish on one and $\lambda$ vanish on the other. That should fail additivity I think. Feb 6, 2017 at 21:57

Let's use $$X = \{x_{1}, x_{2}\}$$, i.e., a set consisting of just two points, $${\cal A}$$ = the set of all (four:) subsets of $$X$$, and the measures defined by the following: $$\mu(\{x_{1}\}) = 1, \quad \mu(\{x_{2}\}) = 0,$$ $$\nu(\{x_{1}\}) = 0, \quad \nu(\{x_{2}\}) = 1.$$
Now, check whether additivity holds: does $$\lambda(\{x_{1}, x_{2}\})$$ equal $$\lambda(\{x_{1}\}) + \lambda(\{x_{2}\})?$$