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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.

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

1 Answer 1

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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 $$ \lambda(\{x_{1}\}) + \lambda(\{x_{2}\})? $$

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