I am trying to solve the exercise below. I have managed to prove that $\mu$ is an outer measure following the definition of outer measures ($\mu(\emptyset)=0$, monotonicity, $\sigma$-subadditivity). However, I am struggling to find $\mathcal{M}^*$, the collection of all $\mu$-measurable sets. It is clear that $\emptyset, X \in \mathcal{M}^*$ but I can't find more sets $A$, which satisfy $ \forall E \subseteq X \; \; \mu(E) = \mu(E \cap A) +\mu(E \cap A^c)$. I also managed to prove that if an outer measure is finitely additive, it is a measure in general, but I can't find the sufficient and necessary conditions to prove that $\mu$ is a measure.

Exercise: If $X \ne \emptyset$ and $\mu$ is a set function defined on $\mathcal{P}(X)$ as \begin{equation} \mu(X) = \begin{cases} \text{Card}(E), & \text{Card}(E) < \aleph_0\\ \infty, & \text{Card(E)} \ge \aleph _0 \end{cases} \end{equation} a) Prove that $\mu$ is an outer measure and find $\mathcal{M}^*$.
b) Prove that if an outer measure is finitely additive, it is also a measure. Find the necessary and sufficient conditions under which the outer measure $\mu$ is also a measure.


1 Answer 1


All subsets $A\subseteq X$ are $\mu$-measurable. If $E\subseteq X$ then $E=(E\cap A)\sqcup(E\setminus A)$. The cardinality of a disjoint union is always the sum of cardinalities, whether they're finite or not $$ \operatorname{Card}(E)=\operatorname{Card}(E\cap A)+\operatorname{Card}(E\setminus A)\tag{1} $$

If $E$ is finite, then all numbers in $(1)$ are finite and you get the conclusion $\mu(E)=\mu(E\cap A)+\mu(E\setminus A)$.

If $E$ is infinite then $\mu(E)=+\infty$. Since an infinite set cannot be a union of two disjoint finite sets, either $E\cap A$ or $E\setminus A$ is an infinite set. Therefore either $\mu(E\cap A)=+\infty$ or $\mu(E\setminus A)=+\infty$ (or both). Either way $\mu(E)=\mu(E\cap A)+\mu(E\setminus A)$.

  • $\begingroup$ Thank you so much, this all makes sense! :) Do I understand correctly that according to this, $\mathcal{M}^* = \mathcal{P}(X)$, so $\mu$ outer measure is always a measure on $\mathcal{P}(X)$? If this is true, what I don't understand is why a) doesn't ask for a proof of $\mu$ being a measure, and why b) is asking for the necessary and sufficient conditions for something that is always true. Is my logic faulty somewhere or do you think this was simply a somewhat misleading question? $\endgroup$
    – wilma72
    Commented Mar 1 at 7:50
  • $\begingroup$ @wilma72: It's possible to prove directly that $\mu$ is $\sigma$-additive on $\mathcal{P}(X)$. I think the question just wanted to exercise the application of $\mu$-measurability on a relatively simple outer measure. $\endgroup$
    – Chad K
    Commented Mar 1 at 10:43

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .