# Prove $\mu$ is an outer measure, find the collection of all $\mu$-measurable sets and the necessary and sufficient conditions s.t. $\mu$ is a measure

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 $$$$\mu(X) = \begin{cases} \text{Card}(E), & \text{Card}(E) < \aleph_0\\ \infty, & \text{Card(E)} \ge \aleph _0 \end{cases}$$$$ a) Prove that $$\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.

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 $$\mu(E)=\mu(E\cap A)+\mu(E\setminus A)$$.
• 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? Commented Mar 1 at 7:50
• @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. Commented Mar 1 at 10:43