$\mu(A) \leq \mu^{*}(A)$ is outer measure bigger than the base measure?

Question

Let $\mu : \mathfrak{A} \to [0,+\infty]$ be a finitely additive measure with $\mathfrak{A}$ ring. Let us recall that $\mathfrak{A} \subseteq \mathbb{P}(X)$ is called ring if

(a) $\mathfrak{A} \neq \emptyset$;

(b) $A, B \in \mathfrak{A}$ implies $A \cup B, A \cap B, A \Delta B, A\setminus B \in \mathfrak{A}$.

Let us define $\mu^{*}\colon \mathbb{P}(X) \to [0,+\infty]$ as \begin{equation} \label{eq:1} \mu^{*}(A) := \inf \left\lbrace \sum_{i = 0}^{+\infty} \mu(A_i) \colon A \subseteq \cup_{i=0}^{+\infty}A_i\right\rbrace. \end{equation}

Is it true that for every $A \in \mathfrak{A}$ we have that $\mu(A) \leq \mu^{*}(A)$ ?

Answer 1

Actually, it's the other way around: $\mu^*(A)\le \mu(A)$. This is because $A$ itself covers $A$.

Specifically, let $A_0=A$ and $A_i=\emptyset$ for $i>0$. Then $A\subseteq\bigcup A_i$.

Now it's easy to see that there are two possibilities for $\mu(\emptyset)$, by finite additivity: either $0$ or $\infty$. If $\mu(\emptyset)=0$, then $\sum\mu(A_i)=\mu(A)$, so $\mu^*(A)\le\mu(A)$. Meanwhile, if $\mu(\emptyset)=\infty$, then $\mu(A)=\infty$ for all sets $A$.

---

For an example of a situation where we can have $\mu^*(A)<\mu(A)$, consider the ring $\mathfrak{A}$ of subsets of a countably infinite base set $X$ which are either finite or cofinite, and let $\mu(A)=\infty$ if $A$ is cofinite, and $0$ if $A$ is finite. Then $\mu$ is a finitely additive measure on $\mathfrak{A}$.

However, I claim that for any $A\subseteq X$, we have $\mu^*(A)=0$! This is because any such $A$ is a union of countably many finite sets ($A$ is the union of its singleton subsets, and since $A\subseteq X$ and $X$ is countable, there are at most countably many of these) and every finite set has $\mu$-measure $0$.

So, for instance, in this case we have $\mu(X)=\infty$ but $\mu^*(X)=0$.

Note that this example implies that the answer to your stated question is "no".