Bartle exercise 9.N from elements of integration The exercise states:
Let $X$ be a set, $\mathbf{A}$ an algebra of subsets of $X$, and $\mu$ a measure on $\mathbf{A}$. If $B\subset X$ is arbitrary, let
$$\mu'(B)=\inf\{\mu(A):B\subset A\in\mathbf{A}\}$$
and
$$\mu^*(B)=\inf\{\sum_{n=1}^\infty \mu(I_n):B\subset \bigcup_{n=1}^\infty I_n,\{I_n\}_{n\in \mathbb{N}}\subset \mathbf{A}\}$$
Show that $\mu'(E)=\mu(E)$ for all $E\in \mathbf{A} $ and that $\mu^*(B)\leq \mu'(B)$. Moreover, $\mu^*=\mu'$ in case $X$ is the countable union of sets with finite $\mu$-measure.
My question: I already proved all but the "Moreover, $\mu^*=\mu'$ in case $X$ is the countable union of sets with finite $\mu$-measure". To prove it one must show that in that case
$$\mu^*(B)\geq\mu'(B)$$
for all $B\subset X$. Any hints or suggestions?
 A: This is not the case. Let $X=\mathbb R$, let $\mathbf A$ be the collection of finite unions of sets of the form $(-\infty,b]$, $(a,b]$, or $(a,\infty)$, and let $\mu$ be the Lebesgue measure. I will show $\mu'(\mathbb Q)=\infty$ but $\mu^*(\mathbb Q)=0$, which provides a counterexample. (A countable collection of finite $\mu$-measure elements of $\mathbf A$ whose union is $X$ is $\{(n,n+1]\}_{n\in\mathbb Z}$, for example.)
Let $A\in \mathbf A$ satisfy $\mathbb Q\subset A$. Such $A$ must be of the form $A=\bigcup_{j=1}^n(a_j,b_j]$, where
$$a_1<b_1<a_2<\ldots<b_{n-1}<a_n<b_n$$
($a_n=-\infty$ and $b_n=\infty$ are permitted; in the latter case, $(a_n,b_n]$ is replaced by $(a_n,\infty)$). If $a_1>-\infty$, there exists an integer $N<a_1$, but then $N\in \mathbb Q\setminus A$, a contradiction. Since $a_1=-\infty$, $\mu(A)\ge\mu((-\infty,a_1])=\infty$. Since this is true for all such $A$, we must have $\mu'(\mathbb Q)=\infty$.
Now fix $\varepsilon>0$, let $\{x_n\}$ be an enumeration of $\mathbb Q$, and let $I_n=(x_n-2^{-n}\varepsilon,x_n+2^{-n}\varepsilon]$. Clearly, $I_n\in\mathbf A$ and $\mathbb Q\subset\bigcup_nI_n$. Moreover, $\mu(I_n)=2^{-(n-1)}\varepsilon$, and so
$$\mu^*(\mathbb Q) \le \sum_{n=1}^\infty \mu(I_n) = 2\varepsilon.$$
It follows that $\mu^*(\mathbb Q)=0$, completing the proof.
