6 Added proof of forward direction
generated $\sigma$-algebra. Nevertheless, the forwardimplication holds.

First, let me explain the forward implication. Suppose that$S$ is a $\sigma$-algebra generated by a countablesubfamily $S_0$ and $\mu$ is a finite measure defined on$S$. The semi-metric on $S$ is defined by$d(A,B)=\mu(A\triangle B)$. Let $S_1$ be the collection offinite Boolean combinations of sets in $S_0$. This is acountable family, and I claim it is dense in thesemi-metric. To see this, let $S_2$ be the closure of $S_1$in the semi-metric, that is, the sets $A\in S$ that areapproximable by sets in $S_1$, in the sense that for any$r\gt 0$ there is $B\in S_1$ such that $d(A,B)\lt r$. Notethat $S_2$ contains $S_1$ and is closed under complementsince the measure was finite. I claim it is also closedunder countable unions: if each $A_n$ is approximable by$B_n$ to within $r/2^n$, then $\cup_n A_n$ is approximatedby $\cup_n B_n$ to within $r$, and so one may find anapproximating finite union. So $S_2$ is actually a$\sigma$-algebra, and since it contains $S_0$, it followsthat $S_2=S$. That is, every set in $S$ is approximable bysets in $S_1$, and so $S_1$ is a countable dense set in thesemi-metric, as desired.

Let's turn now to the reverse implication, which is notgenerally true. The easiest counterexample for this in thestrict sense of the question asked is to let $X$ be an uncountable a set of sizecountably generated on cardinality grounds. Fix any $p\inuncountable a largelarge cardinality set of measure$0$to any space and 5 added 18 characters in body The two notions are not equivalent. Indeed, they are not equivalent even when one considers completing the measure by adding all null sets with respect to any countably generated$\sigma$-algebra. The easiest counterexample for the strict question asked is to let$X$be an uncountable set and$S=P(X)$, the full power set of$X$. This is a$\sigma$-algebra, but it is easily seen not to be countably generated. Fix any$p\in X$and let$\mu$be the measure placing mass$1$at$p$and 0 mass outside {p}. In this case, the family {emptyset, X} is dense in the semi-metric, since every subset is essentially empty or all of$X$, depending on whether it contains$p$. So the semi-metric is separable, but the$\sigma$-algebra is not countably generated. Note that in this counterexample, the$\sigma$algebra is obtained from the counting measure on {p} by adding an uncountable set of measure$0$and taking the completion. Similar counterexamples can be obtained by adding an uncountable set of measure$0$to any space and taking the completion. At first, I thought incorrectly that one could address the issue by considering the completion of the measure, and showing that the$\sigma$-algebra would be contained within the completion of a countably generated$\sigma$-algebra. But I now realize that this is incorrect, and I can provide a counterexample even to this form of the equivalence. To see this, consider the filter$F$of all sets$A\subset \omega_1$that contain a closed unbounded set of countable ordinals. This is known as the club filter, and it is closed under countable intersection. The corresponding ideal$NS$consists of the non-stationary sets, those that omit a club, and these are closed under countable union. It follows that the collection$S=F\cup NS$, which are the sets measured by a club set, forms a$\sigma$-algebra. The natural measure$\mu$on$S$gives every set in$F$measure$1$and every set in$NS$measure$0$. This is a countably additive 2-valued measure on$S$. Note that every set in$S$has measure$0$or$1$; in particular, there are no disjoint positive measure sets. It follows that the family {emptyset,$\omega_1$} is dense in the semi-metric, since every set in$S$either contains or omits a club set, and hence either agrees with emptyset or with the whole set on a club. Thus, the semi-metric is separable. But for any countable subfamily$S_0\subset S$, we may intersect the clubs used to decide the members of$S_0$, and find a single club set$C\subset\omega_1$that decides every member of$S_0$, in the sense that every member of$S_0$either contains or omits$C$. This feature is preserved under complements, countable unions and intersections, and therefore$C$decides every member of the$\sigma$-algebra generated by$S_0$. The completion of the measure on the$\sigma$-algebra generated by$S_0$is therefore contained within the principal filter generated by$C$together with its dual ideal. This is not all of$S$, since there are club sets properly contained within$C$, such as the set of limit points of$C$. Thus, this is a measure space that has a separable semi-metric, but the$\sigma$-algebra is not contained in the completion of any countably generated$\sigma$-algebra. 4 added 562 characters in body; added 59 characters in body The two notions are not equivalentin the generality that is stated, but one gets an equivalence by making a slight change. The change is that rather than saying that the$\sigma$-algebra is countably generatedIndeed, what you get will be that it is contained within the completion of a countably generated$\sigma$-algebra. Note that the second definition you cite is only defined they are notequivalent even when there is also a measure$\mu$on the$\sigma$-algebra, and one considers completing the distance between two measureby adding all null sets in the$\sigma$-algebra is defined with respect to be the measure of their symmetric difference any countably generated$d(A,B)=\mu(A\triangle B)$. For a \sigma$-algebra.

The easiest counterexample to for the strict interpretation of your question , asked isto let $X$ be any an uncountable set and $S$ S=P(X)$, thefull power set of$X$. This is a$\sigma$-algebra, but itis easily seen not to be countably generatedas a$\sigma$-algebra. But fix . Fix any$p\inX$and let$\mu$be the measure placing mass$1$at$p$, with no measure p$ and0 mass outside of {p}. In this case, the family {emptyset, {p}} X}is dense in your the semi-metric, since every subset isessentially empty or {p}. all of $X$, depending on whether itcontains $p$. So the semi-metric is separable, but the$\sigma$-algebra is not countably generated.QED (More interesting examples can easily be constructed without point masses using

Note that in this counterexample, the same idea: add $\sigma$ algebra isobtained from the counting measure on {p} by adding anuncountable set of measure $0$ and use taking the completion.Similar counterexamples can be obtained by adding anuncountable set of the measure $\sigma$-algebra.)

For 0$to any space and taking thepositive resultcompletion. At first, note I thought incorrectly that one could address the counterexample$\sigma$algebra was issue by considering the completion of the finite Boolean algebra generated by {p}. So you should really be speaking of measure, and showing that the$\sigma$-algebra being (would be contained in) the completion of within a countably generated$\sigma$-algebra. And with But I now realize that this version of the questionis incorrect, you get and I can provide a counterexample even to this form of the desired equivalence. To prove the desired equivalencesee this, suppose that a (finite) measure is defined on a consider the filter$\sigma$-algebra F$ of all sets $S$ A\subset \omega_1$that is generated by contain a closedunbounded set of countable subfamily$A_0$,$A_1$,$A_2$, ordinals. This is known as theclub filter, and so onit is closed under countableintersection. Every set The corresponding ideal$A\in S$is obtained by NS$ consists of thenon-stationary sets, those that omit a club, and theseare closed under countable unionof . It follows that theform collection $A=\bigcup_{n\in I} \pm A_n$S=F\cup NS$, where which are the sets measured by aclub set, forms a$+A=A$and \sigma$-algebra. The natural measure$-A$ is the complement of \mu$on$A$S$ gives every set in $F$ measure $1$ and everyset in $I$ NS$measure$0$. This is any a countably additive2-valued measure on$S$. Note that every set of natural numbers. Let in$B_k=\bigcup_{n\in IS$hasmeasure$0$or$1$; in particular, n\lt k} \pm A_n$ be there are no disjointpositive measure sets. It follows that the finite approximationsfamily{emptyset,$\omega_1$} is dense in the semi-metric, so that sinceevery set in $A=\bigcup_k B_{k+1}-B_k$S$either contains or omits a club set, which is andhence either agrees with emptyset or with the whole set ona disjoint unionclub. Thus, the measure of semi-metric is separable. But for anycountable subfamily$B_{k+1}-B_k$must tend S_0\subset S$, we may intersect the clubsused to decide the members of $0$, S_0$, and so we can find some$B_k$as close to a single clubset$A$as desired. Thus, the finite Boolean combinations C\subset\omega_1$ that decides every member of the $A_n$ are dense S_0$, inyour semi-metric space. Conversely, suppose the sense that some countable family every member of$S_0$either contains oromits$C$. This feature is dense in your semi-metric space. Let preserved under complements,countable unions and intersections, and therefore$S$be C$decides every member of the $\sigma$-algebra generated bythese sets. For any $A\in S$, we may find S_0$. The completion of the measure on the$A_n\subset A$with \sigma$-algebra generated by $A_n\in S_0$ and $d(A,A_n)\lt \frac1n$. Thus, is therefore containedwithin the principal filter generated by $\bigcup_n A_n$ C$together withits dual ideal. This is in not all of$S$, and differs from since there areclub sets properly contained within$A$on a C$, such as the set ofmeasure limit points of $0$. So C$. Thus, this is a measure space that has a separablesemi-metric, but the$A$\sigma$-algebra is not contained inthe completion of any countably generated $S$, as desired.\sigma\$-algebra.