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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.

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. Nevertheless, the forward implication holds.

First, let me explain the forward implication. Suppose that $S$ is a $\sigma$-algebra generated by a countable subfamily $S_0$ 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 of finite Boolean combinations of sets in $S_0$. This is a countable family, and I claim it is dense in the semi-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 are approximable by sets in $S_1$, in the sense that for any $r\gt 0$ there is $B\in S_1$ such that $d(A,B)\lt r$. Note that $S_2$ contains $S_1$ and is closed under complement since the measure was finite. I claim it is also closed under countable unions: if each $A_n$ is approximable by $B_n$ to within $r/2^n$, then $\cup_n A_n$ is approximated by $\cup_n B_n$ to within $r$, and so one may find an approximating finite union. So $S_2$ is actually a $\sigma$-algebra, and since it contains $S_0$, it follows that $S_2=S$. That is, every set in $S$ is approximable by sets in $S_1$, and so $S_1$ is a countable dense set in the semi-metric, as desired.

Let's turn now to the reverse implication, which is not generally true. The easiest counterexample for this in the strict sense of the question is to let $X$ be a set of size continuum 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 on cardinality grounds. 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 a large cardinality set of measure $0$ and taking the completion. Similar counterexamples can be obtained by adding such large cardinality 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.

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 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.

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.

The two notions are not equivalent in 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 generated, 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 when there is also a measure $\mu$ on the $\sigma$-algebra, and the distance between two sets in the $\sigma$-algebra is defined to be the measure of their symmetric difference $d(A,B)=\mu(A\triangle B)$.

For a counterexample to the strict interpretation of your question, let $X$ be any uncountable set and $S$ the full power set of $X$. This is not countably generated as a $\sigma$-algebra. But fix any $p\in X$ and let $\mu$ be the measure placing mass $1$ at $p$, with no measure outside of {p}. In this case, the family {emptyset, {p}} is dense in your semi-metric, since every subset is essentially empty or {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 the same idea: add an uncountable set of measure $0$ and use the completion of the $\sigma$-algebra.)

For the positive result, note that the counterexample $\sigma$ algebra was the completion of the finite Boolean algebra generated by {p}. So you should really be speaking of the $\sigma$-algebra being (contained in) the completion of a countably generated $\sigma$-algebra. And with this version of the question, you get the desired equivalence.

To prove the desired equivalence, suppose that a (finite) measure is defined on a $\sigma$-algebra $S$ that is generated by a countable subfamily $A_0$, $A_1$, $A_2$, and so on. Every set $A\in S$ is obtained by a union of the form $A=\bigcup_{n\in I} \pm A_n$, where $+A=A$ and $-A$ is the complement of $A$ and $I$ is any set of natural numbers. Let $B_k=\bigcup_{n\in I, n\lt k} \pm A_n$ be the finite approximations, so that $A=\bigcup_k B_{k+1}-B_k$, which is a disjoint union. Thus, the measure of $B_{k+1}-B_k$ must tend to $0$, and so we can find some $B_k$ as close to $A$ as desired. Thus, the finite Boolean combinations of the $A_n$ are dense in your semi-metric space.

Conversely, suppose that some countable family $S_0$ is dense in your semi-metric space. Let $S$ be the $\sigma$-algebra generated by these sets. For any $A\in S$, we may find $A_n\subset A$ with $A_n\in S_0$ and $d(A,A_n)\lt \frac1n$. Thus, $\bigcup_n A_n$ is in $S$, and differs from $A$ on a set of measure $0$. So $A$ is in the completion of $S$, as desired.

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 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.