Note that to get a true distance, you have to restrict to measurable sets of finite measure, and to quotient over null sets. This way you get a closed subspace $M$ of $L^1(X)$ (the subset of binary functions). If the measure is atomless, then $M$ is even path-connected. A well-known result on atomless measures (essentially due to Sierpinski) is that for any measurable $A$, with measure $\alpha$, there exists a nested family of measurable sets $\{A_t\} _ {t\in[0,\alpha ]}$, such that $A_ 0 = \emptyset$,$A_\alpha=A$ and $m(A_t)=t$ for all $t$. On the other hand, if there is an atom $A\in M$, then $M$ is the union of its non-empty open sets $\{B\in M\\ : \\ B\supset A\}$ and $\{B\in M\\ : \\ B\cap A=\emptyset \}$, so $M$ is not connected.

Note that to get a true distance, you have to restrict to measurable sets of finite measure, and to quotient over null sets. This way you get a closed subspace $M$ of $L^1(X)$ (the subset of binary functions). If the measure is atomless, then $M$ is even path-connected. A well-known result on atomless measures (essentially due to Sierpinski) is that for any measurable $A$, with measure $\alpha$, there exists a nested family of measurable sets $\{A_t\} _ {t\in[0,\alpha ]}$, such that $A_ 0 = \emptyset$,$A_\alpha=A$ and $m(A_t)=t$ for all $t$. On the other hand, if there is an atom $A\in M$, then $M$ is the union of its non-empty open sets $\{B\in M\, : \, B\supset A\}$ and $\{B\in M\, : \, B\cap A=\emptyset \}$, so $M$ is not connected.

Note that to get a true distance, you have to restrict to measurable sets of finite measure, and to quotient over null sets. This way you get a closed subspace $M$ of $L^1(X)$ (the subset of binary functions). If the measure is non-atomic, then $M$ is even path-connected. A well-known result on non-atomic measures (essentially due to Sierpinski) is that for any measurable $A$, with measure $\alpha$, there exists a nested family of measurable sets $\{A_t\} _ {t\in[0,\alpha ]}$, such that $A_ 0 = \emptyset$,$A_\alpha=A$ and $m(A_t)=t$ for all $t$. On the other hand, if there is an atom $A\in M$, then $M$ is the union of its non-empty open sets $\{B\in M\\ : \\ B\supset A\}$ and $\{B\in M\\ : \\ B\cap A=\emptyset \}$, so $M$ is not connected.

Note that to get a true distance, you have to restrict to measurable sets of finite measure, and to quotient over null sets. This way you get a closed subspace $M$ of $L^1(X)$ (the subset of binary functions). If the measure is atomless, then $M$ is even path-connected. A well-known result on atomless measures (essentially due to Sierpinski) is that for any measurable $A$, with measure $\alpha$, there exists a nested family of measurable sets $\{A_t\} _ {t\in[0,\alpha ]}$, such that $A_ 0 = \emptyset$,$A_\alpha=A$ and $m(A_t)=t$ for all $t$. On the other hand, if there is an atom $A\in M$, then $M$ is the union of its non-empty open sets $\{B\in M\\ : \\ B\supset A\}$ and $\{B\in M\\ : \\ B\cap A=\emptyset \}$, so $M$ is not connected.