Let $(X,M, \mu)$ a finite measure space and define a metric space $M'$ as follows: for $A,B \in M$, define: $d(A,B) = \mu(A\triangle B)$.
The space $M'$ is defined as all sets in $M$ where sets $A$ and $B$ are identified if $\mu(A\triangle B) = 0$.
I yet proved that $(M', d)$ is complete metric space.
Question: Show that the space above is not compact when $X = [0, 1]$, $M$ is the family of Borel sets on $[0, 1]$, and $\mu$ is Lebesgue measure.
I know that for any metric space $(X,d)$, $(X,d)$ is compact iff is complete and totally bounded. By the above result the metric space of question is complete then i should show that is not totally bounded. But i have not idea how proceed. This is the best way for to prove? I know also that i could to find a sequence that have not subsequence that converge in $M$, but i have not found. Any tip for the sequence?