## Is this metric space complete?

Let $(X,\mathfrak{M},\mu)$ be a positive finite measure space, then define $\rho(A,B)=\int_X |\chi_A-\chi_B|d\mu$.

Is $(\mathfrak{M},\rho)$ a complete metric space(modulo sets of measure 0)?

I am trying very hard to look for any references, but I cannot find any.

So, if $(\mathfrak{M},\rho)$ is a complete metric space, how can I prove that? Actually, the completeness of this metric space is my major problem.

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Given a Cauchy sequence $A_n$ in this space, the sequence $\chi_{A_n}$ is a Cauchy sequence in $L^1 (\mu)$. Now use the fact that $L^1 (\mu)$. is a complete metric space (it's Banach). – Mark Schwarzmann Mar 25 2011 at 14:41
Of course $\rho(A,B)$ is the measure of the symmetric difference, $\mu(A \Delta B)$, which is how it is also sometimes seen in textbooks. – Gerald Edgar Mar 25 2011 at 16:07
Y., you are just identifying $\mathfrak{M}$ with the closed (hence complete) subset of the Banach space $L^1(X,\mu)$ consisting of all characteristic functions, via $A\mapsto \chi_A$. – Pietro Majer Mar 25 2011 at 18:09

To complement Michael Renardy's response, a reference is: "1.12.6. Theorem" in Bogachev's Measure Theory, Volume 1. The proof goes as follows. Take any Cauchy sequence in the metric, then pass to a subsequence in which the mutual distances converge to zero very fast, then the original sequence converges in the metric to the set-theoretic limsup of the subsequence.

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 Thanks! it seems that this metric is well known. – Y. Fan Mar 25 2011 at 14:37

This is known as the Nikodym metric. It is complete.

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This is a just paraphrasing of my comment, but it may put things in a clearer perspective: $(\mathfrak{M},\rho)$ is embedded in $L^1 (\mu)$ as the subset $\{ \left[\chi_{A}\right]:A\in\mathfrak{M}\} \subset L^{1}\left(\mu\right)$ (where $\left[\chi_{A}\right]$ is the equivalence class of $\chi_{A}$ in $L^1 (\mu)$). So you just need to show that this is closed subset in $L^1 (\mu)$, which is not very difficult.

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