# Total variation distance is complete

For a given measurable space $X$, $\mathcal{P}(X)$ denotes the space of all the probability measures on $X$. The total variation distance $\rho$ on $\mathcal{P}(X)$ is defined by: for $\mu, \nu \in \mathcal{P}(X)$ $$\rho(\mu, \nu) = \sup_{f: \|f\| \leq 1}\int_X f(x)(\mu - \nu)(dx)$$ where $\|f\| := \sup_{x\in X}|f(x)|$.

I read that this total variation distance is complete, but I can't find a proof. Could someone help to give a proof or a reference? I am also wondering if this is true for general $X$ or there should be some restrictions.

Consider $(\mu_n)_{n\geqslant 1}$ a Cauchy sequence for the metric $\rho$. Then for each $f$ (measurable) and bounded by $1$, the sequence $\left( \int_X f(x)\mathrm d\mu_n(x)\right)_{n\geqslant 1}$ is Cauchy. In particular, for each measurable subset $A$ of $X$, the sequence $(\mu_n(A))_{n\geqslant 1}$ is convergent. By the Vitali–Hahn–Saks theorem, we obtain a measure $\mu$ such that $\mu_n(A)=\mu(A)$ for each measurable subset $A$.
It remains to check that $\rho(\mu_n,\mu)$ converges to $0$. Approximating uniformly a bounded measurable by a linear combination of characteristic functions, we can see that the convergence $$\lim_{n\to\infty}\int_Xf(x)\mathrm d\mu_n(x)= \int_Xf(x)\mathrm d\mu(x)$$ takes place for any such $f$.
Fix a positive $\varepsilon$. Then there exists an integer $N$ such that if $m,n\gt N$ and $f$ is a measurable function bounded by $1$, then $$\tag{*} \left|\int_Xf(x)\mathrm d\mu_m(x)-\int_Xf(x)\mathrm d\mu_n(x)\right|\lt \epsilon.$$ Letting $n\to \infty$ in (*), we obtain that $$\left|\int_Xf(x)\mathrm d\mu_m(x)-\int_Xf(x)\mathrm d\mu(x)\right|\leqslant \epsilon$$ for each function $f$ bounded by $1$. Now take the supremum over such $f$.
• @DavideGiraudo Thank you! so the conclusion is that the total variation distance is complete for general $X$ – Petite Etincelle May 27 '15 at 8:01