Proof that the subspace of signed measures integrating d(x,e) is closed Let $\mathcal{M}(S)$ be a space of finite signed measures on a metric space $S$ ($=\mathbb{R}^2$ in my case) equipped with the total variation norm. Let
$\mathcal{M}_1(S)=\{\mu \in \mathcal{M}(S):\int_S d(e,x) \mu(dx)<\infty \}$ where $d$ is a distance. It implicitly follows from this question and its answers that $\mathcal{M}_1$ is a Banach space. However, I could neither find nor work out a proof that it is closed. Probably it is trivial, but I do not see it. Could anyone help?
 A: Let's try this.  I think this is false, since the total variation has no connection with the distance $d$.  (That other question does not seem to mention the total variation norm at all.)
counterexample
Our metric space $S$ is $\mathbb R$ with the usual distance.  For $a \in S$, let $\epsilon_a$ be the unit point mass at $a$.  For $n \in \mathbb N$, define the measure
$$
\mu_n = \sum_{k=1}^n \frac{1}{2^k}\;\epsilon_{2^k}
$$
and
$$
\mu = \sum_{k=1}^\infty \frac{1}{2^k}\;\epsilon_{2^k}
$$
These are in $\mathcal M(S)$ with total variation at most $1$.
Now compute total variation norm
$$
\|\mu-\mu_n\|_{\mathrm{TV}} = \sum_{k=n+1}^\infty \frac{1}{2^k} = \frac{1}{2^n}
$$
so $\mu_n$ converges to $\mu$ in total variation norm.  Finally, take $e=0$ and compute
$$
\int_S d(e,x)\;d\mu_n(x) = \int_{\mathbb R} |x|\;d\mu_n(x) 
= \sum_{k=1}^n \frac{1}{2^k}\left|2^k\right| = n < +\infty
$$
so $\mu_n \in \mathcal M_1(S)$, but
$$
\int_S d(e,x)\;d\mu(x) = \int_{\mathbb R} |x|\;d\mu(x) 
= \sum_{k=1}^\infty \frac{1}{2^k}\left|2^k\right| =+\infty
$$
and $\mu \not\in \mathcal M_1(S)$.
