The bounded Lipschitz metric ($d_{BL}$) metrizes the weak convergence of probability measures on $\mathbb{R}$ with respect to bounded continuous test functions $C^0_b(\mathbb{R})$ $$d(\mu, \nu) = \sup_{f \in \text{Lip}(\mathbb{R})} \Big | \int_{\mathbb{R}} f d \nu - \int_{\mathbb{R}} f d \mu \Big |$$ where $$\text{Lip}(\mathbb{R}) = \Big \{ f \in C_b(\mathbb{R}) : \sup_x |f(x) | \leq 1, \sup_{x \neq y} \frac{| f(x) - f(y) |}{|x-y|} \leq 1 \Big \}.$$


(1) Does bounded Lipschitz metric also do the same (i.e. metrize) the space of non-negative finite measures ($\mathcal{M}^+(\mathbb{R})$)?

(2) In addition to being a metric does it also define a norm such that $(\mathcal{M}^+(\mathbb{R}),d_{BL})$ is a complete normed space?

Note: As far as I know, $d_{BL}$ induces a norm on the space of finite signed measures ($\mathcal{M}(\mathbb{R})$), however this space is not complete.

Motivation: I need the space of $\mathcal{M}^+(\mathbb{R})$ to be a Banach space so that I can apply classical Parabolic compactness technique by Jacques Simon (1987), which requires the underlying spaces to be Banach spaces, to a problem living on the space of measures.

  • $\begingroup$ For the first question, notice that for a sequence of non-negative measures, convergence in law is the same as $\int fd\mu_n\to \int f\mu$ for all $f$ continuous with compact support and $\mu_n(\mathbb R)\to \mu(\mathbb R)$. For the second question, are you sure that $\mathcal M^+$ is a vector space? $\endgroup$ Commented Aug 19, 2013 at 21:22
  • $\begingroup$ Thanks @DavideGiraudo for the prompt reply. So $\mathcal{M}^+$ is not a vector space over $(\mathbb{R},+)$. However I am not too sure of it being one over the space $(\mathbb{R}^+,\cdot)$. $\endgroup$
    – UPS
    Commented Aug 20, 2013 at 9:02
  • $\begingroup$ You can consider instead the vector space of finite signed measures. $\endgroup$ Commented Aug 20, 2013 at 9:10
  • $\begingroup$ The vector space of finite signed measures with the bounded Lipschitz metric is not complete (as far as I know, though I am still looking for a counter example), hence not a Banach space, which is what I need. $\endgroup$
    – UPS
    Commented Aug 20, 2013 at 9:34

1 Answer 1


The answer is only partially YES. However $\mathcal{M}^+(\mathbb{\mathbb R})$ obviously cannot be a vector space due to the positivity constraint. So this rules out both questions as currently written. What is true, though, is that the metric space $(\mathcal{M}^+(\mathbb{R}),d_{BL})$ is complete and metrizes the weak convergence. I will only prove rigorously the completeness, see my final remark for how to get the "metrization". The proof below actually works in any dimension, and also in any domain $\Omega\subset \mathbb R^d$.

Let $\{\mu_n\}$ be a sequence of positive measures, let me denote the mass $m_n:=\mu_n(\mathbb R)\geq 0$, and assume that the sequence is Cauchy $$ d_{BL}(\mu_p,\mu_q)\to 0 \qquad \mbox{as }p,q\to\infty. $$

  1. As a first step, it is easy to see that $\{m_n\}$ is a (real) Cauchy sequence: for, testing $ f \equiv 1$ in the definition of $d_{BL}$, we get $$ |m_p-m_q|=\left|\int 1 d\mu_p -\int 1 d\mu_q \right|\leq d_{BL}(\mu_p,\mu_q). $$ Since the real line is complete, there is $m\geq 0$ such that $m_n\to m$. If $m=0$ then it is immediate to see that, for any $f\in \mathcal C_b$, there holds $|\int f d \mu_n|\leq ||f||_\infty m_n\to 0$, which proves that $\mu_n\to 0$ weakly (narrowly).

  2. If $m>0$ then we can assume that $m/2\leq m_n\leq 2m$ for $n$ large enough, and the renormalized sequence $\tilde \mu_n:=\frac {\mu_n}{m_n}\in\mathcal P(\mathbb R)$ is well-defined. I claim that $\{\tilde\mu_n\}$ is $d_{BL}$-Cauchy as well. Indeed, for $p,q$ large enough we have by triangular inequality \begin{multline*} \left| \int f d\tilde\mu_p -\int f d\tilde\mu_q\right| = \left| \int f \frac{1}{m_p}d\mu_p -\int f \frac{1}{m_q}d\mu_q \right| \\ \leq \frac 1m \left| \int f d\mu_p -\int f d\mu_q \right| \\ + \left|\left(\frac 1{m_p}-\frac 1m \right)\int f d\mu_p\right| + \left|\left(\frac 1{m_q}-\frac 1m \right)\int f d\mu_q\right| \\ \leq \frac 1m \left| \int f d\mu_p -\int f d\mu_q \right| \\ + \left|\frac 1{m_p}-\frac 1m \right| \|f\|_\infty 2m + \left|\frac 1{m_q}-\frac 1m \right| \|f\|_\infty 2m. \end{multline*} Taking the supremum over $f$ such that $\|f\|,Lip(f)\leq 1$ gives $$ d_{BL}(\tilde\mu_p,\tilde\mu_q)\leq \frac 1md_{BL}(\mu_p,\mu_q) + 2m\left|\frac 1{m_p}-\frac 1m \right| +2m\left|\frac 1{m_q}-\frac 1m \right| $$ and entails my claim.

  3. Since $(\mathcal P(\mathbb R),d_{BL})$ is complete there is a proabability measure $\tilde \mu\in \mathcal P(\mathbb R)$ such that $d_{BL}(\tilde\mu_n,\tilde \mu)\to 0$. Because we already proved that $m_n\to m$, it is then easy to check that $ \mu_n=m_n\tilde\mu_n$ converges (in the Bounded-Lipschitz distance) to the limit $\mu:=m\tilde\mu$. Indeed for fixed $f$ \begin{multline*} \left|\int f d\mu_n- \int f d\mu \right| =\left|m_n\int f d\tilde\mu_n- m\int f d\tilde\mu \right| \\ \leq |m_n-m|\cdot \left|\int f d\tilde \mu_n\right| + m\left|\int f d\tilde\mu_n-\int f d\tilde\mu \right| \\ \leq |m_n-m|\cdot\|f\|_\infty+ m\left|\int f d\tilde\mu_n-\int f d\tilde\mu \right|. \end{multline*} Taking one last time the supremum over $f$'s gives $d_{BL}(\mu_n,\mu)\leq |m_n-m| + md_{BL}(\tilde\mu_n,\tilde\mu)\to 0$ and the proof is complete.

Final remark: following the same lines it is easy to see that $d_{BL}$ does indeed metrize the weak convergence. The strategy of proof is identical: show that the masses converge, use this to suitably renormalize $\tilde\mu_n:=\frac{1}{m_n}\mu$, and exploit that the statement is already known for probability measures. (The case of vanishing mass $m_n\to 0$ must be treated separately.)


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