Is there a strong topology on the set of probability measure? Let $\mathcal P$ the set of probability measure on $\mathbb R$. then $\mu_n\to \mu$ weakly if for all $f\in \mathcal C_b(\mathbb R)$ (bounded continuous), $$\lim_{n\to \infty }\int fd\mu_n=\int fd\mu.$$
I was wondering : is there a strong topology on $\mathcal P$ ?
 A: The idea to the following is from: https://en.wikipedia.org/wiki/Convergence_of_measures?wprov=sfla1
Let $(X, \mathcal{A})$ be a measurable space and let $\mathcal{P}$ denote the set of probability measures on $(X, \mathcal{A})$. Define the function $d \colon \mathcal{P} \times \mathcal{P} \to [0, \infty)$ given by $$d(\mu, \nu) = \sup\limits_{f \in \mathcal{F}} \bigg\{\int f d\mu - \int f d\nu\bigg\}$$ where $\mathcal{F}$ is the set of measurable funtions $f \colon X \to [-1,1]$. Note, that $d$ is well defined, since $d(\mu, \nu) \geq 0$, since the constant zero function is a measurable function from $X \to [-1,1]$. Moreover, $d$ is a metric. Indeed, let $\mu, \nu \in \mathcal{P}$ and note that $$d(\mu, \nu) = 0 \iff \forall f \in \mathcal{F} \colon \int f d\mu - \int f d\nu = 0 \implies \forall A \in \mathcal{A} \colon \mu(A)=\nu(A)$$ so $\mu=\nu$. Moreover we clearly have $d(\mu, \nu) = d(\nu, \mu)$ and in particular if $\sigma \in \mathcal{P}$ is another measure, then
\begin{eqnarray}
d(\mu, \nu) = \sup\limits_{f \in \mathcal{F}} \bigg\{\int f d\mu - \int f d\nu\bigg\} = \sup\limits_{f \in \mathcal{F}} \bigg\{\big(\int f d\mu - \int f d\sigma \big) + \big(\int f d\sigma - \int f d\nu \big)\bigg\} \\ \leq d(\mu, \sigma) + d(\sigma, \nu)
\end{eqnarray} 
So $d$ is a metric and thus $(\mathcal{P}, d)$ is a metric space. In particular the topology thus generated is stronger, than the one you mention in your question, since if $0 \neq f \in \mathcal{C}_b(\mathbb{R})$ (the case $f = 0$ is trivial) and $\{\mu_n\}_n \subset \mathcal{P}$ is a sequence such that $d(\mu_n, \mu) \to 0$ for some $\mu \in \mathcal{P}$, then $$ \bigg\{ \int f d\mu_n - \int f d\mu \bigg\} = \|f\|_\infty \bigg\{ \int \frac{f}{\|f\|_\infty} d\mu_n - \int \frac{f}{\|f\|_\infty} d\mu \bigg\} \leq \underbrace{\|f\|_\infty}_{< \infty} d(\mu, \mu_n) \to 0$$ since $\frac{f}{\|f\|_\infty} \in \mathcal{F}$. Thus $d(\mu, \mu_n) \to 0$ implies $$\int f d\mu_n \to \int f d\mu$$ for all $f \in \mathcal{C}_b(\mathbb{R})$ as wanted.
