Skip to main content

Compactness with respect to topology induced by TVtotal-variation distance

I've been working on a problem and at some point in the proof I need to show that the following set $$\left\{\mu \in \mathcal{P}_{ac}(\mathbb R^d): \int \varphi(x)\mu(\mathrm{d}x)\leq C\right\}$$ is a compact with respect to the topology induced by TVtotal-variation distance on $\mathcal{P}_{ac}(\mathbb R^d).$ Here, $\mathcal{P}_{ac}(\mathbb R^d)$ is the set of probability measures absolutely continuous with respect to Lebesgue measure on $\mathbb R^d$, $\varphi$ is a real-valued bounded measurable function on $\mathbb R^d$ and $C > 0$ is a constant.

What I've thought about so far: Prokhorov's theorem is not helpful here since it only gives compactness with respect to the weak convergence of measures with is a weaker topology than the one induced by TVtotal-variation distance. I know that for Wasserstein spaces there is this result: for any $p'> p \geq 1,$ the set $$\left\{\mu \in \mathcal{P}_{p}(\mathbb R^d): \int |x|^p\mu(\mathrm{d}x)\leq C\right\}$$ is $W_p$-compact, $W_p$ is known to metrize weak convergence on $\mathcal{P}_{p}(\mathbb R^d).$

I would greatly appreciate if someone could point out a reference for what I'm trying to prove or to give me some ideas how to prove it. Thank you!

Compactness with respect to topology induced by TV distance

I've been working on a problem and at some point in the proof I need to show that the following set $$\left\{\mu \in \mathcal{P}_{ac}(\mathbb R^d): \int \varphi(x)\mu(\mathrm{d}x)\leq C\right\}$$ is a compact with respect to the topology induced by TV distance on $\mathcal{P}_{ac}(\mathbb R^d).$ Here, $\mathcal{P}_{ac}(\mathbb R^d)$ is the set of probability measures absolutely continuous with respect to Lebesgue measure on $\mathbb R^d$, $\varphi$ is a real-valued bounded measurable function on $\mathbb R^d$ and $C > 0$ is a constant.

What I've thought about so far: Prokhorov's theorem is not helpful here since it only gives compactness with respect to the weak convergence of measures with is a weaker topology than the one induced by TV distance. I know that for Wasserstein spaces there is this result: for any $p'> p \geq 1,$ the set $$\left\{\mu \in \mathcal{P}_{p}(\mathbb R^d): \int |x|^p\mu(\mathrm{d}x)\leq C\right\}$$ is $W_p$-compact, $W_p$ is known to metrize weak convergence on $\mathcal{P}_{p}(\mathbb R^d).$

I would greatly appreciate if someone could point out a reference for what I'm trying to prove or to give me some ideas how to prove it. Thank you!

Compactness with respect to topology induced by total-variation distance

I've been working on a problem and at some point in the proof I need to show that the following set $$\left\{\mu \in \mathcal{P}_{ac}(\mathbb R^d): \int \varphi(x)\mu(\mathrm{d}x)\leq C\right\}$$ is compact with respect to the topology induced by total-variation distance on $\mathcal{P}_{ac}(\mathbb R^d).$ Here, $\mathcal{P}_{ac}(\mathbb R^d)$ is the set of probability measures absolutely continuous with respect to Lebesgue measure on $\mathbb R^d$, $\varphi$ is a real-valued bounded measurable function on $\mathbb R^d$ and $C > 0$ is a constant.

What I've thought about so far: Prokhorov's theorem is not helpful here since it only gives compactness with respect to the weak convergence of measures with is a weaker topology than the one induced by total-variation distance. I know that for Wasserstein spaces there is this result: for any $p'> p \geq 1,$ the set $$\left\{\mu \in \mathcal{P}_{p}(\mathbb R^d): \int |x|^p\mu(\mathrm{d}x)\leq C\right\}$$ is $W_p$-compact, $W_p$ is known to metrize weak convergence on $\mathcal{P}_{p}(\mathbb R^d).$

I would greatly appreciate if someone could point out a reference for what I'm trying to prove or to give me some ideas how to prove it. Thank you!

Became Hot Network Question
Source Link

Compactness with respect to topology induced by TV distance

I've been working on a problem and at some point in the proof I need to show that the following set $$\left\{\mu \in \mathcal{P}_{ac}(\mathbb R^d): \int \varphi(x)\mu(\mathrm{d}x)\leq C\right\}$$ is a compact with respect to the topology induced by TV distance on $\mathcal{P}_{ac}(\mathbb R^d).$ Here, $\mathcal{P}_{ac}(\mathbb R^d)$ is the set of probability measures absolutely continuous with respect to Lebesgue measure on $\mathbb R^d$, $\varphi$ is a real-valued bounded measurable function on $\mathbb R^d$ and $C > 0$ is a constant.

What I've thought about so far: Prokhorov's theorem is not helpful here since it only gives compactness with respect to the weak convergence of measures with is a weaker topology than the one induced by TV distance. I know that for Wasserstein spaces there is this result: for any $p'> p \geq 1,$ the set $$\left\{\mu \in \mathcal{P}_{p}(\mathbb R^d): \int |x|^p\mu(\mathrm{d}x)\leq C\right\}$$ is $W_p$-compact, $W_p$ is known to metrize weak convergence on $\mathcal{P}_{p}(\mathbb R^d).$

I would greatly appreciate if someone could point out a reference for what I'm trying to prove or to give me some ideas how to prove it. Thank you!