All Questions
5 questions
15
votes
2
answers
2k
views
Where does the Lebesgue differentiation theorem fail?
The Lebesgue differentiation theorem says that for certain metric spaces $X$ (see below), any Borel measure $\mu$ that is finite on bounded sets and any $f: X \rightarrow \mathbb{R}$ locally $\mu$-...
2
votes
0
answers
116
views
For Polish $X,Y$, $L^p(X,Y)$ is separable
Let $X$ and $Y$ be Polish spaces. Equip $X$ with a Borel probability measure $\mu_X$ and $Y$ with a metric $d_Y$. We can define the $L^p$ space as follows:
Definition. Define
$\begin{align}L^p(X,Y) = \...
2
votes
0
answers
95
views
Can we control the Wasserstein metric between $\mu$ and $\nu$ by their moment difference?
Fix $p \in [1, \infty)$. Let $(\mathcal P_p(\mathbb R^d), W_p)$ be the Wasserstein space of all Borel probability measures on $\mathbb R^d$ with finite $p$-th moment. Let $D_p$ be the collection of ...
1
vote
0
answers
52
views
A local base for space of probability measures with Prohorov metric
Let $S$ be a Polish space. Let $P(S)$ denote the space of probability measures on $(S,\mathcal{B})$, where $\mathcal B$ is the Borel-$\sigma$-algebra over $S$. Equip $P(S)$ with the Prohorov metric. I ...
0
votes
0
answers
618
views
The set of continuous bounded functions $f:X\to Y$ is dense in $L^p(X,Y)$ where $X,Y$ are Polish
It is well known that the set of real-valued continuous functions with compact support is dense in $L^p(\mu)$ where $\mu$ is a Radon measure (see e.g. [Folland, Proposition 7.9]) Clearly, the set of ...