Search Results
| Search type | Search syntax |
|---|---|
| Tags | [tag] |
| Exact | "words here" |
| Author |
user:1234 user:me (yours) |
| Score |
score:3 (3+) score:0 (none) |
| Answers |
answers:3 (3+) answers:0 (none) isaccepted:yes hasaccepted:no inquestion:1234 |
| Views | views:250 |
| Code | code:"if (foo != bar)" |
| Sections |
title:apples body:"apples oranges" |
| URL | url:"*.example.com" |
| Saves | in:saves |
| Status |
closed:yes duplicate:no migrated:no wiki:no |
| Types |
is:question is:answer |
| Exclude |
-[tag] -apples |
| For more details on advanced search visit our help page | |
Results tagged with measure-concentration
Search options answers only
not deleted
user 454
13
votes
Accepted
Violating the Lebesgue density theorem
It is a theorem of Besicovitch that measures on $\mathbb R^d$ do satisfy the density theorem.
Fremlin, Measure Theory, Chap. 47
added
Besicovitch, around 1930, extended his density pr …
- 41.1k
3
votes
Doob Martingale: Where is the catch?
What if we work it out? All of $X_j$ are equal to $X_0$, in particular measurable with respect to $X_0$. So
$$
\sum_{i=0}^n X_i = (n+1)X_0,
\\
Y_i = (n+1)X_0\qquad\text{for all } i
\\
Y_{i+1}-Y_i = …
- 41.1k
4
votes
Accepted
Counterexample of non-negative sequence weakly converging in $\mathscr{M}^1$ but not $L^1$
Let's build a "fat Cantor set". Start with $A_0 = [0,1]$ with measure $\alpha_0=1$.
Then remove a short open interval centered at $1/2$, leaving a set $A_1 \subset A_0$ of measure $\alpha_1 < \alpha_ …
- 41.1k