Timeline for Notion of convergence on a dense subset
Current License: CC BY-SA 3.0
11 events
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Oct 29, 2015 at 16:58 | history | edited | yada | CC BY-SA 3.0 |
edited body
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Oct 29, 2015 at 10:44 | history | edited | yada | CC BY-SA 3.0 |
adding an EDIT with a further question
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Oct 29, 2015 at 8:17 | comment | added | yada | I think you mean that the closure $\overline{Y}$ of any full measure set $Y$ (i.e. $\mu(Y) = \mu(X)$) contains the support $supp(\mu)$ of the measure $\mu$? In case $\mu$ has full support ($supp(\mu) = X$) we have $\overline{Y} = X$. Moreover, $X$ first-countable seems to be enough: Take $x \in supp(\mu)$ and a countable neighborhood base $N_x^n$, $n \in \mathbb{N}$. Since $\mu(Y) = \mu(X)$ we have that $N_x^n \cap Y \neq \emptyset$ and we can create a sequence $x_n \in N^x_n \cap Y$ that converges to $x$, thus $x \in \overline{Y}$. | |
Oct 28, 2015 at 16:27 | comment | added | Nate Eldredge | At least for second countable spaces $X$, the closure of any full measure set is the support of the measure. So the fact we are really using is that Lebesgue measure has full support. | |
Oct 28, 2015 at 16:24 | comment | added | yada | Nice, thanks. The fact that on $X = \mathbb{R}$ a set with full measure is dense is important (I think for general spaces $X$ such a connection between measure theory and topology does not hold). | |
Oct 28, 2015 at 16:01 | comment | added | Nate Eldredge | And note that the typewriter functions can be defined to be cadlag. | |
Oct 28, 2015 at 15:57 | comment | added | Nate Eldredge | So, doesn't a.e. convergence also give a counterexample for your question? Consider the typewriter sequence $f_n$ converging to $0$ in measure but diverging pointwise everywhere. Every subsequence $f_{n_k}$ contains a further subsequence $f_{n_{k_l}}$ which converges a.e. to 0. Sets of full measure are dense, so $f_{n_{k_l}} \to' 0$. Since we do not have $f_n \to' 0$, (iii) fails. | |
Oct 28, 2015 at 15:26 | history | edited | yada | CC BY-SA 3.0 |
Only (i) and (ii) are necessary to define a limit space.
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Oct 28, 2015 at 15:18 | comment | added | yada | You are right, I think to have mixed things up: Properties (i) and (ii) are enough to define a limit space. Every limit space defines a sequential topology by defining a set $F$ as closed if it contains all limits of $\to'$-convergent sequences in $F$. This sequential topology in turn induces another notion of convergence $\to$ for which (iii) holds. If (iii) holds already for $\to'$ then $\to'$-convergence and $\to$-convergence are equivalent. I edit the thread correspondingly. | |
Oct 28, 2015 at 15:05 | comment | added | Nate Eldredge | Your last comment about a.e. convergence seems relevant. But how does a.e. convergence define a limit space? It seems to me that property (iii) is not satisfied (consider a sequence which converges in measure but not a.e.). | |
Oct 28, 2015 at 14:43 | history | asked | yada | CC BY-SA 3.0 |