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Oct 29, 2015 at 16:58 history edited yada CC BY-SA 3.0
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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
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.
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