T. Eskin
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 Jun 12 asked Vitali set of outer-measure exactly $1$. Jun 12 revised Interior, exterior and boundary of sets in $\mathbb R^2$ added 10 characters in body Jun 12 comment Interior, exterior and boundary of sets in $\mathbb R^2$ Your answer on $B$ is not correct. Jun 12 answered Interior, exterior and boundary of sets in $\mathbb R^2$ Jun 12 comment Limits of Subsequences With the same idea it actually follows that for any subsequence of $\{t_{n}\}$ we find such a subsequence of $\{s_{n}\}$ that decreases more rapidly. And I don't think we needed the fact that $s_{n}\leq t_{n}$ at all. Jun 12 comment subsets probability question What have you tried so far? And is this a homework? Jun 11 comment True if sigma-compact @SamL.: Thank you. Jun 11 answered True if sigma-compact Jun 11 comment Form of $\sigma(X_n)$ and $\sigma(X_n)$-measurability @Justin: Topology is just the family of open sets. If it confuses you, you can ignore that part and simply think of $\mathscr{B}(X)$ being the Borel $\sigma$-algebra of $X$ the way you have it defined. Jun 11 comment True if sigma-compact Found it now. And it's true that this is very questionable since no proof is given. If $X$ is $\sigma$-compact then the Lemma is true, as $E$ is a countable union of Borel sets. Let me see if I'm able to do something with the locally compact Hausdorff case. Jun 11 comment True if sigma-compact I couldn't find this Lemma from Royden's book. What page is it? Jun 11 comment is there any one one and onto function from $A$ to $B$ That is correct. Jun 11 comment is there any one one and onto function from $A$ to $B$ You could start by writing down the sets $A$ and $B$ in simpler forms. They are both open intervals. Jun 11 comment True if sigma-compact What do we know about the topology of $X$? Jun 11 revised Form of $\sigma(X_n)$ and $\sigma(X_n)$-measurability added 1 characters in body Jun 11 answered Form of $\sigma(X_n)$ and $\sigma(X_n)$-measurability Jun 10 comment Evaluating $\int \frac{dx}{x^2 - 2x} dx$ @MarkBennet: The latter part to the answer was included after all the previous comments took place. And the first part, it is purely unnecessary steps to obtain $\frac{1}{x(x-2)}$ from $\frac{1}{x^{2}-2x}$. Jun 10 comment Evaluating $\int \frac{dx}{x^2 - 2x} dx$ My intention is not to complain, if it seemed like that then I am glad I can correct the misunderstanding. I'm just having hard time seeing why this answer is useful. Do you really think that $x^{2}-2x=x(x-2)$ is what the OP is having hard time to figure out? Jun 10 comment Evaluating $\int \frac{dx}{x^2 - 2x} dx$ You start from $\frac{1}{x^{2}-2x}$ and end up with $\frac{1}{x(x-2)}$ with 2 unnecessary middle-steps. You could just take $x$ as common multiplier in the first place. Jun 10 comment Evaluating $\int \frac{dx}{x^2 - 2x} dx$ Why didn't you just take $x$ as a common multiplier? Why did you need all those middle steps?