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Doing math.


May
30
answered Hausdorff dimension of Cantor set
May
30
revised $\sigma$-field from $(X,Y)$ and Probability measure comparison
Added set brackets where necessary.
May
30
awarded  Suffrage
May
30
suggested suggested edit on $\sigma$-field from $(X,Y)$ and Probability measure comparison
May
30
comment $\sigma$-field from $(X,Y)$ and Probability measure comparison
In that case, the sets are correct. Btw notice that while inside dollars you have to type \{ to make the regular { symbol appear. I edited the set brackets accordingly, if you want you can review it after it's visible.
May
30
comment $\sigma$-field from $(X,Y)$ and Probability measure comparison
Note that in your attempt the sets $\sigma(X)$ and $\sigma(Y)$ are not $\sigma$-fields to start with.
May
30
revised Equivalent definition of absolutely continuous
added spacing in LaTeX for easier readability.
May
30
suggested suggested edit on Equivalent definition of absolutely continuous
May
30
revised $\sigma$-algebras of $\{1,2,3\}$
Added latex symboling.
May
30
suggested suggested edit on $\sigma$-algebras of $\{1,2,3\}$
May
29
answered $\sigma$-algebras of $\{1,2,3\}$
May
29
comment $\sigma$-algebras of $\{1,2,3\}$
The given example is actually closed under complements; so this is not where it fails to be a $\sigma$-algebra.
May
29
revised Compactness properties imply continuity
deleted 4 characters in body
May
29
comment Compactness properties imply continuity
Thanks @MTurgeon, I updated my answer:)
May
29
revised Compactness properties imply continuity
added 1405 characters in body
May
25
comment Prove that $f_n$ converges to $f$ in $L_1$ norm given $\int f_n \to \int f$
Sure, didn't know that. Thanks.
May
25
revised Prove that $f_n$ converges to $f$ in $L_1$ norm given $\int f_n \to \int f$
added 41 characters in body
May
25
answered Prove that $f_n$ converges to $f$ in $L_1$ norm given $\int f_n \to \int f$
May
25
comment Measurability of a simple Set-Valued Map
Do you mean that a set-valued map is measurable if preimage of the singleton $S^{-1}(\{O\})$ is measurable for all open sets $O$? Because $S$ is set-valued, then $O$ is an element of its co-domain not a subset, and as far as I'm concerned we don't know a priori that $S$ would be a bijection. Also, what does $\bar{\mathbb{B}}$ represent?
May
25
comment Limit points of sets
The set of limit points for $B$ is not $\{x^{2}+y^{2}=1\}$: that's its boundary. The set of limit points is $\{x^{2}+y^{2}\geq 1\}$.