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May
30
revised Equivalent definition of absolutely continuous
added spacing in LaTeX for easier readability.
May
30
suggested approved edit on Equivalent definition of absolutely continuous
May
30
revised $\sigma$-algebras of $\{1,2,3\}$
Added latex symboling.
May
30
suggested approved 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
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May
29
comment Compactness properties imply continuity
Thanks @MTurgeon, I updated my answer:)
May
29
revised Compactness properties imply continuity
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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 $\int f_n \to \int f$
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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 $\{x^{2}+y^{2}\geq 1\}$.
May
24
comment Integration from 0 to 0 - why does my calculator say “undefined” in one case, and “0” in another?
The one-point set $\{0\}$ over which you are integrating is zero measurable: so any measurable function should yield $0$ when integrating over it. However, neither of these functions are defined at $0$, so I think the answer should be "undefined" for both integrals.
May
24
answered Limit points of sets
May
24
comment Question about proving something using reverse Fatou's lemma
@DavideGiraudo: I assumed that $\epsilon$ is the collection of measurable functions, as it is the only reasonable explanation that I could think of. Considering, that this requirement is one of the two assumptions in the Fatou's lemma, other which is stated right after the first verse.
May
24
revised Question about proving something using reverse Fatou's lemma
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May
24
revised Question about proving something using reverse Fatou's lemma
added 240 characters in body
May
24
revised Question about proving something using reverse Fatou's lemma
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