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seen Dec 17 at 20:22

Doing math.


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
added 240 characters in body
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
added 6 characters in body
May
24
answered Question about proving something using reverse Fatou's lemma
May
24
revised neighborhood and topological space
Added LaTeX.
May
24
suggested approved edit on neighborhood and topological space
May
24
comment A basic question on diameter of a metric space
@froggie: In my opinion that is a rather strict requirement, and I'm not quite sure what you mean by the 'easiest'. It is sufficient for sure but it is far away from being necessary: a set having finite diameter is very far away from its closure being compact. A set with a compact closure is called relatively compact. This would in a sense identify relative compact sets with bounded sets, while the first property is very strong and the latter very weak.
May
24
comment Prove that $\mu(\bigcup_{i=1}^{\infty}\bigcap_{n=i}^{\infty}B_{n}) \le \liminf_{n\to \infty} \mu (B_{n})$
What Davide uses is that $\Big(\bigcup_{i=1}^{n}\bigcap_{k=i}^{\infty}B_{k}\Big)_{n=1}^{\infty}$ is an increasing sequence of sets and applies then convergence of measure. Also, since $\mu(\bigcap_{k\geq n}B_{k})\leq \mu(B_{j})$ for all $j\geq k$, then $\mu(\bigcap_{k\geq n}B_{k})\leq \inf_{k\geq n}\mu(B_{k})$, which yields the result since $\bigcup_{i=1}^{n}\bigcap_{k=i}^{\infty}B_{k}=\bigcap_{k\geq n}B_{k}$ for all $n\in\mathbb{N}$.
May
24
comment Analyze the convergence or divergence of $\{1/n^2\}$
@AndréNicolas: Some authors use this notation instead of $(\cdot)$ for a sequence, and it is not even that unusual. See for example Falconer's book, the geometry of fractal sets, which is a classic piece. He uses $\{\cdot\}$ notion for sequences, and there are plenty of others. Hence I would not conclude that this question is not about sequences based on that notation.
May
23
comment Prove that $\mu(\bigcup_{i=1}^{\infty}\bigcap_{n=i}^{\infty}B_{n}) \le \liminf_{n\to \infty} \mu (B_{n})$
The same question was also answered some two weeks ago in here: math.stackexchange.com/questions/141697/…
May
23
comment Looking for some function such that $\lim\limits_{x\to\infty}f(x) \ne \infty$
@AmihaiZivan: Your example doesn't work since $\frac{1}{1+x^{2}}\leq 1$ for all $x\in\mathbb{R}$ (instead of $\geq 1$!). As Arturo Magidin demonstrated below, a function with such properties does not exist.
May
22
comment prove $f'(x)=f(x)$
In fact, what happens at $0$ isn't even necessarily to comment separately since $\frac{f(y)-1}{y}=\frac{1+y\,g(y)-1}{y}=g(y)\to 1$ as $y\to 0$.
May
22
answered Generated $\sigma(X)$ where $\Omega \neq \mathbb{R}$
May
22
answered monotonicity of inf