I do find that as there are more and more crappy "do my homework" questions and questions where the OP makes no effort to write anything that makes sense, I'm losing patience almost entirely.
One mod was removed by SE staff for what appears to be a potential to violate a future (but not current) code of conduct, and a lot of other mods didn't take kindly to that
I will repost this one because I am still struggling with measure theory. :(
In the comments of math.stackexchange.com/… Daniel basically instructs the OP to show that any measurable null set $E$ is sandwiched by two other sets $A\subseteq E\subseteq B$ such that $\mu(B\setminus A) = 0$. Thus $\mu(A)\in\mathcal M^*$. I don't see how this necessitates that all subsets of $E$ are in $\mathcal M^*$.
@ÉricoMeloSilva In the comments of math.stackexchange.com/… Daniel basically instructs the OP to show that any measurable null set $E$ is sandwiched by two other sets $A\subseteq E\subseteq B$ such that $\mu(B\setminus A) = 0$. Thus $\mu(A)\in\mathcal M^*$. I don't see how this necessitates that all subsets of $E$ are in $\mathcal M^*$.
This is a homework problem and I need some guidance on a proof.
Let $(X,\mathcal{M},\mu)$ be a measure space, $\mu^*$ the outer measure induced by $\mu$ according to (1.12), $\mathcal{M}^*$ the $\sigma$-algebra of $\mu^*$-measurable sets, and $\overline{\mu}=\mu^*\mid_{\mathcal{M}^*}$.
(...
@TedShifrin two questions: (1) do/did you know of any Deaf mathematicians, and (2) do you know why Thurston spoke in ASL with his family in his later days (was it due to what caused his death?).
I don't know the details of Thurston's demise. I have had deaf students, but I don't think I know any deaf mathematicians. The famous mathematician who figured out how to invert the sphere was blind !!
I've seen Giroux's lectures on video. Are you sure he's blind? I don't believe that. No, the guy who figured out how to invert the sphere was Morin (French). Yes, deaf students had (university-provided) interpreters. No big deal.
Hi. Is it appropriate to ask what algorithm the ti-84 uses on mathstackexchange to calculate, say the row reduced echelon form of the matrix [1 , 1 , 1 , 2; 4 , 2 , 1 , 5 ; 9 , 3 , 1 , 10]. I was curious because the second row, last column is something super ridiculously close to 0 that should actually be 0. I'm just curious why it is super close and why the calculator doesn't show 0 there instead. I know it is a calculator and can mostly be trusted about alot of calculations.
Yeah, I've done the calculation by hand, @randomgirl. I don't understand what it did. The "intelligent" way to do Gauss-Jordan would not have any roundoff error at all, because everything is multiples of $.5$. I'm guessing the calculator divided by $9$ in the last row rather than subtracting $9$ times the first row, etc. This would introduce roundoff error.
Thanks Ted. The calculator was super super close. I understand to expect that sometimes. It is interesting though what the algorithm is when something like this does happen.
Mind you, I could be misunderstanding the hint. But to me it looks like he is saying that given $B$ you can find an $A$ such that [...], moreover $A$ is measurable.
And grand, that works for that particular $A$, but I don't see how it solves the question.
$\mu(F) \geq \mu^{\ast}(F \backslash A) \geq \mu^{\ast}(F \backslash E) = \mu(F)$ because $\mathcal{M} \subset \mathcal{M}^{\ast}$ trivially
i meant to put stars on the $\mu(F)$s
F arbitrary
@anakhro $\mu^{}(B\A) \leq \mu^{}(B\E) + \mu^{}(E\A) = 0$ so it does though, i also don't know if writing $\mu(B\A)$ was a typo but it seems like one to me bc we're proving $\mu^{}$ measurability, not $\mu$-measurability, which isn't true a priori anyway
this is true by what i said, this inner regularity thing is strictly stronger than you need to reach the conclusion of the question but is a response to something OP tried to argue in the comments clearly
I think its propagation is due mainly to Bourbaki, but you define relations as subsets of product sets AxB. A function is a relation satisfying certain properties (most notably, single valued and "total").
Try with 44=42+2; then:
$(42+2)^n=42^n+ 42(....)+2^n= 2^n(mod7)$
Then you just need to find the least n with $2^n=1(mod7)$ , i.e., you just need to find
an n so that $7|k(2^n -1)$ . Note that $2^1-1=1, 2^2-1=3,...$ and notice the remainders of $2^n-1$ when you divide by 7
I'm curious, how can I approach this problem via Pigeon Hole principle? Is it even possible?
@user709833 Yes, a set $f$ is called a function if all of its elements are ordered pairs and if $(a,b)$ and $(a,c)$ are both in $f$ then $b=c$. The interpretation here is that $f(a)=b$
@AlessandroCodenotti My question though is that if we are using functions does it mean that we must also be using a set theory underlying it first / have one defined?
Say that I know that sp(4,C) is a semisimple Lie algebra. To deduce that it is a simple Lie algebra I can consider its root system. If the root system is irreducible, then it is a simple Lie algebra.
To prove that the root system is irreducible, I have to show that I cannot find two orthogonal subsets of the roots, which I indeed cannot do just looking at the root system drawn with the correct angles. However, say that I do not know what the root system looks like when drawn in R^2. Do I have to 1) Compute all of the roots, 2) Take each of the Cartan generators to their dual, with respect …
@user709833 You don't have to, there are foundational systems in which functions are primitive notions, but the most common choice is to start with set theory and build everything from there
@schn: That thing is horribly written and confuses basic ideas. I would avoid that website in the future. (Just with a cursory glance I notice two things wrong.)
It's supposed to be a local parametrization of the level curve (not of $f$ itself — garbage), which, by the implicit function theorem, a priori exists only locally.
So, I have a discrete probability distribution depenedent on $m$. So $n\in\{0,1,\ldots,m\}$ and appears to have marginal mass distribution of $p_N(n)=\frac{m}{15}$, where $m\in\{1,2,3,4,5\}$. However, this confuses me, because when $m=5$ we get that the probability of the six possible values are each $\frac{1}{3}$, and so the total probability appears to be 2.
Am I misinterpreting something? Is this mass function possible? Or am I correct in thinking it is impossible?
Try with 44=42+2; then:
$(42+2)^n=42^n+ 42(....)+2^n= 2^n(mod7)$
Then you just need to find the least n with $2^n=1(mod7)$ , i.e., you just need to find
an n so that $7|k(2^n -1)$ . Note that $2^1-1=1, 2^2-1=3,...$ and notice the remainders of $2^n-1$ when you divide by 7
@Semiclassical The random vector $(X,Y)$ has the following joint distribution $$P(X=m,Y=n)=\binom{m}{n}\frac{1}{2^m}\frac{m}{15}$$ where $m=1,2,\ldots,5$ and $n=0,1,\ldots,m$. Derive the conditional pmf of $Y$
The approach was to sum over $n$ and then pull the $\frac{m}{15}$ out front, to notice that this is then $\frac{m}{15}$ multiplied by the sum of the probabilities of a binomial distribution or it's entire domain (which is equal to 1).
Try with 44=42+2; then:
$(42+2)^n=42^n+ 42(....)+2^n= 2^n(mod7)$
Then you just need to find the least n with $2^n=1(mod7)$ , i.e., you just need to find
an n so that $7|k(2^n -1)$ . Note that $2^1-1=1, 2^2-1=3,...$ and notice the remainders of $2^n-1$ when you divide by 7
This is a homework problem and I need some guidance on a proof.
Let $(X,\mathcal{M},\mu)$ be a measure space, $\mu^*$ the outer measure induced by $\mu$ according to (1.12), $\mathcal{M}^*$ the $\sigma$-algebra of $\mu^*$-measurable sets, and $\overline{\mu}=\mu^*\mid_{\mathcal{M}^*}$.
(...
Try with 44=42+2; then:
$(42+2)^n=42^n+ 42(....)+2^n= 2^n(mod7)$
Then you just need to find the least n with $2^n=1(mod7)$ , i.e., you just need to find
an n so that $7|k(2^n -1)$ . Note that $2^1-1=1, 2^2-1=3,...$ and notice the remainders of $2^n-1$ when you divide by 7
