421 reputation
18
bio website msemac.redwoods.edu/~darnold/…
location Eureka, California
age
visits member for 1 year, 3 months
seen 2 days ago

Apr
13
comment Case C: Euler's equation in Simmon's textbook
I've adjusted my work above to reflect what I learned from your comment. Am I understanding this properly?
Jul
13
comment Folland, Chapter 1 Problem 17
Eric, nope, you have it incorrect. It might be because we're using different letters and it's getting you confused. Here is the definition, direct from Folland: If $\mu^*$ is an outer measure on $X$, a set $A\subset X$ is called $\mu^*$-measurable if $\mu^*(E)=\mu^*(E\cap A)+\mu^*(E\cap A^C)$ for all $E\subset X$. Now if I change my letters it would read: A set $E$ is called $\mu^*$-measurable if $\mu^*(A)=\mu^*(A\cap E)+\mu^*(A\cap E^C)$ for all $A\subset X$. Hope this helps.
Jul
13
comment Folland, Chapter 1 Problem 17
$B$ is the union of a countable number of disjoint $\mu^*$-measurable sets, so $B$ is measurable, meaning $\mu^*(E)=\mu^*(E\cap B)+\mu^*(E\cap B^C)$ for any $E\subset X$.
Jul
9
comment Constructing a complete measure space
In your latest proof, why is $\mu(U\cup G)=0$?
Jul
9
comment Constructing a complete measure space
Nope, still a problem. You have $E=\phi\cup E$. To show that this is an element in $B_0$, you have to show that $\phi\in B$ (which is true) and $E\in B_0$, which hasn't been shown. $F\in B_0$ and $E\subset F$ does not necessarily mean that $E\in B$.
Jul
9
comment Constructing a complete measure space
Using these characters and the hypotheses set up, I believe what is needed is to prove that $E\in B_0$, which I don't see that this is saying.
Jul
9
comment $f$ measurable, $f=g$ almost everywhere, complete measure space
Very well written. Thank you for the response.
Jul
9
comment $f$ measurable, $f=g$ almost everywhere, complete measure space
Nice example, but there is part of it that I don't understand. If your measure space is not complete, how do you prove that there is a non-measurable set contained in a set of measure zero.
Jul
9
comment $f$ integrable but $f^2$ not integrable
I'd like to thank folks for these extremely helpful examples. This continues to open doors for me.
Jul
8
comment Show there exists an N such that $n\ge N$ implies $\int|f^+-\phi_n|\,d\mu<\epsilon/2$
I think it does help. Because $\int \varphi_n d\mu \nearrow \int f \, d\mu$, there exists an $N$ such that $n\ge N$ implies that $\int f^+\,d\mu-\epsilon/2<\int \phi_n\,d\mu\le \int f^+\,d\mu$. Then $\int f^+\,d\mu-\int \phi_n\,d\mu<\epsilon/2$ for all $n\ge N$, which is saying the same thing as $\int(f^+-\phi_n)\,d\mu<\epsilon/2$ for all $n\ge N$, which is what I need.
Jul
8
comment Show there exists an N such that $n\ge N$ implies $\int|f^+-\phi_n|\,d\mu<\epsilon/2$
At this point (Chapter 5) in Bartle's Elements of Integration, if $X$ is a nonempty set, $\mathcal{X}$ is the $\sigma$-algebra of subsets of $X$, and $\mu$ is a measure defined on $\mathcal{X}$ (i.e., $\mu(E)\ge 0$ for any $E\in\mathcal{X}$, $\mu(\phi)=0$, and $\mu$ is countably additive). If $f\in L(X,\mathcal{X},\mu)$, that means that $f:X\to R$, $f$ is measurable, and $\int f^+\,d\mu<0$ and $\int f^-\,d\mu<0$. The integral of $f$ is defined as $\int f\,d\mu=\int f^+\,d\mu-\int f^-\,d\mu$. The space $L^p$ has not been discussed as yet.
Jul
7
comment $f$ integrable, $g$ measurable, $f = g$ almost everywhere implies $g$ integrable
Hope this is a good proof. I would appreciate any comments.
Jul
7
comment $f$ integrable, $g$ measurable, $f = g$ almost everywhere implies $g$ integrable
I don't think either hint will work as I can't say that $\int g\,d\mu$ exists until I first show that $\int g^+\,d\mu<+\infty$ and $\int g^-\,d\mu<+\infty$.
Jun
28
comment $f, g$ measurable function on $E$ that are finite a.e. on $E$
Thanks to everyone for their helpful ideas. I've posted a possible route of solution in my question above. Can folks look at it and tell me if my argument is valid?
Jun
28
comment Open set as a countable union of open bounded intervals
OK. If $x\in\cup(p,q)$, then $x\in(p,q)$ for some rational numbers $p$ and $q$. Because $(p,q)\subset A$, then $x\in A$. Hence, $\cup(p,q)\subset A$. On the other hand, suppose $x\in A$. Because $A$ is open, there exists and $\epsilon>0$ such that $(a-\epsilon,a+\epsilon)\subset A$. Now because of the density of $Q$, there exists rational $p$ and $q$ such that $a-\epsilon<p<x<q<a+\epsilon$. Hence, $x\in\cup (p,q)$ and $A\subset\cup(p,q)$. Therefore, $\cup(p,q)=A$. Have I got it right?
Jun
28
comment Open set as a countable union of open bounded intervals
Hmmm... Is this a collection of intervals? A collection of counterexamples?
Jun
28
comment $f, g$ measurable function on $E$ that are finite a.e. on $E$
True, but $f+g$ may not even be defined on $E\backslash E_0$.
Jun
26
comment If $m^*(E)=\infty$, then $E=\bigcup_{k=1}^{\infty}E_k$, $E_k$ measurable and $m^*(E_k)<+\infty$
I've added this fact (E is measurable) to the question above.
Jun
26
comment If $m^*(E)=\infty$, then $E=\bigcup_{k=1}^{\infty}E_k$, $E_k$ measurable and $m^*(E_k)<+\infty$
Yes, E is measurable.
Jun
24
comment Outer measure fundamental question
Ah, of course, and then you cover any subset of R.