Indeed, this is true from the definition, but I don't see how given ##\phi=\sum_{i=1}^N c_i\chi_{E_i}=\sum_{i=1}^M b_i\chi_{F_i}##, we have ##\sum_{i=1}^N c_i\mu(E_i)=\sum_{i=1}^M b_i\mu(F_i)##. How would you show this?
I wonder, how does one show that the integral is independent of the representation of the simple function? Suppose $$\phi=\sum_{i=1}^N c_i\chi_{E_i}=\sum_{i=1}^M b_i\chi_{F_i}.$$ How does it follow then that $$\sum_{i=1}^N c_i\mu(E_i)=\sum_{i=1}^M b_i\mu(F_i)?$$
I have discussed this problem...
Ok, I think I understand @pasmith, thank you. It doesn't matter here to specify whether the finite union is disjoint or not, i.e. Folland writes that a collection of finite disjoint unions of half-open intervals is an algebra, but I don't think the word disjoint is necessary, since we'll get an...
I'm reading in these notes the following passage (I only have a question about the last two sentences):
The last two sentences confuse me. Which sets does the author have in mind? I know what an algebra is (basically a sigma algebra, but not closed under countable infinite operations, but...
If by sketch you mean a drawing, then I don't see how this could prove the lemma, even if one were to only consider ##n=2##. What I meant by sketch is an outline of how one would go about proving the lemma.
Hence, I want to append some background material to my above post.
The author notes...
I am reading these notes on measure theory. On page 27, in chapter 2 on the construction of the Lebesgue measure, in section 2.8 on linear transformations, the author presents a lemma which is not proved. I wonder, how can one prove this?
The author uses the following terminology; a (closed)...
Thanks. I guess 3. is not so hard after all.
If ##\mathcal{F}_\alpha\subset\mathcal{G}## for all ##\alpha<\lambda##, then ##\mathcal{F}_\lambda=\cup_{\alpha<\lambda}\mathcal{F}_\alpha\subset\mathcal{G}##. The conclusion follows from the following implications: $$E\in...
Some definitions:
The following statement has been left as an exercise in transfinite induction in a handout.
I'm looking at Wikipedia and am trying to follow their outline:
1. Show it for the base case, i.e. that ##\mathcal{F}_{0}\subset\mathcal{G}##.
This is, however, trivial, since we...
Hmm, I don't think I understand you.
The first part of definition 3 specifies that the neighborhood be compact, i.e. that there is a compact set ##V\subset X## such that it contains an open set ##U\subset X## that contains ##x##.
The second part of definition 3 specifies simply that there is an...
I have a hard time accepting definitions that are inequivalent. So the main point of my post is to ask for confirmation that it does not matter having inequivalent definitions, but I'm not sure about this. Maybe these two definitions being inequivalent actually have some consequences.
First...
Upon closer thought, I think I have figured out the answers to my questions:
1. ##R## is an almost disjoint union of rectangles ##\{R_1,\ldots,R_N\}##, each which is in turn an almost disjoint union of rectangles from the collection ##\left\{S_{j_{1} j_{2} \ldots j_{n}}: 1 \leq j_{i} \leq N_{i}...
The notes I'm reading are from here. But I have summarized all the necessary details in this post. My question concerns the proposition, but it uses the definition below and the lemma.
We say two rectangles are almost disjoint if they intersect at most along their boundaries.
I omit the...
Consider the following theorem:
First, I don't know how to take the contrapositive of this statement. I'm not sure if the opening hypothesis, that is, ##f## is continuous, remains outside. Because the way this is proved in the text I'm reading is by assuming ##f(X)## is disconnected and then...