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I'm trying to do the following problem, and I'm getting hung up on one part. Here is the problem:

Let $f: X \longrightarrow \mathfrak{X}$ be a continuous map from a compact Hausdorff space $X$ into a Banach space $\mathfrak{X}$, with $\mu$ a Radon measure on $X$. Consider elements of the form: $$I_{\lambda}(f) = \sum_{k=1}^n f(s_k)\mu(E_k)$$ where the $E_k$'s are Borel sets partitioning $X$ and $s_k \in E_k \subset \{s \in X \: | \: \lvert f(s)-f(s_k) \rVert \leq \epsilon \}$ for $\epsilon > 0$. With $\lambda = \{E_1,\dots,E_n,\epsilon\}$, prove that $(I_{\lambda}(f))_{\lambda \in \Lambda}$ is a convergent net. We denote the limit by: $$\int_{X} f(s) \: d\mu(s).$$

My main issue I'm having is what the ordering needs to be on the set $\Lambda$. I figured at first that I would want something like $\lambda \leq \mu$ if and only if $\mu$ contains a refinement of the partition in $\lambda$, and $\epsilon_{\mu} \leq \epsilon_{\lambda}$. However, this doesn't seem to work, as the smaller the $\epsilon$ I stipulate, this seems to directly affect how many $E_k$'s I would need in my partition since $E_k \subset \{s \in X \: | \: \lVert f(s)-f(s_k) \rVert \leq \epsilon\}$. Hence if $\epsilon > 0$ was really small, I should expect there to be many more $E_k$'s as $\mu(\{s \in X \: | \: \lVert f(s)-f(s_k) \rVert \leq \epsilon\}) \longrightarrow 0$ as $\epsilon \rightarrow 0^{+}$.

Thus, my point is that I shouldn't be able to just change $\epsilon$ independently of changing the $E_1,\dots,E_n$. But this goes against what my ordering is supposedly okay with.

Is there another ordering I'm supposed to be using?

Isochron
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  • You could show that the elements $I_{\lambda}$ converge to the Bochner integral of $f$. Though, it seems like this exercise is meant to define the Bochner integral in this special case. – Mason Dec 01 '23 at 04:26
  • What would the ordering on the set $\Lambda$ be in order to discuss convergence? – Isochron Dec 01 '23 at 04:28
  • I don't understand the problem with the ordering you've described. What's wrong with $\epsilon$ effecting the number of $E_k$'s you need? – M W Dec 06 '23 at 18:55
  • Well I ideally want to prove that for any $\delta > 0$ that there exists $\lambda' \in \Lambda$ such that $\left\lVert I_{\lambda}(f) -I_{\mu}(f)\right\rVert \leq \sum_{k=m+1}^n \lVert f(s_k) \rVert \mu(E_k) < \delta$ whenever $\lambda \geq \mu \geq \lambda'$. However to do this, I need to make $\mu(E_k)$ small for each $k$. This can be done by making the $\epsilon > 0$ small so that $\mu(E_k) < \epsilon$. However for $\epsilon > 0$ small, the sum $\sum_{k=m+1}^n \lVert f(s_k) \rVert \mu(E_k)$ now contains a ton of terms since the refinement is very large. – Isochron Dec 06 '23 at 19:03
  • @MW The issue I see with the refinement being very large is that I can't choose $\epsilon > 0$ to seemingly make up for how many terms are in the sum, as the smaller the $\epsilon$ is, the more terms I have. So I cant do like an $\epsilon/2^n$ type of argument. – Isochron Dec 06 '23 at 19:04
  • It is a bit confusing to use $\mu$ for both your measure and for your partition. – M W Dec 06 '23 at 19:07
  • I do see what you're saying about the proof, though, I misinterpreted your question as being whether the ordering is valid, I will think some more on the proof. – M W Dec 06 '23 at 19:12
  • I think everything should still work fine, though, if you just take $\epsilon'<\epsilon/\mu(X)$, right? It shouldn't matter how many terms there are in the sum. – M W Dec 06 '23 at 19:17
  • Then you would just get $\sum_{k=m+1}^n \lVert f(s_k) \rVert \mu(E_k) \leq (n-m-1)\lVert f \rVert_{\infty} \frac{\delta}{\mu(X)}$ which is still tending to $\infty$ for larger and larger $n$. So the measure $\mu(E_k)$ must be less than something like $\delta/2^k$ – Isochron Dec 06 '23 at 20:07

1 Answers1

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Your proposed ordering works fine. To be precise, we should say that $\Lambda$ consists of triples $(\{E_1,\dots,E_n\},\epsilon,\{s_1,\dots,s_n\})$ with each $s_k\in E_k$, and $(\{E_1,\dots,E_n\},\epsilon,\{s_1,\dots,s_n\})\leq (\{F_1,\dots,F_m\},\eta,\{t_1,\dots,t_m\})$ if $\eta\leq \epsilon$ and each $F_l$ is contained in $E_k$ for some $k$.

Then given $\epsilon>0$, choose a partition $\lambda_0=(\{E_1,\dots,E_n\},\epsilon_0,\{s_1,\dots,s_n\})$ with $\epsilon_0<\frac{\epsilon}{\mu(X)}$.

(To see that such a partition exists, note that at each $x\in X$, we have by continuity a neighborhood $U$ where $\|f(y)-f(x)\|<\epsilon/2$ for $y\in U$. By compactness there is a finite cover $U_1,\dots,U_n$ of $X$ of such neighborhoods, and so we may take $E_1=U_1$, $E_k=U_k\backslash\bigcup_{j<k}E_j$ as our partition, and may choose any points in $E_k$ as our $s_k$.)

Then for any partition $\lambda=(\{F_1,\dots,F_m\},\eta,\{t_1,\dots,t_m\})$ with $\lambda\geq \lambda_0$, we have each $F_l$ contained in some $E_{k_l}$, so we can estimate

\begin{align*} \|I_\lambda(f)-I_{\lambda_0}(f)\| &=\left\|\sum_{k=1}^nf(s_k)\mu(E_k) -\sum_{l=1}^m f(t_l)\mu(F_l) \right\| \\ &= \left\|\sum_{l=1}^mf(s_{k_l})\mu(F_l) -\sum_{l=1}^m f(t_l)\mu(F_l) \right\|\\ &= \left\|\sum_{l=1}^m(f(s_{k_l})-f(t_l))\mu(F_l) \right\|\\ &\leq \sum_{l=1}^m\left\|(f(s_{k_l})-f(t_l)) \right\|\mu(F_l)\\ &\leq \epsilon_0\sum_{l=1}^m\mu(F_l) = \epsilon_0\mu(X)<\epsilon\text{.} \end{align*}

It quickly follows that the net is a Cauchy net, hence by completeness of $\mathfrak X$ the net converges.

M W
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  • This looks correct, but I'm not sure what uniform continuity means for a general top. space. – Isochron Dec 06 '23 at 20:22
  • Yeah sorry not sure what I was thinking with that. Let me adjust – M W Dec 06 '23 at 20:24
  • @Isochron I've edited to make it work now. – M W Dec 06 '23 at 20:33
  • What are you doing in the second equality? I see that you have some subcollection $E_1,E_2,\dots,E_{n,1}$ whose union gives you $F_1$, but what is happening to the points $s_1,\dots,s_{n,1}$ to be replaced with $s_{k_1}$? – Isochron Dec 06 '23 at 20:46
  • Each $E_k$ is partitioned into a finite union of the $F_l$'s, so we can sum over each $F_l$, and then $k_l$ is the index $k$ corresponding to the element $E_k$ that contains $F_l$. – M W Dec 06 '23 at 20:49
  • I think in your comment you have it backwards, ${F_l}$ is refining ${E_k}$, not the other way around. – M W Dec 06 '23 at 20:53
  • I'm getting hung up on the fact that there are now more $s_{i}$'s in the second line, since m is presumably larger than n, so unless there are repeats, I don't see how that works. – Isochron Dec 06 '23 at 21:00
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    There are indeed repeats. So really, each $E_k$ is split into $F_{l_1}\cup F_{l_2}\dots \cup F_{l_p}$, then $f(s_k)\mu(E_k)$ becomes $f(s_k)\mu(F_{l_1})+\dots+f(s_k)\mu(F_{l_p})$. So we define $k_{l_i}:=k$ for each $i=1,\dots,p$. – M W Dec 06 '23 at 21:01
  • I see, thank you – Isochron Dec 06 '23 at 21:01