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Another question in my studies (of Lebesgue integration):

I'm given a continuous function $f:\mathbb{R}\to\mathbb{R}$, a nonempty compact set $E\subseteq\mathbb{R}$, and a sequence of nonempty compact sets $E_i$ such that $$\lim d(E_i , E) = 0$$where this $d$ is Hausdorff distance.

I am asked to prove that $$\lim \int_{E_i} f=\int_E f$$

Visually, this seems pretty straightforward, the $E_i$ are being forced to look more and more like $E$, and the continuity of $f$ ensures that the farther along the $E_i$ we pick, $f$ has to vary less and less where $E_i$ and $E$ don't coincide.

I've been at this for a while and seem to be stuck. For a while though I hadn't noticed the compactness assumed of $E$ and the $E_i$. Putting the continuity and convergence in Hausdorff distance together, I see that for any $x\in E$, $\varepsilon>0$, and $\delta > 0$, there is eventually some $E_i$ for which I can find a $y\in E_i$ within $\delta$ of $x$ and with $|f(x)-f(y)|<\varepsilon$. I guess using compactness, I can make this uniform over all $x\in E$ (so that my picture described above is actually as nicely behaved as I was probably picturing it to begin with).

At this point though I'm not sure how to proceed. Should I try to bound the integral over the symmetric difference of $E_i$ and $E$ or something like that? I'm not sure how I would control it, despite knowing it would be constrained somehow by "nearby" values of $f$. Should I cut up $E$ somehow, or select some dense subset of $E$ to use representative values of $f$ on? Is there some subtlety my visual intuition is overlooking?


It seems like there is a counterexample to this theorem as stated:

Let $E=[0,1]$ and let $E_i = \{ k 2^{-i} : k=0,1,\dots,2^i \}$ so that $$\array{ E_1 & = & \{0,\frac{1}{2}, 1\} \\ E_2 & = & \{0,\frac{1}{4}, \frac{2}{4}, \frac{3}{4}, 1\} \\ & \vdots }$$

Then, for example, with $f$ taken to be the constant $1$ function on $[0,1]$, we have $\int_{E_i} f = 0$ for all $i$, yet $\int_{E} f = 1$.

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Maybe you could replace the metric by $d(A,B) = m(A \Delta B)$. Or do you want to work with the same metric? –  Jonas Teuwen May 1 '11 at 21:21
That would certainly work with the problem as otherwise stated. I guess I'm mostly interested in what the intention of the author of the question was. I feel a bit bad about asking it now myself. I don't know whether they made a mistake on the metric or on the hypotheses on the sets. I'm not sure how good of a question this is for math.SE now, since it's somewhat speculation at this point. I guess it still has merit as a "what conditions are needed" question though. –  matt May 1 '11 at 21:29
I think the question is now too open (but others might not agree) and because of that it is hard to decide what a good answer may be. You have your counterexample now ;-). –  Jonas Teuwen May 1 '11 at 21:38
My opinion is that with the counterexample, this question was satisfactorily answered, and should have been left like that. Then, a new question could be opened (and linked to this one) to ask about appropriate hypotheses so that the conclusion holds. –  Glen Wheeler May 2 '11 at 19:14
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2 Answers

up vote 2 down vote accepted

I might have an idea, don't shoot me if it is wrong.

Given compact $E$ we can find for every $\epsilon > 0$ a finite number of closed sets $Q_j$ such that

$$F = \bigcup_{j = 1}^N Q_j \text{ and } m(E \Delta F) \leq \epsilon$$

So define a function $f_\epsilon$ which is equal to $\sup f|_{Q_j}$ on $Q_j$.

Now we we want to estimate

$$\int_E |f - f_\epsilon| \, dm$$

We can split integral in three parts

$$\int_E |f - f_\epsilon| \, dm = \left (\int_{E \Delta F} + \int_{E \cap F} - \int_{F \setminus E} \right ) |f - f_\epsilon| \, dm$$

So this can be made smaller than $C \epsilon$ (by uniform continuity, maybe adjust $f_\epsilon$ a bit).

Now we need to estimate

$$\int_{E_i} |f - f_\epsilon| \, dm,$$ we note that $E_i \subset E \cup [\inf E - d(E, E_i), \inf E] \cup [\sup E + d(E, E_i)]$. The integral over $E$ is treated the same and the other two follow by Lebesgue dominated convergence.


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Hmm, I think a problem is that $E \Delta F$ does not have to be compact, probably some others too. Others? –  Jonas Teuwen May 1 '11 at 20:58
I think this is the right idea -- as mentioned by Robert, any correct argument has to use convergence in measure. One question: is this equivalent to adding $\mu(E_i) \rightarrow \mu(E)$ as an additional hypothesis? (so that we have convergence in the Hausdorff metric and in measure) I would think so, and then it seems like it should definitely work. –  Glen Wheeler May 2 '11 at 19:20
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Let $\varepsilon >0$ and choose $\delta >0$ so that whenever $m(A)<\delta$, it follows that $\int _A|f|dx<\varepsilon$.

Then, choose $N$ sufficiently large so that $d(E_N,E)<\delta$. Then,

$$ \left| \int _{E_N}fdx-\int _Efdx\right| =\left| \int _{E_N-E}fdx-\int _{E-E_N}fdx\right| \leq \int _{E_N-E}|f|dx+\int _{E-E_N}|f|dx. $$

Now, from here you have to shown, because $d(E_N,E)<\delta$, that $m(E_N-E),m(E-E_N)<\delta$ (up to a constant factor anyways, so that you might have to rescale the choice of $\delta$). Then, by the above inequality, you would have

$$ \left| \int _{E_N}fdx-\int _Efdx\right| <2\varepsilon , $$

and hence $\lim \int _{E_N}fdx=\int _Efdx$.

Simplifying the picture of Hausdorff distance given on Wikipedia to one dimension, I think that the argument that I left out shouldn't be too bad. If I have time I'll try to help with the details later.

-Jonny Gleason

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If you have that $m(E_n\backslash E),m(E\backslash E_n)\to0$, can't you just apply dominated convergence to $f_n=\chi_{E_n}f$ (dominated by $\chi_{W}|f|$, where $W=\overline{\bigcup_n E_n}$, which should be compact)? I could be wrong, but it seems like the verification of the limits $m(E_n\backslash E),m(E\backslash E_n)\to0$ is the crux of the problem. –  Nick Strehlke Apr 30 '11 at 23:45
That would work, I just don't see why $W$ would necessarily be compact, and hence I don't see why $\chi _W|f|$ would necessarily be integrable. –  Jonathan Gleason Apr 30 '11 at 23:56
@GleasSpty The reason I believe $W$ has to be compact is that $\alpha=\sup_i d(E_i,E)$ has to be finite. If $E\subset [a,b]$ ($a,b$ finite), we have to have $E_i\subset[a-\alpha,b+\alpha]$, in which case $W$ is bounded and closed. Of course I could be wrong. –  Nick Strehlke May 1 '11 at 0:07
I went back to think about this again, and now I'm not sure on the $m(E_n - E)\to 0$ part. Why is this not a counter-example?: $E=[0,1]$, and $E_0 = \{0,\frac{1}{2},1\}$, $E_1 = \{0,\frac{1}{4},\frac{1}{2},\frac{3}{4},1\}$, and so on, subdividing each time. We will always have $E_n$ a countable union of singletons, hence measure 0, but the measure of $E$ is 1, and $m(E-E_n )=1$. Still, $d(E_n ,E)\to 0$. –  matt May 1 '11 at 5:55
The point of the counterexamples is that convergence in Hausdorff metric does not imply convergence of Lebesgue measure. More generally, any compact set can be approximated in the Hausdorff metric by finite sets, and in any nonatomic measure those finite sets will have measure 0. You need a hypothesis that has something to do with measure, not just metric. –  Robert Israel May 1 '11 at 22:54
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