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I like the Lebesgue integral a lot more than the other alternatives, because of its connections to measure theory; it's a way of thinking about integration that's as "liberated" from the structure of the real line as possible, which cannot be said of, for example, the Henstock-Kurzweil integral. Sometimes, however, the improper version is needed. For example, here is a famous example of a function that has an improper Riemann integral, but no Lebesgue integral - but we can solve the problem by taking a limit of Lebesgue integrals, of course. And, if I understand correctly, the Henstock-Kurzweil integrals were invented to integrate functions that have neither a Lebesgue integral, nor an improper Riemann integral. Again, I'm under the impression that we can usually get away with taking a limit of Lebesgue integrals most of the time.

But improper Lebesgue integrals aren't a purely measure-theoretic concept. So my question is:

Question. Is there a notion of "measure space equipped with further structure" such that each such thing is associated with an improper Lebesgue integral?

When applied to the real line, this ought to define an integral that equals or supersedes both the Henstock and ordinary Lebesgue integrals insofar as the space of functions that can be integrated.

I was thinking, in particular, of measure spaces equipped with a notion of "small" subset, perhaps they form a bornology or something. We proceed to define that the improper Lebesgue integral of a function is a limit of its Lebesgue integrals on small subsets, and hopefully we can prove that every function that is Henstock-Kurzeil integrable is improperly Lebesgue integrable.

Is anything like this out there?

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    $\begingroup$ Every "improperly Riemann integrable function" is also "improperly Lebesgue integrable". That Lebesgue doesn't know how to handle this as much as Riemann (not counting H-K) is a commonly wide-spread myth: for instance, you just take $\lim_{x \to \infty} \int_{[c,x]} f$ in the same way as Riemann. The difference is that Lebesgue has a meaning on itself for the value $\int_{[c,\infty]} f$, whereas Riemann doesn't, and this can differ from the "improper". This is not a disadvantage over Riemann, since Riemann can't even properly evaluate $\int_{[c,\infty]}$ at first place. $\endgroup$ – Aloizio Macedo Apr 2 '17 at 13:45
  • $\begingroup$ @AloizioMacedo, but note that $[c,x]$ isn't a measure-theoretic concept. $\endgroup$ – goblin Apr 2 '17 at 13:46
  • $\begingroup$ ? I don't see your point. $\endgroup$ – Aloizio Macedo Apr 2 '17 at 13:46
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    $\begingroup$ @goblin I don't know any literature on expanding the concept of improper integration. I guess one could consider locally compact spaces and measures a la Riesz representation and then a limit of a net indexed on the compact subsets. This could be somewhat useful, but that is a wild guess. I will expand in an answer if there is none in a while. $\endgroup$ – Aloizio Macedo Apr 2 '17 at 14:08
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    $\begingroup$ As a side thought, one might be interested in using $$\int_0^\infty\mu\{x~|~f(x)>t\}-\mu\{x~|~f(x)<-t\}~\mathrm dt$$ as an improper Lebesgue integral. $\endgroup$ – Simply Beautiful Art Jun 18 at 2:10
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Okay, after writing an answer for a long time I retract my comment: I don't think it is possible to give a canonical (in a sense which I'll explain soon) useful meaning to an "improper integral" in a general setting. I'll leave the point where the my answer broke as a reference (the previous answer can be seen in the end of the post).

Firstly, one important observation (in everything that follows, "integrable" means with respect to the Lebesgue sense): if $f: \mathbb{R}^n \to \mathbb{R}$ is integrable, then it is improperly integrable and to the same value (also, it doesn't depend on how you go to infinity). More precisely, let $A_i$ be any increasing sequence of sets such that $\bigcup A_i=\mathbb{R}^n$. Then - if $f$ is integrable - we have $$\lim \int_{A_i}f=\int_{\mathbb{R}^n}f .$$ This is a direct consequence of the dominated convergence theorem. The problem is when $f$ is not integrable (this is exactly like the contrast absolutely convergent/conditionally convergent. Even more so as we shall see.)

The case in $\mathbb{R}$ already shows that there is a huge issue: the "way" you take the limit matters (c.f. Cauchy principal value). Therefore, what makes sense to define is the following: having chosen an increasing sequence of sets $A_n$ such that $\bigcup A_n=X$ on a measure space $X$, let $$\int_X^{\mathrm{imp}(A_n)}f:=\lim \int_{A_n}f.$$ This depends on the choice of $A_n$, as it should. It is in this sense that I said that the definition is not "canonical". It coincides with the Lebesgue integral of $f$ if it is integrable (again due to the dominated covnergence theorem). Note however that the above definition cannot even define the improper integral $\int_{-\infty}^{\infty} f$ (in its usual definition $\int_a^{\infty}f+\int_{-\infty}^bf$) properly.

One important observation is that the way the sequence is conceived is important, since we are taking a limit. This is in strong relation to the way a series can conditionally converge and not absolutely converge (indeed, a series is just an integral on a countable set). Note however that there is the general concept of a summable family (c.f. here) which if inspected closely may resemble the definition I assembled below: in fact, it is precisely the definition below in the case where we take the discrete topology (the compact sets on the discrete topology are the finite sets) and the counting measure. The concept below seems to enjoy the same good property as the improper integral: it coincides (in $\mathbb{R}$, and I believe that also in any $\sigma$-compact space) with the integral if the function is integrable. However, it does not generalize the concept of improper integral as you wanted, since they don't coincide even in $\mathbb{R}$. It is "another" integral.


Consider a locally compact Hausdorff space $X$ and a regular measure $\mu$ on $X$. Considering $\leq$ the inclusion on the set $\mathcal{K}$ of compact subsets of $X$, we have that $\mathcal{K}$ is directed, since finite union of compact sets is compact. Hence, given a real function $f: X \to \mathbb{R}$, we have a net $\lambda: \mathcal{K} \to \mathbb{R}$ given by $$\lambda_K:=\int_K f d\mu.$$ Then, define $$\int_X^{\mathrm{imp}}f:=\lim \lambda. $$


This coincides with the improper Lebesgue integral in $\mathbb{R}$ (!!no!!): for example, Let $\int_{[0,\infty)}^\mathrm{impR}$ denote the improper Lebesgue integral. Suppose $\int_{[0,\infty)}^\mathrm{impR}f=L$. Let $\epsilon>0$. There exists $A>0$ such that if $x>A$ then $|L-\int_{[0,x)} f|<\epsilon/4$. Note that this implies that $|\int_{[x,y]} f| <\epsilon/2$ for every $x,y>A$, since $$|\int_{[x,y]} f| = |\int_{[0,y]}f-\int_{[0,x]}f|=|L-\int_{[0,x]} f+\int_{[0,y]} f-L| < \epsilon/2. $$ Take $K=[0,A+1]$. Given any $K'>K$, then $K' \subset [0,B]$ for some $B>A+1.$ It follows that $$|L-\int_{K'} f|=|L-\int_Kf+\int_K f-\int_{K'} f| \leq \epsilon/2+|\int_{K'-K}f|$$ Then, if $K'> K$.... the argument broke here. This won't work. Just take a non-integrable function and consider $K'$ a compact set big enough adjoining only portions where $f$ is positive. So, the notions don't coincide necessarily. However, it coincides if $f$ is integrable.

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  • $\begingroup$ Thanks, I really appreciate your thoughts! $\endgroup$ – goblin Apr 2 '17 at 23:05
  • $\begingroup$ Can you explain why it depends on the choice of $A_n$? $\endgroup$ – Keshav Srinivasan Oct 16 '18 at 3:19
  • $\begingroup$ Couldn't it be considered "canonical" to take the most symmetric sequences possible toward infinity, i.e., $A_i:=B_i(0):=\{x\in\mathbb{R}^n\;|\;d(x,0)\leqslant i\}$? $\endgroup$ – WillG May 22 at 16:26

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