After struggling for some time to find an appropriate definition for the notion of integration I came across another attempt for which I would need your help deciding to what extend this can or cannot be reasonable.


The Lebesgue integral as most of you know can be interpreted as some sort of absolute integrability. Very much the same holds true for the Bochner integral that - being equivalent to the Lebesgue integral - makes this fact even more light up. While studying the concept of summation I noticed similarities of this technique within infinite sums. That is in order to handle infinite sums and their conditional convergence one introduces a Banach space structure. Then by absolute convergence or similar square summability these can be controlled so to guarantee unconditional convergence. However, taking a closer look on the concept of summation itself one understands that these problems merely arise as a missuse of sums by series giving all sort of problems of conditional convergence. So I started wondering is the analogues problem happening within the Lebesgue integral...

Starting from sums one might guess that nets can handle the problem (of course one has to be very careful as measures are considered as 'only' countably additive). Now the big problem stands in the middle: If an appropriate concept can give a satisfactory notion of integral what will be the ordering giving nets then?

A first try will give the Riemann integral. Passing from halfopen partitions to measurable partitions pathological examples like the Dirichlet function are managed while keeping continuous functions over compact sets integrable. However unbounded functions like the reciprocal squareroot stay untouched. So one has to work harder. Thinking about the problems one understands that there is two conceptual problems, unboundedness of domain and unboundedness of range. The first one can be handled rather easily by first allowing naturally only finite measure spaces as for example in the case of complex measures and then extending by improper integrals by the use of the net of all finite subspaces (not only some sequence). This immediately rules out the possibility to salvage the reciprocal squareroot as an improper integral as other unbounded functions. Now, any way of introducing an ordering on tagged partitions of the domain fails to handle general unbounded functions (a control of unbounded functions taking values in a topological vector space becomes even harder).

Another attempt is to introduce a gauge. Though being very very powerful over compact intervals they heavily rely on the compactness of the interval and moreover on the measure space being Borel. So one has hard chance to generalize the gauge integral for example to the situation faced in the spectral theorem. The very much the same problem arises for any kind of gauge integral...

So one must find another way to attack unboundedness of the range. A guess then is to of course as Lebesgue already did so somehow partition the range - this time however conceptually similar to the case of summability as described above.

Once more said: The idea is to find a concept similar to summability without absolute convergence!


Here comes the second attempt...

Let $\Omega$ be a finite measure space and $V$ a topological vector space and $f:\Omega\to V$.

Call a triple $\rho:=(S,U;A_s)$ a resolution w.r.t. $f$ with:
$S$ a finite subset of $V$, $U$ an open neighborhood of zero and $A_s$ disjoint measurable sets of $\Omega$ if:

  • $A_s\subseteq f^{-1}(U+s)$
  • $\bigcup_{s\in S}A_s=f^{-1}(U+S)$

(This should not be interpreted as a gauge but rather as a partition.)

Introduce an ordering $\rho\leq\rho'$ on the resolutions by:

  • $S\subseteq S'$
  • $U\supseteq U'$
  • $f^{-1}(U+S)\subseteq f^{-1}(U'+S')$

(The ordering is directed.)

Construct the net of sums by: $$I(f;\rho):=\sum_{s\in S}\mu(A_s)s$$ (The neighborhood $U$ takes the role of relating measures to vectors.)

Define $f$ to be integrable with integral $I(f)$ if: $$I(f;\rho)\to I(f)$$ (The integral being unique provided $V$ being Hausdorff.)

Now these definitions seem horribly complicated but they follow a few simple ideas:

  • The central idea here is to focus on vectors as the major object within the sums also since the integral itself is one, so start with a finite sum of vectors $s\in S$
  • The problem now is to relate vectors to sets, this is basically done by the neighborhood of zero $U$. Having a topological vector space makes $U+s$ a neighborhood of $s$.
  • The integral of a function should involve the function itself of course, that is the preimage of the function under consideration relates the neighborhoods $U+s$ to measurable subsets $A_s$. The requirement that $U$ be better open becomes evident here.
  • Finally these measurable sets weight the individual summands by their mass $\mu(A_s)$ and closes the circle vector-set-set-number-vector. Here again having a vector space makes it evident to allow/restrict to complex measures rather than positive ones (for nonfinite measure spaces see doscussion in the context)

The rest is technicalities...

Moreover, these can be interpreted as an approximation by simple functions. The nice thing in here is that it automatically sorts out bad steps like blocks escaping to infinity or towers blowing up as peaks...

Lebesgue Integral


As a first check:

  • Is $f(x)=c$ for some $c\in V$ integrable over $[0,1]$ with integral $I(f)=c\cdot 1=c$?
  • Is $f(x)=x$ integrable over $[0,1]$ with integral $I(f)=\frac12\cdot1^2=\frac12$?
  • Is $f(x)=\frac{1}{\sqrt{x}}$ integrable over $[0,1]$ with integral $I(f)=2\cdot\sqrt{1}=2$?

Now comes the dry (interesting) part:

  • Does it include simple functions resp. continuous functions over compact sets?
  • Does it include/extend the Riemann integral resp. Lebesgue integral resp. McShane integral?
  • Does it allow convergence theorems resp. the fundamental theorem of calculus?
  • Does it allow computational methods?
  • Etc. ...

Here I really could need your help =S

  • 7
    $\begingroup$ +1 Even though this is clearly a well-thought question, perhaps dividing it into different parts (or trying to ask a more specific point) would increase the chances of you getting an answer. Not many people would go over your definitions, take care of the technicalities that you mentioned, and finally perhaps answer your arguably very broad questions. This was just a point I wanted to make so as to perhaps increase the chances of you eventually resolving your questions on your integral. Good luck! $\endgroup$
    – Lord Soth
    Aug 3, 2014 at 20:52
  • $\begingroup$ @LordSoth: First of all thanks for your compliments :) That's a good point: Will definitely think about it! (I could keep this as the thread with the full question and add threads focusing on the several points with a reference hereto.) $\endgroup$ Aug 3, 2014 at 21:02
  • 1
    $\begingroup$ Where is Part ${\tt I}$ ?. $\endgroup$ Aug 24, 2014 at 1:46
  • $\begingroup$ @FelixMarin: Part 1 was closed after i found a lack in it (maybe you can still find it via google) $\endgroup$ Aug 24, 2014 at 7:30
  • $\begingroup$ @FelixMarin: Here's the link: math.stackexchange.com/questions/788301/… $\endgroup$ Aug 24, 2014 at 7:41

1 Answer 1


Without loss of generality, assume that we are dealing with a probability space $(\Omega,\Sigma,\mu)$. Let's call a function F-integrable if it satisfies the definition.

Every simple function in a Hausdorff locally convex topological vector space is F-integrable.

Proof: Let $f$ be simple and $V$ be a neigborhood of $0$. Let $n=|f(\Omega)|$ and let $U$ be a balanced convex neighborhood of zero such that $U\subseteq V$ and such that $x-y\notin U+U$ for all $x,y\in f(\Omega)$. Let $S=f(\Omega)$, let $A_s=f^{-1}(s+U)$. I claim that for $\rho\geq(S,U,(A_s))$, $$I(f,\rho)\in\sum_{s\in f(\Omega)}\mu\Big(f^{-1}\big(\{s\}\big)\Big)+V.$$ So let $\rho=(S',U'(B_s))$. For each $s\in f(\Omega)$, let $G(s)=\{y\in S':y\in s+U\}$. For all $y\notin S'\backslash\bigcup_{s\in f(\Omega)}G(s)$, we have $\emptyset=f^{-1}(y+U)\supseteq f^{-1}(y+V)$ and hence $B_y=\emptyset$. Also, note that $G(s)\cap G(s')=\emptyset$ for $s\neq s'$.

We have therefore $$I(f,\rho)=\sum_{s\in f(\Omega)}\sum_{y\in G(s)}\mu(B_y)y.$$ Now, for each $s\in f(\Omega)$, we have $\bigcup_{y\in G(s)}B_y\subseteq A_s=f^{-1}(s+U)$ and hence $$\mu\bigg(\bigcup_{y\in G(s)}B_y\bigg)\leq\mu(A_s).$$ If one of these inequalities were strict, the $B_y$ could not form a partition of $\Omega=f^{-1}(S'+U')$. So they are all equalities and for each $s\in f(\Omega)$, the point $\sum_{y\in G(s)}\mu(B_y)y$ is a convex combination of a point $y_s\in s+U$ multiplied by $\mu(A_s)=\mu\big(f^{-1}(s+U)\big)$. So $$I(f,\rho)=\sum_{s\in f(\Omega)}\mu\Big(f^{-1}\big(\{s\}\big)\Big)y_s.$$ Now $y_s-s\in U$, so $$I(f,\rho)-\sum_{s\in f(\Omega)}\mu\Big(f^{-1}\big(\{s\}\big)\Big)s$$

$$=\sum_{s\in f(\Omega)}\mu\Big(f^{-1}\big(\{s\}\big)\Big) (y_s-s),$$ which is a convex combination of elements of $U$, since $\mu$ is a probability measure. $\square$

To be continued...

  • $\begingroup$ Me terrible: I forgot to upvote the nice answer.. :O Thank you!!! :) $\endgroup$ Dec 12, 2016 at 22:49

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