8
$\begingroup$

TL:DR; For arbitrary sets (not necessarily countable) we have two notions: Lebesgue integral w.r.t. the counting measure and sum of family indexed by this set. Are these two notions equivalent?


For an arbitrary set $X$ we can define counting measure simply by putting $\mu(A)=|A|$. (I.e., if $A$ is finite then $\mu(A)$ is simply number of elements of $A$; otherwise it is $+\infty$.) In this way we get a $\sigma$-additive measure on $\mathcal P(X)$ and it is possible to work with Lebesgue integral with respect to this measure.

If the integral $$\int f \;\mathrm{d} \mu$$ of a function $f\colon X\to\mathbb R$ exists, it is natural to interpret this integral as a sum of the values $f(x)$ over all $x\in X$.


There is also a (more-or-less standard) notion of a sum of values on a given set which includes uncountable sets. Let me briefly recall the definition. (Below I will add a few links to other posts on this site where this definition can be found.)

Definition. Let $f\colon X\to\mathbb R$ be a function and $S\in\mathbb R$. We say that $$\sum_{x\in X} f(x) = S$$ if and only if for every $\varepsilon>0$ there exists a finite set $F_0$ such that for all finite sets $F\supseteq F_0$ we have $\left| \sum\limits_{x\in F} f(x) - S \right| < \varepsilon$. $$(\forall \varepsilon>0) (\exists F_0\text{ finite }) \left(F\text{ is finite and }F\supseteq F_0 \Rightarrow \left| \sum\limits_{x\in F} f(x) - S \right| < \varepsilon \right)$$

Some further remarks:

  • We can modify the above definition in a natural way to be able to say when $\sum f(x)=+\infty$ and $\sum f(x)=-\infty$.
  • If we work with non-negative values, i.e., $f(x)\ge0$, then we get a much simpler equivalent definition $$\sum_{x\in X} f(x) = \sup \{\sum_{x\in F} f(x); F\text{ is finite}\}.$$
  • This type of sum is also defined in the Wikipedia article about series: Summations over arbitrary index sets (current revision).
  • This definition can be interpreted nicely using convergence of nets. We take the directed set consisting of finite subsets of $X$ ordered by inclusion. For every such finite set we have the value $s_F=\sum_{x\in F} f(x)$. The sum as defined above is equal to $S$ iff $S$ is the limit of this net.
  • With this definition, the distinction between conditional and absolute convergence no longer makes sense. (Which is natural, since we do not take any kind of ordering on $X$ into account.) In particular, in the case $X=\mathbb N$ this correspond to definition of sum of absolutely convergent series. (At least if we work with real values. In more general contexts, it can happen that unconditional convergence and absolute convergence might be different. This definition of sum corresponds to unconditional convergence.)
  • A variant of Cauchy's criterion for such sums can be shown.
  • The same definition can be used in more general settings. (You still probably need the structure to be at least a topological group if you expect the sum to behave reasonably.)

Question. Is the sum as defined above equivalent to the notion of Lebesgue integral with respect to the counting measure?

In particular I would like to know:

  • Are there some problems that arise if we work with uncountable sets, rather than just with countable ones?
  • Are there any specific problems if I also allow negative values?

I would be grateful for both references to some texts which deal with relationship between these two notions. And, of course, for a proof (or sketch of a proof) if this is sufficiently simple to fit into a post on this site.


I have checked whether something about this is mentioned in the Wikipedia article Counting measure. This kind of sum is mentioned there, but in a slightly different context. The current revision of the Wikipedia article says that:

The counting measure is a special case of a more general construct. With the notation as above, any function $f \colon X \to [0, \infty)$ defines a measure $\mu$ on $(X, \Sigma)$ via $$\mu(A):=\sum_{a \in A} f(a)\, \forall A\subseteq X,$$ where the possibly uncountable sum of real numbers is defined to be the sup of the sums over all finite subsets, i.e., $$\sum_{y \in Y \subseteq \mathbb R} y := \sup_{F \subseteq Y, |F| < \infty} \left\{ \sum_{y \in F} y \right\}.$$ Taking $f(x)=1$ for all $x$' in $X$ produces the counting measure.


Some related links:

$\endgroup$
  • $\begingroup$ I've found myself wondering about this from time to time. Great post! Hopefully someone can answer in the affirmative. Here's a possibly dumb question. Could you consider some Banach space and work this there since absolute and unconditional convergence might be different? $\endgroup$ – Cameron Williams Apr 28 '18 at 15:12
  • $\begingroup$ @CameronWilliams Definitely the notion of sum over arbitrary set can be defined in the same way in Banach spaces. (In fact, this is done in some of the references given in this answer. For example, Dixmier works with normed spaces.) Since I do not know much about integration of vector-valued functions, I can't really say whether there is a chance that the question I asked really makes sense in that context, too. I.e., whether we could treat the sum as some kind of integral. (But my guess would be that probably yes.) $\endgroup$ – Martin Sleziak Apr 28 '18 at 15:21
2
$\begingroup$

If $f$ is non-negative, then the fact that $$\sum_{x\in X} f(x) = \sup \{\sum_{x\in F} f(x); F\text{ is finite}\}$$ tells us that $\int_X f=\sum\limits_{x \in X} f$ by definition of the Lebesgue integral as the $\sup$ of the integral of lower simple functions (not exactly by definition, but every simple function $\sum\limits_{x \in F}s(x)\chi_{\{x\}}$ which is lower than $f$ is lower than $\sum\limits_{x \in F}f(x)\chi_{\{x\}}$, and thus the sup of the definition of the Lebesgue integral in this case is equal to the $\sup$ above).

Thus, if $f$ is integrable, it holds that $\int_X f=\sum\limits_{x \in X} f$.*

If $f$ is not integrable, then wlog we can suppose that $\int_X f^+=+\infty$. This implies that there exists a countable set $C$ inside $X$ on which $f$ is positive and $\sum\limits_{x \in C}f(x)=+\infty$.

Thus, given any $S \in \mathbb{R}$, given $\epsilon=1$ and any finite $F_0 \subset X$ we can take a big enough finite $F'$ (which can be taken disjoint from $F_0$) inside $C$ such that $\sum\limits_{x \in F'}f(x) >S+1-\sum\limits_{x \in F_0} f(x)$, and thus \begin{align*} \sum_{x \in F' \cup F_0} f(x)&=\sum_{x \in F'} f(x)+\sum_{x \in F_0} f(x)>S+1, \end{align*} contradicting the definition of $\sum\limits_{x \in X} f(x)=S$.

It follows that $f$ is integrable if and only if it is "summable", in the sense you define in your Definition.

*For this to hold, we need to prove that if $\sum\limits_{x \in X} f^+=S_1$ and $\sum\limits_{x \in X} f^-=S_2$, then $\sum\limits_{x \in X} f=S_1-S_2$.

Let $\epsilon>0$. Take a finite set $F_0$ for $f^+$ and $F_1$ for $f^-$, corresponding to $\epsilon/2$ with respect to the definition of summable. Note that we can take those to be disjoint, since they come from the positive and negative parts. Now, let $F \supset F_0 \cup F_1$ be finite. Then \begin{align*} |\sum_{x \in F}f-S_1+S_2|&= |\sum_{x \in F-F_1}f-S_1+\sum_{x \in F_1}f+S_2|\\ &< \epsilon, \end{align*} as desired.


Alternatively, we can prove that $f \in L^1(X)$ implies that $f$ is summable by reducing to the countable case.

Note that $\operatorname{supp} f :=\{x \in X \mid f(x) \neq 0\}$ is at most countable. This follows by considering $X_n:=\{x \in X \mid |f(x)| \geq 1/n\}$, for then $$\operatorname{card} X_n \leq \int_{X_n} |nf| = n\int_X |f|<\infty. $$ Since $\operatorname{supp} f=\bigcup X_n,$ it follows that it is countable. Thus, $f$ being integrable allows us to reduce to the countable case and to infer the result that $f \in L^1(X) \implies f$ is summable.

$\endgroup$
  • $\begingroup$ Thanks for the answer. Just to clarify: In the whole answer you work only with non-negative $f$? By summable/integrable you mean that there exists finite sum/integral? $\endgroup$ – Martin Sleziak Apr 28 '18 at 16:21
  • 1
    $\begingroup$ @Martin Sleziak No, in the first part of the answer I work with non-negative $f$. In the last part (when I suppose that $f$ is not integrable), I don't restrict to non-negative $f$. However, I now see a point which I didn't see before: the step "Thus, if $f$ is integrable, it holds that $\int_X f=\sum\limits_{x \in X} f$" is not so justified. We would need to prove a similar separation of the summation you define in positive and negative parts. I will try to address this now. $\endgroup$ – Aloizio Macedo Apr 28 '18 at 16:25
  • $\begingroup$ I don't understand why $\operatorname{card}X_n=\int_{X_n}\vert nf\vert.$ If you use counting measure, we have $\operatorname{card}X_n=\int_{X}1_{X_n}(x)d\mu(x).$ Can you explain ? Thanks $\endgroup$ – user546361 Apr 29 '18 at 16:32
  • 1
    $\begingroup$ @AlexHF Yes, it was supposed to be an inequality, since $|nf| \geq 1$ on $X_n$. I made the correction, thanks. $\endgroup$ – Aloizio Macedo Apr 29 '18 at 16:43

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.