Is Lebesgue integral w.r.t. counting measure the same thing as sum (on an arbitrary set)?

Solution 1:

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.