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Write $L^1$ for the Banach space $L^1([0, 1])$. Given $f \in L^1$, define $f_1, f_2 \in L^1$ by $$ f_1(x) = f(x/2), \qquad f_2(x) = f((x + 1)/2). $$ Let $I = \int_0^1$. Then $I$ is the unique bounded linear functional on $L^1$ satisfying the equations $$ I(\text{constant function at }1) = 1, \qquad I(f) = (I(f_1) + I(f_2))/2. $$ This, then, is a simple characterization of integration.

Presumably this has been known for a really long time -- maybe for a century? -- but I'm having trouble tracking it down in the literature. I simply can't find it anywhere.

Does anyone know anything about the history of this result?

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The only writeup on this topic I've seen is yours. Probably you know about that one already... – Pete L. Clark Jan 3 '11 at 15:31
Well, the integral is already been used to define what $L^1$ is. A simpler way to state this would be to say that the Lebesgue measure is the unique one such that the measure of a set $A\subseteq[0,1]$ is the same as the average of those of $A/2$ and $(A+1)/2$. – Mariano Suárez-Alvarez Jan 3 '11 at 15:32
Pete: yes, that's what I thought you might have meant. It's here: Guess I should tidy up my web page a bit... – Tom Leinster Jan 3 '11 at 16:07
I don't know about that result, but the the same result for continuous functions is due to Freyd ( If I learned about that paper from a blog post of yours on the n-Category Café, I apologize. – Omar Antolín-Camarena Jan 3 '11 at 16:14
@Mariano: L^1([0, 1]) can be defined as the abstract Banach space completion of, say, the step functions on [0, 1] with the L^1 norm (which has a purely finitary description). This definition technically doesn't require any knowledge of integration. – Qiaochu Yuan Jan 3 '11 at 19:27
up vote 3 down vote accepted

I don't know if this is of help, but I have seen this idea for defining integration elsewhere, specifically on pages 10-11 of Reed and Simon's Functional Analysis. It would go something like this: let $S$ be the space of step functions obtained as linear combinations of characteristic functions of half-open intervals $[k/2^n, (k+1)/2^n)$. Then $S$ is dense in the space $C$ of bounded piecewise continuous functions continuous to the right, with respect to sup norm. By the two conditions, the definition of $I$ is uniquely determined on $S$ and defines a bounded linear functional on $S$ with respect to sup norm, so it uniquely extends to a bounded linear functional on $C$. This in fact defines the Riemann integral $I$, and it's pretty much exactly what Reed and Simon do, except they don't restrict to dyadic rational endpoints.)

(Then, we can go on and define $L^1[0, 1]$ to be the completion of $C$ with respect to the norm given by $f \mapsto I(|f|)$, and since $I$ is a bounded linear functional also with respect to this norm, it extends uniquely to the completion $L^1[0, 1]$.)

Edit:: Ah, Reed and Simon mention that this approach to the Riemann integral can be found especifically in Dieudonné, Foundations of Modern Analysis, or in Loomis and Sternberg, Advanced Calculus, Addison-Wesley, 1968. Perhaps one can consult these for further history.

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Thanks, Todd! That's an excellent start, and Dieudonné sounds especially likely to be useful. – Tom Leinster Jan 3 '11 at 23:02

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