## Indefinite Integrals and Convergence I

Let’s see how the notion of an indefinite integral plays with sequences of simple functions in the norm.

If is a mean Cauchy sequence of integrable simple functions, and if each has indefinite integral , then the limit exists for all measurable sets . Indeed, for each we have a sequence of real numbers . We compare

and find that since the sequence of simple functions is mean Cauchy the sequence of real numbers is Cauchy. And thus it must converge to a limiting value, which we define to be . In fact, the convergence is uniform, since the last step of our inequality had nothing to do with the particular set !

Now, this set function is finite-valued as the uniform limit of a sequence of finite-valued functions. Since limits commute with finite sums, and since each is finitely additive, we see that is finitely additive as well; it turns out that it’s actually countable additive.

If is a disjoint sequence of measurable sets whose (countable) union is , then for every pair of positive integers and the triangle inequality tells us that

Choosing a large enough we can make the first and third terms arbitrarily small, and then we can choose a large enough to make the second term arbitrarily small. And thus we establish that

We can say something about the sequence of set functions : each of them is — as an indefinite integral — absolutely continuous, but in fact the sequence is *uniformly* absolutely continuous. That is, for every there is a independent of so that for every measurable set with .

Let be a sufficiently large integer so that for we have

which exists by the fact that is mean Cauchy. Then we can pick a so that

for all and with . We know that such a exists for each by absolute continuity, and so we just pick the smallest of them for .

This will then work for all , but what if ? Well, then we can write

and so the same works for all as well.

[…] we can define the limiting […]

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[…] know that the indefinite integrals of the are uniformly absolutely continuous, so we have control over the size of the first term on the right. The second term on the right can […]

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[…] also have a couple more results whose proofs don’t really depend on the simplicity of , and so these carry across without […]

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