## Indefinite Integrals and Convergence II

Unlike our recent results, today’s proposition is specifically stated and proved for integrable *simple* functions, and won’t be generalized later.

If and are mean Cauchy sequences of integrable simple functions, then they’re both also Cauchy in measure, which implies that they each converge in measure to some function. If they converge to the *same* function (a.e.) , then their indefinite integrals converge to the same limiting set function. That is, if and are the indefinite integrals of and :

then we can define the limiting functions

and we assert that for all measurable .

For every and positive integer we define the set

And using our usual technique we find

Since and both converge in measure to , the measures of both terms here go to zero as gets large, and so .

If is a measurable set with , we have the inequality

Here, the first term on the right is bounded above by . The other two terms can be made arbitrarily small by choosing a large enough , by the uniform absolute continuity we showed yesterday. We can also see that

and so it follows that , and thus that for every measurable set with finite measure. Since and are countably additive, we immediately extend this result to all -finite sets .

Okay, now here’s where our assumption really comes in: since each of the and is an integrable simple function, each one takes a nonzero value on a finite number of sets, each of which has a finite measure. Thus we can take the union of *all* these (countably many) sets, which is a -finite set. For any measurable set , then, we have

because all the functions and are identically zero off of .

Therefore we conclude that . This, then, implies that and , and each of these sets is then -finite (as subsets of ). And so the proof is complete.

[…] we know that since and both converge in measure to the same function, the limiting set functions and […]

Pingback by Integrable Functions « The Unapologetic Mathematician | June 2, 2010 |