The Unapologetic Mathematician

Mathematics for the interested outsider

Indefinite Integrals and Convergence I

Let’s see how the notion of an indefinite integral plays with sequences of simple functions in the L^1 norm.

If \{f_n\} is a mean Cauchy sequence of integrable simple functions, and if each f_n has indefinite integral \nu_n, then the limit \nu(E)=\lim_n\nu_n(E) exists for all measurable sets E\subseteq X. Indeed, for each E we have a sequence of real numbers \{\nu_n(E)\}. We compare

\displaystyle\begin{aligned}\lvert\nu_n(E)-\nu_m(E)\rvert&=\left\lvert\int\limits_Ef_n\,d\mu-\int\limits_Ef_m\,d\mu\right\rvert\\&=\left\lvert\int\limits_Ef_n-f_m\,d\mu\right\rvert\\&\leq\int\limits_E\lvert f_n-f_m\rvert\,d\mu\\&\leq\int\limits\lvert f_n-f_m\rvert\,d\mu\end{aligned}

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

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

If \{E_n\} is a disjoint sequence of measurable sets whose (countable) union is E, then for every pair of positive integers n and k the triangle inequality tells us that

\displaystyle\left\lvert\nu(E)-\sum\limits_{i=1}^k\nu(E_i)\right\rvert\leq\left\lvert\nu(E)-\nu_n(E)\right\rvert+\left\lvert\nu_n(E)-\sum\limits_{i=1}^k\nu_n(E_i)\right\rvert+\left\lvert\nu_n\left(\bigcup\limits_{i=1}^kE_i\right)-\nu\left(\bigcup\limits_{i=1}^kE_i\right)\right\rvert

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

\displaystyle\nu(E)=\lim\limits_{k\to\infty}\sum\limits_{i=1}^k\nu(E_i)=\sum\limits_{i=1}^\infty\nu(E_i)

We can say something about the sequence of set functions \{\nu_n\}: each of them is — as an indefinite integral — absolutely continuous, but in fact the sequence is uniformly absolutely continuous. That is, for every \epsilon>0 there is a \delta independent of n so that \lvert\nu_n(E)\rvert<\epsilon for every measurable set E with \mu(E)<\delta.

Let N be a sufficiently large integer so that for n,m\geq N we have

\displaystyle\int\lvert f_n-f_m\rvert\,d\mu<\frac{\epsilon}{2}

which exists by the fact that \{f_n\} is mean Cauchy. Then we can pick a \delta so that

\displaystyle\lvert\nu_n(E)\rvert=\left\lvert\int\limits_Ef_n\,d\mu\right\rvert\leq\int\limits_E\lvert f_n\rvert\,d\mu<\frac{\epsilon}{2}

for all n\leq N and E with \mu(E)<\delta. We know that such a \delta exists for each f_n by absolute continuity, and so we just pick the smallest of them for n\leq N.

This \delta will then work for all n\leq N, but what if n>N? Well, then we can write

\displaystyle\begin{aligned}\lvert\nu_n(E)\rvert&=\left\lvert\int\limits_Ef_n\,d\mu\right\rvert\\&\leq\int\limits_E\lvert f_n\rvert\,d\mu\\&\leq\int\limits_E\lvert f_n-f_N\rvert+\lvert f_N\rvert\,d\mu\\&\leq\int\limits_E\lvert f_n-f_N\rvert\,d\mu+\int\limits_E\lvert f_N\rvert\,d\mu\\&<\frac{\epsilon}{2}+\frac{\epsilon}{2}=\epsilon\end{aligned}

and so the same \delta works for all n>N as well.

May 31, 2010 Posted by | Analysis, Measure Theory | 3 Comments

   

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