# The Unapologetic Mathematician

## 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