# The Unapologetic Mathematician

## 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 $\{f_n\}$ and $\{g_n\}$ 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.) $f$, then their indefinite integrals converge to the same limiting set function. That is, if $\nu_n$ and $\lambda_n$ are the indefinite integrals of $f_n$ and $g_n$:

\displaystyle\begin{aligned}\nu_n(E)&=\int\limits_Ef_n\,d\mu\\\lambda_n(E)&=\int\limits_Eg_n\,d\mu\end{aligned}

then we can define the limiting functions

\displaystyle\begin{aligned}\nu(E)&=\lim\limits_{n\to\infty}\nu_n(E)\\\lambda(E)&=\lim\limits_{n\to\infty}\lambda_n(E)\end{aligned}

and we assert that $\nu(E)=\lambda(E)$ for all measurable $E\subseteq X$.

For every $\epsilon>0$ and positive integer $n$ we define the set

$\displaystyle E_n=\left\{x\in X\big\vert\lvert f_n(x)-g_n(x)\rvert\geq\epsilon\right\}$

And using our usual technique we find

$\displaystyle E_n\subseteq\left\{x\in X\bigg\vert\lvert f_n(x)-f(x)\rvert\geq\frac{\epsilon}{2}\right\}\cup\left\{x\in X\bigg\vert\lvert f(x)-g_n(x)\rvert\geq\frac{\epsilon}{2}\right\}$

Since $\{f_n\}$ and $\{g_n\}$ both converge in measure to $f$, the measures of both terms here go to zero as $n$ gets large, and so $\lim_n\mu(E_n)=0$.

If $E$ is a measurable set with $\mu(E)<\infty$, we have the inequality

$\displaystyle\int\limits_E\lvert f_n-g_n\rvert\,d\mu\leq\int\limits_{E\setminus E_n}\lvert f_n-g_n\rvert\,d\mu+\int\limits_{E\cap E_n}\lvert f_n\rvert\,d\mu+\int\limits_{E\cap E_n}\lvert g_n\rvert\,d\mu$

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

$\displaystyle\lvert\nu_n(E)-\lambda_n(E)\rvert=\left\lvert\int_Ef_n\,d\mu-\int_Eg_n\,d\mu\right\rvert=\left\lvert\int_Ef_n-g_n\,d\mu\right\rvert\leq\int\limits_E\lvert f_n-g_n\rvert\,d\mu$

and so it follows that $\lim_n\lvert\nu_n(E)-\lambda_n(E)\rvert=0$, and thus that $\nu(E)=\lambda(E)$ for every measurable set $E$ with finite measure. Since $\nu$ and $\lambda$ are countably additive, we immediately extend this result to all $\sigma$-finite sets $E$.

Okay, now here’s where our assumption really comes in: since each of the $f_n$ and $g_n$ 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 $E_0$ of all these (countably many) sets, which is a $\sigma$-finite set. For any measurable set $E$, then, we have

$\displaystyle\nu_n(E\setminus E_0)=\int\limits_{E\setminus E_0}f_n\,d\mu=0=\int\limits_{E\setminus E_0}g_n\,d\mu=\lambda_n(E\setminus E_0)$

because all the functions $f_n$ and $g_n$ are identically zero off of $E_0$.

Therefore we conclude that $\nu_n(E\setminus E_0)=0=\lambda_n(E\setminus E_0)$. This, then, implies that $\nu(E)=\nu(E\cap E_0)$ and $\lambda(E)=\lambda(E\cap E_0)$, and each of these sets is then $\sigma$-finite (as subsets of $E_0$). And so the proof is complete.

June 1, 2010 Posted by | Analysis, Measure Theory | 1 Comment