The Unapologetic Mathematician

Convergence in Measure I

Suppose that $f$ and all the $f_n$ (for positive integers $n$) are real-valued measurable functions on a set $E$ of finite measure. For every $\epsilon>0$ we define

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

That is, $E_n(\epsilon)$ is the set where the value of $f_n$ is at least $\epsilon$ away from the value of $f$. I say that $\{f_n\}$ converges a.e. to $f$ if and only if

$\displaystyle\lim\limits_{n\to\infty}\mu\left(E\cap\bigcup\limits_{m=n}^\infty E_n(\epsilon)\right)=0$

for every $\epsilon>0$.

Given a point $x\in X$, the sequence $\{f_n(x)\}$ fails to converge to $f$ if and only if there is some positive number $\epsilon$ so that $x\in E_n(\epsilon)$ for infinitely many values of $n$. That is, if $D$ is the set of points where $f_n$ doesn’t converge to $f$, then

$\displaystyle D=\bigcup\limits_{\epsilon>0}\limsup\limits_{n\to\infty}E_n(\epsilon)=\bigcup\limits_{k=1}^\infty\limsup\limits_{n\to\infty}E_n\left(\frac{1}{k}\right)$

Of course, if $\{f_n\}$ is to converge to $f$ a.e., we need $\mu(E\cap D)=0$. A necessary and sufficient condition is that $\mu(E\cap\limsup\limits_{n\to\infty}E_n(\epsilon))=0$ for all $\epsilon>0$. Then we can calculate

\displaystyle\begin{aligned}\mu(E\cap\limsup\limits_{n\to\infty}E_n(\epsilon))&=\mu\left(E\cap\bigcap\limits_{n=1}^\infty\bigcup_{m=n}^\infty E_n(\epsilon)\right)\\&=\lim\limits_{n\to\infty}\mu\left(E\cap\bigcup\limits_{m=n}^\infty E_n(\epsilon)\right)\end{aligned}

Our necessary and sufficient condition is thus equivalent to the one we stated at the outset.

We’ve shown that over a set of finite measure, a.e. convergence is equivalent to this other condition. Extracting it a bit, we get a new notion of convergence which will (as we just showed) be equivalent to a.e. convergence over sets of finite measure, but may not be in general. We say that a sequence $\{f_n\}$ of a.e. finite-valued measurable functions “converges in measure” to a measurable function $f$ if for every $\epsilon>0$ we have

$\displaystyle\lim\limits_{n\to\infty}\mu\left(\left\{x\in X\big\vert\lvert f_n(x)-f(x)\rvert\geq\epsilon\right\}\right)=0$

Now, it turns out that there is no metric which gives this sense of convergence, but we still refer to a sequence as being “Cauchy in measure” if for every $\epsilon>0$ we have

$\displaystyle\mu\left(\left\{x\in X\big\vert\lvert f_m(x)-f_n(x)\rvert\geq\epsilon\right\}\right)\to0\quad\text{as }n,m\to\infty$

May 19, 2010 - Posted by | Analysis, Measure Theory

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7. among all the modes of convergence, which one is the strongest and which one is the weakest?

Comment by cyprian | February 25, 2011 | Reply

8. In what context? Not all modes of convergence are defined at the same time.

Comment by John Armstrong | February 25, 2011 | Reply