The Unapologetic Mathematician

Mathematics for the interested outsider

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

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