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

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

8 Comments »

  1. [...] proposition we started with yesterday shows us that on a set of finite measure, a.e. convergence is equivalent to convergence in [...]

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  4. [...] and 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 [...]

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  6. [...] of characteristic functions must converge in mean to some function . But mean convergence implies convergence in measure, which is equivalent to a.e. convergence on sets of finite measure, which is what we’re [...]

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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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