## Convergence in Measure I

Suppose that and all the (for positive integers ) are real-valued measurable functions on a set of finite measure. For every we define

That is, is the set where the value of is at least away from the value of . I say that converges a.e. to if and only if

for every .

Given a point , the sequence fails to converge to if and only if there is some positive number so that for infinitely many values of . That is, if is the set of points where doesn’t converge to , then

Of course, if is to converge to a.e., we need . A necessary and sufficient condition is that for all . Then we can calculate

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 of a.e. finite-valued measurable functions “converges in measure” to a measurable function if for every we have

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 we have

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Pingback by Completeness of the Metric Space of a Measure Space « The Unapologetic Mathematician | August 9, 2010 |

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

Comment by cyprian | February 25, 2011 |

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

Comment by John Armstrong | February 25, 2011 |