# Existence of limit for convergence by measure for Cauchy-in-measure sequence+completeness of metric space?

Sorry if this is the wrong place to put it. But this question come from a graduate level textbook and seems pretty hard to me, so I hope this is a good place.

Anyway, this come from the book Real analysis for graduate students (Bass). Note that in this book they deal mainly with real-valued function.

From exercise 10.1:

A sequence of measurable function $f_{n}$ is Cauchy in measure if for any $a,\epsilon>0$ then there exist an $N$ such that for all $m,n\geq N$ then $\mu(\{x:|f_{n}(x)-f_{m}(x)|>a\}<\epsilon$.

A sequence of measurable function $f_{n}$ converge in measure to measurable $f$ if for any $a,\epsilon>0$ then there exist an $N$ such that for all $n\geq N$ then $\mu(x:|f_{n}(x)-f(x)|>a\}<\epsilon$.

Now this question ask that, given any sequence that is Cauchy in measure, prove that it must converge in measure.

As you can see, the question do not specify any function $f$ to converge in measure to, so I assume that "converge in measure" means "converge in measure to $f$ for some measurable $f$". Since the definition of converge in measure require an $f$, the first step would be to acquire one somehow.

So here is what I got so far:

Constructing $f$ through finite arithmetic operation on these $f_{n}$ also fail. These $f_{n}$ can stay in the realm of rationals and $f$ can still be irrational.

Constructing $f$ through pointwise limit fail. The sequence can diverge pointwise everywhere and still have an $f$ to converge-by-measure to.

Every theorem in the book already assume existence of $f$, so they are of no help.

These $f_{n}$ can be unbounded, or even not integrable, yet an $f$ still exist. And $\mu$ is not required to be $\sigma$-finite(edit: figured out that the problem is reducible to finite measure case; but still can't bound the functions). So this basically throw a wrench into any method that try to use integration.

"Average out" (that is $f(x)=\lim\limits_{n\rightarrow\infty}\frac{\sum\limits_{i=1}^{n}f_{i}(x)}{n}$) also fail, as it can produce the wrong $f$.

If $f$ exist, it is unique under the equivalent relation of almost everywhere equal.

I am at my wit's end here, what's with all the technique rendered useless.

Right now I am trying this direction, but I don't have much hope for this:

Define $d([f],[g])=\min\{\epsilon:\mu(\{x:|f(x)-g(x)|>\epsilon\})\leq\epsilon\}$ where the equivalence class is due to equal-almost-everywhere relation. This is not completely well-defined, as the set taking the minimum over might be empty. However, using Cauchy-in-measure property allow us to get to a point in the sequence where we can generate a metric space containing all the remaining function in the sequence; also, the sequence would then be Cauchy in this metric too. Now if I were to be able to prove that this metric space is complete in a way that does not require me to prove the original question, then I can use it to do the problem. However, this space is clearly not compact, so that's one proof technique out (edit: wait, this space is totally bounded; so if it are to be complete, it should be compact. Still, not sure how to prove that it is compact)(edit: nevermind it's not actually totally bounded, I just defined it too small).

EDIT: I am investigating the possibility of filtering out a subsequence that converge pointwise almost everywhere. However, it seems like there are many possible subsequence that can converge to many different function. So not much hope here, but I am trying to see if it can be proved that all these function are equal almost everywhere.

So can someone shed a light on this? Any helps would be appreciated. Thank you very much.

## migrated from mathoverflow.netOct 20 '13 at 22:11

This question came from our site for professional mathematicians.

• Oh sorry, I was not so sure if it should belong there or here, but it just seems pretty hard. – Gina Oct 21 '13 at 4:44