What is the relation between completeness in a measure space $(\Omega,\mathcal{A},\mu)$ (i.e. every subset of a nullset is measurable, i.e. in $\mathcal{A})$ and completeness in a normed space $(X, \lVert\cdot\rVert)$ (i.e. every Cauchy sequence in $X$ has a limit in $X)$?

I am unable to make that link. Thank you for your help!

My question is very much like this question without an answer.

Here people are saying the two concepts are unrelated without reasoning, which seems strange to me.

My try:

By definition, every Cauchy sequence in $X$ has a limit in $X$ means if $(x_n)_{n\in\mathbb{N}}$ is a sequence in $X$ s.t. for all $\varepsilon>0$, there exists some $n_0(\varepsilon)$ with $\lVert x_n-x\rVert<\varepsilon$ for $m,n\ge n_0(\varepsilon)$, then there exists some $x\in X$ with $\lim\limits_{x\to\infty}\lVert x_n\to x\rVert=0$.

Now I think, that I should somehow link the norm to the measure...

From an other perspective, on Wikipedia: Complete metric space it says:

Intuitively, a space is complete if there are no "points missing" from it (inside or at the boundary). For instance, the set of rational numbers is not complete, because e.g. $\sqrt{2}$ is "missing" from it, even though one can construct a Cauchy sequence of rational numbers that converges to it. It is always possible to "fill all the holes", leading to the completion of a given space.

Missing points would be subsets of a nullset. Now, if these points are not in $\mathcal{A}$, the limit $x\in X$ might not exist.

But if the measure space is complete, these missing points, subsets of nullsets are measurable and thus in the sigma-algebra and the limit always exists.

(Not a satisfactory argument, seems very handwaving to me.)


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.