MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Given a sequence of continuous functions $f_n(x)$, all defined on a compact set $D$ and assuming $f_n(x)$ is uniformly bounded. Let $f(x) = sup_n f_n(x)$.

It is clear that $f(x)$ is not necessarily continuous. For example, $f_n(x) = 1-x^n, D=[0,1]$. But my questions is can $f(x)$ be discontinuous on a set with positive measure? In the example I give above, $f(x)$ is discontinuous at only $x=1$.


share|cite|improve this question

closed as too localized by Bill Johnson, George Lowther, Denis Serre, Gerald Edgar, Ryan Budney Oct 19 '11 at 14:40

This question is unlikely to help any future visitors; it is only relevant to a small geographic area, a specific moment in time, or an extraordinarily narrow situation that is not generally applicable to the worldwide audience of the internet. For help making this question more broadly applicable, visit the help center.If this question can be reworded to fit the rules in the help center, please edit the question.

Is this a homework question? (Not really a research level Q so you need to provide context to convince anyone to answer it) – Anthony Quas Oct 19 '11 at 1:02
To me, at least, the question is a fairly natural one and doesn't come across as being a homework question; if it is, then the phrasing does at least indicate some thought by the OP. – Yemon Choi Oct 19 '11 at 2:45
It doesn't seem obvious to me. I don't even recall how to construct a function whose points of discontinuity form a set of positive measure. But then I found Smith-volterra-Cantor sets on wiki. So one should be able to cook up a sequence that converges to one minus its indicator function. – John Jiang Oct 19 '11 at 3:16
Look for a counterexample where $f$ is the characteristic function of a dense open set. – Gerald Edgar Oct 19 '11 at 3:20
Clarification of Gerald's comment: take disjoint intervals $\Delta_1,\Delta_2,\dots$ on $(0,1)$ so that $|\Delta_i|<3^{-i}$ but $\cup \Delta_i$ contains all rational points of $(0,1)$. Then define non-negative functions $f_n$ such that $f_n$ is supported on $\cup_{i=1}^n\Delta_i$ and $\lim f_n=1$ for any $x\in \cup \Delta_i$. Then $f$ is continuous exactly on the set $\cup \Delta_i$. – Fedor Petrov Oct 19 '11 at 7:20
up vote 3 down vote accepted

Given a closed set $E$, define the distance $d(x,E)$ from $x$ to $E$ in the usual way. Let $K_n$ denote the set of $x$ so that $d(x,E)\ge 1/n$. Observe that the set $K_n$ is closed and disjoint from $E$.

Urysohn's Lemma now says that there is a continuous function $f_n:\mathbb{R}\rightarrow[0,1]$ which is 0 on $K_n$ and 1 on $E$. The infimum over $f_n$ is then simply the characteristic function of $E$. To translate this to a supremum, simply observe that $\sup (-f_n)=-\inf(f_n)$

Now you merely need to concern yourself with producing a closed set $E$ whose boundary has positive measure. This can be done using a Cantor-type construction, as mentioned in the comments.

share|cite|improve this answer
Hm, I suppose you could also have just made it 1 on $K_n$ and 0 on $E$ and avoided that $\sup(-f_n)$ business... – Peter Luthy Oct 19 '11 at 10:33
I think that works. Thanks a lot for the construction. And Smith–Volterra–Cantor set works as E. – user18629 Oct 19 '11 at 23:19

Not the answer you're looking for? Browse other questions tagged or ask your own question.