# Can the supremum of continuous functions be discontinuous on a set of positive measure? [closed]

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 $x=1$.

Thanks!

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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. –  Captain Oates 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

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

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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.
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