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!



Can the supremum of continuous functions be discontinuous on a set of positive measure?
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.


