TITLE: Function always continuous in a Sobolev Space?
QUESTION [5 upvotes]: Hy everybody got a quick question.
I know that all function F in a Sobolev Space has a continuous representative called U such as U=F almost everywhere.
Lets take for example: The Sobolev space on ]-1,1[ 

the function F(x)=0 on x<0 and F(x)=1 x>0  we can't find a continuous representative because it would be on a larger interval let say ]-eps ,eps[ so not in a Sobolev space
the function F(x)=x on x<0 and x>0 and F(0)=1 where we could find the representative U=x

My question is: if F isn't continuous is it sure that it isn't an element of a Sobolev Space?
It was given was given as a necessary condition in my course.
Thanks for the help.

REPLY [8 votes]: Your reasoning is correct.
If $f$ cannot be redefined on a set of measure zero to become a continuous function, then $f$ is not in $H^1(-1,1)$. 
Chances are that sooner or later you will encounter Sobolev functions of more than one variable.
Those need not have a continuous representative.
For example,
$$
f(x,y)
:= \sqrt{- \log(x^2+y^2)}
$$
is in $H^1(D)$ where $D$ is the disk in two dimensions.
It cannot be made continuous, or even bounded, by changing its values on a set of measure zero.