# Function always continuous in a Sobolev Space?

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.

#### Solutions Collecting From Web of "Function always continuous in a Sobolev Space?"

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 zerp.