Why does it not make sense to talk about functions on sets with measure zero? Trace Theorem In the context of Sobolev spaces, there is a theorem called Trace Theorem. In Evans (page257) he says that

Since $\partial U$ (the boundary of $U$, I believe it is safe to assume this is open) has n-dimensional Lebesgue measure zero, there is no direct meaning we can give to the expression $u$ restricted to $\partial U$.

Here $u \in C(cl(U))$ the space of uniformly continuous functions.
Now the Trace Theorem defines a linear map $T: W^{1,p}(U) \to L^p(\partial U)$ such that
(1) $Tu = u|_{\partial U}$
(2) $T$ is a bounded linear map (and hence continuous).
Now what I don't get is

*

*Isn't $L^p(\partial U)$ equipped with the norm $\int_{\partial U}$? He said the boundary has measure zero, so isn't the norm $0$?


*Again he still writes $u$ restricted to the boundary for $T$, is this just notation? I just don't get what this means.


*He also talks about zero traces after the theorem. Is this just a theorem about when PDEs have homogenous boundary conditions?
 A: The problem that Evans identifies is that when we have a function $u$ defined on an open set $U$ and belonging to the Sobolev space $W^{1,p}(U)$ and we want to extend that to the boundary we run into problems.  As he notes, if $u\in C(\bar U)$ additionally then we are fine: $u$ is well defined on $\partial U$ in this case and we can proceed.  But if not, then because $u$ is only defined a.e. in $U$ (i.e. is a representative of an equivalence class of functions that agree everywhere except, perhaps, on sets of measure zero) then we don't have an obvious strategy for determining what $u$ looks like on $\partial U$.
Note that $\partial U$ is only a set of measure zero here in terms of $W^{1,p}(U)$ because the boundary has codimension $1$ with respect to the Sobolev norm.  That means that we can consider $\partial U$ as a Sobolev space in its own right, with its own norm, and then we can look for a way to extend $u$ (as uniquely as possible) to $\partial U$.  The conditions under which we can do that are given by the Trace theorem as you've identified.
So, we look for a continuous linear operator $T: W^{1,p}(U) \rightarrow L^p(\partial U)$.  Each space is equipped with its own norm, so answering your question $1$, no, the norm is not identically zero on $L^p(\partial U)$.  For the sake of clarity, let's say that $Tu = v$.  Our specific requirement is that $u=v$ on $\partial U$.  Should $v$ somehow have values on $U$ outside of $\partial U$ we don't care.  We will never look at that.  So we can write $Tu = v = u_{\mid \partial U}$ in order to emphasize where our $v$ actually lives and why we're talking about it.  So your second question is answered as: yes, this is notational convenience to help you understand what's going on and where this value on the boundary has come from.
You haven't, for whatever reason, transcribed the statement of the theorem correctly though: condition (i) is:
$$ Tu = u_{\mid \partial U} \mbox { if } u \in W^{1,p}\cap C(\bar{U}) $$
-- we are forcing the boundary values to be taken from those functions that do exist on the boundary.  Since these functions are chosen from $W^{1,p}$ they are still part of the equivalence class, so they still equal $u$ a.e. on $U$.  So this is a sensible choice, and allows us to state what we're doing here:
we define $u$ on the boundary by looking at the subset of functions that are continuous of the closure of $U$ and accepting the values they have in common on the boundary as the boundary values for any $u \in U$
We call $Tu$ the trace of $u$ on $U$ because it's a kind of remnant or left-over -- it's not what every $u \in U$ would give, but it's the best we'll get as several such $u$ agree there.  So the values are like a tracery over the boundary.  (That might be rather too poetic for many people, feel free to ignore it.)
Finally for your third question: typically when a function takes infinitely many zero values we start suspecting that it's actually zero everywhere, and if $Tu=0$ that's what we might suspect.  But we're dealing with a subset of all $u \in U$ now, and equivalence classes to boot, so we need to check our intuition.  Theorem $2$ then looks at what it means to have $Tu=0$ on $\partial U$ and concludes that $u$ does not have to be identically zero, but only that needs to be compactly supported on $U$ (i.e. it can vanish everywhere outside of $\bar U$).  That is what Evans means by $u \in W^{1,p}_0(U)$.
Compactly supported is awesome -- compactness is one of those "we can't have finite but this is almost as good" properties that we should grab whenever we can get it.  If all we have to do to prove $u\in W^{1,p}_0(U)$ is show $Tu=0$ on $\partial U$ that's a pretty good offer.
