Can anyone tell me why the arclength integral is a lower semicontinuous function on the set of continuously differentiable real-valued functions? I posted the question stating that it was upper semicontinuous, but that was definitely wrong.  I am trying to prove lower semicontinuity.
 A: The result you're looking for can be found in the book Mathematical analysis: linear and metric structures and continuity by Mariano Giaquinta, Giuseppe Modica as Theorem 11.3,
p.396.
I will copy their proof here.
The settings in which they are working is a metric space $(X,d)$.

Proposition Let $f_i:X\to\overline{\mathbb R}$, $i\in I$, be a family of lower semicontinuous functions on a metric space $X$. Then $f := \sup f_i $ is a lower semicontinuous function.
Theorem.  The length functional $L(\varphi)$ is lower
  semicontinuous in $C^0([a,b],X)$.
Proof. Recall that we have $$L(\varphi)=\sup_{S\in\mathcal S} V_S(f),$$
  where $V_S(f)=\sum_i d(f(t_i),f(t_{i+1}))$, $S=\{t_0=a<t_1<\dots<t_N=b\}$. Since the functional $f\to V_S(f)$ is continuous for every fixed subdivision $S$ of $[a,b]$, the result follows.

They also provide a simple example showing that length is not continuous: Example 6.25, p.204.

This result for functions defined on a closed interval $[0,1]$ is shown in First course in functional analysis by Casper Goffman, George Pedrick
p.40.
BTW I was first trying to prove this myself. When I was unsuccessful, I tried googling for curve length semicontinuous and I found the above two references. (And you will probably find some others if you go through the search results or if you try similar searches.)
