When we learn about differential forms and Stokes' theorem, we are told that $\partial [a,b] = \{-a,+b\}$ so that $\int_{[a,b]} f^{\prime} = \int_{\{-a,+b\}}f = f(b) - f(a)$ so that Stokes' theorem is a generalization of the FTC.
Now, $\int_{\{-a,+b\}}f$ appears to be an integration over a signed measure (See this stack exchange post) and so does not satisfy certain properties: e.g. that $f\leq 0$ implies $\int_{\{-a,+b\}}f\leq 0$ (e.g suppose $f(b)=0,f(a)=-1$).
Now, in Theorem 6 of section 2.5, Evan proves the finite propagation property for the wave equation for all $n$.
In this proof, the last idea is that $\frac{\partial u}{\partial \nu} u_{t} - \frac{1}{2} u_{t}^{2} - \frac{1}{2}|Du|^{2} \leq 0 $ so then $\int_{\partial B(x_{0},t_{0}-t)} \frac{\partial u}{\partial \nu} u_{t} - \frac{1}{2} u_{t}^{2} - \frac{1}{2}|Du|^{2}dS \leq 0$.
Based on my above analysis, this implication does not seem to be hold if $B(x_{0},t_{0}-t)$ is some interval.
Is this analysis accurate? Where exactly is my mistake? If the analysis is accurate, how do you include the $n=1$ case in the energy method proof for this fact?
So I think we can solve this by properly differentiating under the integral sign. If we define $E$ by: $$ E(t) = \int_{x_0 - t_0 +t}^{x_0 +t_0 -t} u_t^2 + u_{x}^2 \mathrm{d}x$$
Then $E'(t)$ is given by: $$ E'(t) = \frac{d}{dt}(x_0 + t_0 -t)(u_t^2 + u_x^2)(t,x_0 + t_0 -t) -\frac{d}{dt}(x_0 - t_0 +t)(u_t^2 + u_x^2)(t,x_0 - t_0 -t) + \int_{x_0 -t_0 +t}^{x_0 +t_0 -t} \frac{d}{dt} \left(u_t^2 + u_{x}^2 \right) \mathrm{d}x $$ Computing the outside terms gives us: $$ -(u_t^2 + u_x^2)(t,x_0 + t_0 -t) - (u_t^2 + u_x^2)(t,x_0 - t_0 -t) \leq 0 $$ And the inner term is given by: $$ \int_{x_0 -t_0 +t}^{x_0 +t_0 -t} \frac{d}{dt} \left(u_t^2 + u_{x}^2 \right) = \int_{x_0 -t_0 +t}^{x_0 +t_0 -t} 2u_tu_{tt} +2u_{xx}u_t \mathrm{d}x = 0 $$ So the term is less than or equal to $0$. I'm not sure if this answers your question about integrating over signed measures, but I've brute forced it for the 1-d case, just to verify.