Sorry this seems like a basic question, but I'm having trouble figuring out the answer.
Let g(x) be the step function over [-1,1] and f(x) a function with $f\in H^1[-1,1]$, that is it has a square integrable first derivative.
Can I write: $\int_{[-1,1]}g(x)f(x)dx=\int_{[-1,1]}g(x)(f(x_0)+f'(x)(x-x_0))dx$
This is a stripped down version of what I'm trying to do. Basically I want to know if I can use a mean value theorem type result under the integral if I know my function is $H^1$.
Maybe you should not have stripped it down; the formula you gave looks like a mess. The Mean Value Theorem says that if $f$ is differentiable, then between any two points $x,x_0$ we can find $\xi$ such that $$f(x)=f(x_0)+f'(\xi)(x-x_0)\tag{MVT}$$ I don't know why you have $f'(x)$ in there. Furthermore, (MVT) requires pointwise differentiability: for example, it fails for the Lipschitz function $f(x)=|x|$ which is of course in $H^1$: $$|1|=|-1|+\color{red}{f'(\xi)}(1-(-1)) \tag{Uh-oh}$$
Even when it exists, $\xi$ depends on $x$ in some infathomable way (is it even measurable?), making the appearance of (MVT) under an integral look odd.
Given the presence of an integral, I suspect you meant to use some form of MVT for integrals. But I can't tell what you really want to do with the integral, and whether the $H^1$ assumption is going to help.