Often in physics we integrate by parts $$\int_{x_0}^{x_1} f(x) \frac{d}{dx}( \delta(x-y))dx$$
by:
$$=[f(x) \delta(x-y)]_{x_0}^{x_1} - \int_{x_0}^{x_1} \delta(x-y) \frac{df}{dx} dx$$.
I have a really simple question, how can we assume that $[f(x) \delta(x-y)]_{x_0}^{x_1}=0$?
Intuitively the delta function is zero except for at $x=y$, but what if either $x_0$ or $x_1$ was equal to y?
Is the answer simply 'we must assume separately that $x_0,x_1 \neq y$, or is there something obvious that I'm missing, or is there some measure theory reason why we can say it is zero?
I think the issue is mostly one of notation. The delta function is actually a distribution. If $a \in \mathbf R$ and $f$ is a smooth compactly supported function then $\delta_a(f) = f(a)$.
The derivative of a distribution $T$ is defined to make the integration by parts formula work. That is, $T'(f) = -T(f')$ for any smooth compactly supported function. In the particular case of the delta function you get $$\delta_a'(f) = - \delta_a(f') = - f'(a).$$
People have found it notationally convenient to denote $\delta_a(f)$ using the notation $$\delta_a(f) = \int f(x) \delta_a(x) \, dx = \int f(x) \delta(x-a) \, dx$$ and similarly $$\delta_a'(f) = \int f(x) \delta_a'(x) \, dx = \int f(x) \delta'(x-a) \, dx.$$
Thus by definition $$\int f(x) \delta'(x-a) \, dx = \delta_a'(f) = - \delta_a(f') = - \int f'(x) \delta(x-a) \, dx.$$