Consider the piecewise constant function $\psi:I=[-1,1] \rightarrow \mathbb{R}$ given by
$$\psi(x) = \begin{cases} \psi_1 & x \leq 0, \ \psi_2 & x > 0 \end{cases}$$
for some constants $\psi_1, \psi_2 \in \mathbb{R}$. I would like to evaluate the integral
$$\int_I \delta(x) \psi(x) dx$$
where $\delta$ is the Dirac delta distribution centered at $x=0$. Of course, if we think about distributions in the usual way then you might say that $\delta(\psi) = \langle \delta, \psi \rangle = \psi(0) = \psi_1.$ But then the result depends on a fairly arbitrary choice when defining $\psi$: should the left- or right- half of the interval be closed? This question doesn't seem to have a meaningful answer when dealing with a problem that arises from a physical system (say).
Instead, consider a family of distributions $\phi_\epsilon(x)$ such that
- $\phi_\epsilon(x)=\phi_\epsilon(-x)$,
- $\int_I \phi_\epsilon(x)=1$ for all $\epsilon$, and
- $\lim_{\epsilon \rightarrow 0} \phi_\epsilon = \delta$,
i.e., any family of even distributions with unit mass that approaches the Dirac delta distribution as $\epsilon$ approaches zero. (For instance, you could use the family of Gaussians $\phi_\epsilon(x) = \frac{1}{\epsilon\sqrt{\pi}}e^{-x^2/\epsilon^2}$.)
I can now think of my integral as
$$\lim_{\epsilon \rightarrow 0} \int_I \phi_\epsilon(x) \psi(x) dx.$$
For some $\epsilon > 0$, the integral inside the limit can be expressed as
$$u(\epsilon) = \psi_1 \int_{-1}^0 \phi_\epsilon(x) dx + \psi_2 \int_0^1 \phi_\epsilon(x) dx = \frac{1}{2}\left( \psi_1 + \psi_2 \right) = \bar{\psi},$$
where $\bar{\psi}$ is the mean of the constant values. Can we say, then, that $\lim_{\epsilon \rightarrow 0} u(\epsilon) = \bar{\psi}$? It would seem so: for any $\mu > 0$ there exists an $\epsilon_0$ such that $\epsilon < \epsilon_0$ implies $|u(\epsilon) - \bar{\psi}|<\mu$ (namely, $\epsilon$ is any positive constant!). But clearly I've got a problem somewhere, because $\bar{\psi} \ne \psi_1$, i.e., this result does not agree with my earlier interpretation.
So what's the right thing to do here? The latter answer ($\bar{\psi}$) agrees more with my "physical" intuition (because it's invariant with respect to deciding which half-interval is open), but I'm concerned about rigor.
Edit: Since the problem as stated is not well-posed ($\delta$ cannot be evaluated on discontinuous functions), let me give some motivation. Imagine that I have a pair of piecewise linear functions $f,g:I^2 \rightarrow \mathbb{R}$, which are again discontinuous only at $x=0$. I would like to integrate the wedge product of $df$ and $dg$ over the domain:
$$\int_{I^2} df \wedge dg = \int_I \int_I \frac{\partial f}{\partial x} \frac{\partial g}{\partial y} - \frac{\partial f}{\partial y}\frac{\partial g}{\partial x} dx dy.$$
Consider just the first term $(\partial f/\partial x)(\partial g/\partial y)$ and consider just the inner integral $\int_I \cdot dx$. We now have (almost) the original problem: $\partial f/\partial x$ can be thought of as a $\delta$ (plus a piecewise constant), and $\partial g/\partial y$ is simply piecewise constant along the $x$-direction.
So, the problem could be restated as: how do I integrate the wedge product $df \wedge dg$ of piecewise linear 0-forms $f$ and $g$ defined over a planar region? Formally this problem may again be ill-posed, yet it is a real problem that comes up in the context of finite element analysis where basis functions are nonsmooth or even discontinuous.