Why is Delta Function Not in Same Equivalence Class as $f(x) = 0$?

Question

In "Measure and Integration" by Dietmar A. Salamon, P4, the author states: "... it is often convenient to identify two measurable functions if they agree almost everywhere, i.e. on the complement of a set of measure zero. This defines an equivalence relation."

In "Generalized Functions" V.1, by Gel'Fand and Shilov, P5, the authors state (regarding the Dirac Delta Function $\delta (x - x_1)$: "The singular function $\delta (x - x_1)$ vanishes in a neighborhood of every point $x_0 \neq x_1$". So, $\delta (x)$ vanishes for all x, except on a single interval of measure zero.

In Rigged Hilbert Space theory, as described in "Generalized Functions" V.2 by Gel'Fand and Shilov, there is a linear topological space $\Phi'$ and $\delta (x) \in \Phi'$. $\Phi'$ is also the conjugate to the Schwartz ($S$) space ($\Phi$). Both are countably normed (but not normed) and $\Phi \subset \Phi'$. $\delta (x)$ operates on functions in $\Phi$

Therefore, in $\Phi'$ isn't $\delta (x)$ in the same equivalence class as $f(x) = 0$? I know that cannot be because the two integrate differently with $\Phi$ test functions but, by the above definition of equivalence relation and the fact that the delta function vanishes at all x except on an interval of measure $0$, it looks like one could say that $\delta (x)$ is in the same equivalence class as f(x) = 0 in $\Phi'$. Please show me what I am overlooking.

Answer 1

The Dirac delta is not a function, it is a distribution.

Answer 2

Short answer — because the delta function isn't actually a function, i.e. it's not a function on $\mathbb{R}$. People often informally think of it as a function, but it isn't. Even writing it as "$\delta(x)$" is a nod to this heuristic informal point of view, but it's not actually meaningful. It's a generalized function, a.k.a. a distribution.

So in a space of functions on the real line, be it all measurable functions or bounded or whatnot, we can't even pose the question of whether $\delta(x)$ and $f(x)=0$ are in the same equivalence class — because the delta function isn't even an element of any such space at all.

And in the space of distributions $\Phi'$, they are distinct because they clearly define different linear functionals. Note that $\Phi\subset\Phi'$, and it's a strict inclusion. $f(x)=0$ is an element of $\Phi'$ that comes from $\Phi$, but $\delta(x)$ does NOT come from $\Phi$.