Dirac delta and non-test functions Normalization of the delta function (distribution) is often informally written as an integral
$$\int_{-\infty}^{+\infty} \delta(x) \, dx = 1$$
An attempt to write this formally would be expression like $\delta[1] = 1$. However, the constant function $f(x) = 1$ has no compact support, therefore it is not a test function. Is there a precise way to translate such integral expression into the language of distributions?
Also, a related problem occurs with expressions like $\delta[\chi_{[a,b]}]$, where $\chi_{[a,b]}$ is a characteristic function over $[a,b] \subset \mathbb{R}$ (defined such that $\chi_{[a,b]}(x) = 1$ for $x \in [a,b]$ and $\chi_{[a,b]}(x) = 0$ for $x \notin [a,b]$). Informally, this would correspond to the integral
$$\int_a^b \delta(x) \, dx$$
and, intuitively, one would like to get result $1$ for $0 \in [a,b]$ and $0$ otherwise. However, $\chi_{[a,b]}$ is not a test function (since it has discontinuities at $a$ and $b$), so we again have a formal problem as the one mentioned above.
 A: Suppose $d$ is a distribution with compact support $S$ and $f$ is a smooth function, not necessarily with compact support. Define a smooth, compactly supported cutoff function $k$ such that $k=1$ on a neighborhood of $S$. Then $\langle d,kf \rangle$ makes proper sense. One can prove that the value of this expression does not depend on $k$, under the preceding assumptions. So this defines a meaningful extension of $d$ to all of $C^\infty$. Now that we know the key idea is that of a compactly supported distribution, a quick google search finds a proof: http://www.math.chalmers.se/~hasse/distributioner_eng.pdf page 31.
Now the Dirac delta has support $\{ 0 \}$, so this applies to it.
The principle is the same if you want to apply $d$ to a function like $\chi_{[a,b]}$ which is smooth on a neighborhood of the support of $d$ but not globally smooth: use a smooth cutoff which keeps the function the same on a neighborhood of the support of $d$ but such that the result is globally smooth. So this procedure works for $\delta[\chi_{[a,b]}]$ provided $a,b \neq 0$.
A: You can see the Dirac not only as a distribution, but also as an element of the dual of continuous functions : a Radon measure. 
This point of view gives sense to $\delta[1]$.
And if a measure $\mu$ is absolutely continuous in respect to the Lebesgue measure,
$$\mu( \phi ) = \int_{\mathbb{R}} \phi(x) d\mu(x) = \int_{\mathbb{R}} \phi(x) \mu(x) dx$$
(the first equality is Riesz representation theorem)
This gives, by abuse of notation (as $\delta$ is not absolutely continuous in respect to the Lebesgue measure)
$$\delta( 1 ) = \int_{\mathbb{R}} 1 d\delta(x) = \int_{\mathbb{R}} \delta(x) dx$$
For $\chi_{[a,b]}$, I'm not sure what could be the correct point of view.
