Integrating Dirac delta distribution from $0$ to $1$.

Question

Let $\delta$ be rigorously defined as a generalized function (lim of a function).

I am guessing that $\int_{-1}^0\delta(x)d x=\int_0^1\delta(x)d x=\frac{1}{2}$?

Also, let $E$ denote a set contains 1/3 of the randomly chosen points on $[-1,1]$

Is it true that $\int_E\delta(x)d x=\frac{1}3$?

I must be wrong somewhere. For example it is possible that in a uncountable set I cannot just randomly choose points.

Answer 1

As discussed several times in this community (for example, see this), the value of $\int_{0}^{a} \delta(x) \, \mathrm{d}x$ is not well-defined. There are several different choices of giving value to this expression, depending on the ambient physical process.

Also, there is no well-defined notion of a "set containing $\frac{1}{3}$ of the randomly chosen points on $[-1, 1]$". One may try to realize such a set via approximation: Let $E_n$ be a random set that is the union of "about $\frac{1}{3}$ of the subintervals of $[-1, 1]$ of equal length $2/n$. For example,

• We may let each subinterval to be included in $E_n$ with probability $\frac{1}{3}$, or
• We may choose $E_n$ to be one of $\binom{n}{\lfloor n/3\rfloor}$ possible unions of $\lfloor n/3\rfloor$ subintervals with equal probability.

Now, for any such sensible choices of approximations $E_n$, the corresponding indicator function $\mathbf{1}_{E_n}$ converges weakly to $\frac{1}{3}\mathbf{1}_{[-1, 1]}$. Of course, this is not an indicator function of a subset of $\mathbb{R}$.