Multivariate Dirac delta function

Question

1. I was told that the multivariate Dirac delta function $\delta(\mathbf x), \mathbf x \in \mathbb R^n$ with

\begin{align} 0 &= \delta(\mathbf x) \quad \forall \mathbf x \neq (0,0,...,0) \\ 1 &= \int_{-\infty}^\infty \cdots \int_{-\infty}^\infty \delta(\mathbf x) \; \text d x_1 \cdots \text d x_n \end{align}

is equivalent to the product of the it's marginals, so

\begin{align} \delta(\mathbf x) &= \delta(x_1,x_2,...,x_n) = \delta(x_1) \cdot \delta(x_2) \cdots \delta(x_n) = \prod_{i=1}^n \delta(x_i) \end{align}

Why is that the case?

2. Also, I was told that a Dirac delta function that not only non-zero for the origin

\begin{align} 0 &= \delta(x_1, x_2, ..., x_m, ... , x_n) = \prod_{i=1}^m \delta(x_i) \quad \forall x_1, x_2, ... , x_m \neq 0 \\ 1 &= \int_{-\infty}^\infty \cdots \int_{-\infty}^\infty \prod_{i=1}^m \delta(x_i) \; \text d x_1 \cdots \text d x_n \end{align}

for $m<n$. Why does the second equation (still) hold (while still integrating over the entire n-dimensional set/space)?

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EDIT: For illustrative purposes I imagine

For illustration I image the full case in 1 to be equivalent to

\begin{align} \delta(\mathbf x) = \begin{cases} \lim\limits_{a\rightarrow0} \quad \dfrac{1}{a^n} & \forall x_i \in [-\frac a2,\frac a2], 1\le i\le n \\[6pt] \quad 0 & \text{otherwise} \end{cases} \end{align}

Thais how it was introduced to me at university.

So, I was guessing I would need for the second case something like

\begin{align} \delta'(\mathbf x) = \begin{cases} \lim\limits_{a\rightarrow0} \quad \dfrac{a^{n-m}}{a^m} & \forall x_i \in [-\frac a2,\frac a2], x_j, \in [-\frac2a,\frac2a], 1\le i\le m < j\le n, \\[6pt] \quad 0 & \text{otherwise} \end{cases} \end{align}

so that the "volume" / integral over $\delta'$ would be the support multiplied by the "height" of the Dirac:

$$ \frac{a^m}{a^{n-m}} \cdot \frac{a^{n-m}}{a^m} = 1 $$

independent of $a$.

So, if we consider $\delta'$ to be measures for a measure space $(X, \Sigma, \delta')$, we take the $\delta'$-measure for an $S_1\subset X$ and then translate $S_1$ only in directions of $x_i, m<i\le n$ to $S_2$, then the $\delta'$-measure of $S_2$ should be the same as for $S_1$.

But I do not see how this relates to what I wrote in 2 above (before this edit). My main issue is, that $\delta(x_1) \cdot \delta(x_2) \cdots \delta(x_m)$ seems to "lack" the "normalization" so that the integral becomes 1: For example if we have $m=1, n=2$, then shouldn't the "height" of the Dirac delta function be 1? Because we have $a$ in one direction and $1/a$ in the other, so to get 1 for the integral, we need 1 as the height. But this does not seem to be incorporated into $\delta(x_1) \cdot \delta(x_2) \cdots \delta(x_m)$.

Answer 1

1. Observe that \begin{align} \delta(x_1) \cdots \delta(x_n)=0 \Leftrightarrow x_1,...,x_n\neq0. \end{align} Just like it should be for $\delta(x_1,...,x_n)$. Also: \begin{align} \int_{-\infty}^\infty \cdots \int_{-\infty}^\infty \delta(x_1) \cdots \delta(x_n) dx_1\cdots dx_n= 1\cdots 1=1=\int_{-\infty}^\infty \cdots \int_{-\infty}^\infty \delta(x_1,...,x_n) dx_1\cdots dx_n \end{align} (they are equivalent functions/distributions)

2. As the function $\Pi_{i=0}^m \delta(x_i)$ is not defined in the last n-m Variabels it is also integrated over, the integration is an integration over a Null Set. I think this is what is bothering you. But note that the function $\delta(x)$ itself only has a value other than zero at x=0, which also is a Null Set in the domain of the function (e.g. reel Numbers). One must conclude that the value of $\delta(0)$ is something like infinity (just for intuition), or else the integral would be zero. This is how you can think about why integrating over the dirac delta function is always going to evaluate to 1, regardless if you are integrating over Null Sets.