Dirac delta function and dirac measure I want to know the relationship between these two things. e.g. What's the results of following integrals? (Let $\mu$ Lebesgue measure, $\nu$ Dirac measure)
$$(1)\int\delta_c(x)\mu(dx)$$
$$(2)\int\delta_c(x)\nu(dx)$$
$$(3)\int f(x)\delta_c(x)\mu(dx)$$
$$(4)\int f(x)\delta_c(x)\nu(dx)$$
 A: So first the Dirac delta is not and can not be a function but it is a distribution of order $0$, which can be seen as a measure. This is why it is better to use the notation
$$
\int f(x)\,\delta_c(\mathrm{d}x) = f(c)
$$
or $\langle \delta_c,f\rangle = f(c)$. In particular, there is no general way to multiply two measures.  However, one can multiply a measure by a continuous function. Moreover, one can associate to every measure $\lambda$ that is absolutely continuous with respect to the Lebesgue measure a function $f_\lambda$ and so one can identify these measures with functions by writing $\lambda(\mathrm{d}x) = f_\lambda(x)\,\mu(\mathrm{d}x)$ (and one usually denotes just by $\mathrm{d}x$ the Lebesgue measure so that the associated function is $1$). So one could define the product of $\delta_c$ with $\lambda$ by $(\lambda\delta_c)(\mathrm{d}x) = f_{\lambda}(x)\,\delta_c(\mathrm{d}x)$. In any case, one cannot multiply the Dirac delta with itself.
Knowing all that, lets try to give a meaning to your very formal expressions:
$$\begin{align*}
(1)&& \int 1\, \delta_c(\mathrm{d}x) &= 1
\\
(2)&& \int \delta_c(\mathrm{d}x)^2 &\ \textbf{ has no meaning!} 
\\
(3)&& \int f(x)\,1\, \delta_c(\mathrm{d}x) &= f(c)
\\
(4)&& \int f(x)\,\delta_c(\mathrm{d}x)^2 &\ \textbf{ has no meaning!} 
\end{align*}$$
As usual, this comes from the misconception that $\delta_c$ is a function ...
Edit: I see in the comments you are asking about probability distributions. In general, probability distributions can be measures that are not absolutely with respect to the Lebesgue measure. As an example, $\mathbf{1}_{[0,1]}$ is the distribution of the uniform law in $[0,1]$, $\delta_0$ is the distribution of the constant law $\mathbb{P}(X=0)=1$. You can for example then get $f = \frac{1}{2} \mathbf{1}_{[0,1]} + \frac{1}{2} \delta_0$ to get half a chance to get $0$ and half the chance to be in $[0,1]$. But there is no meaning to multiply probability distributions. The product of random variables will imply the convolution product of their probability distributions, and the law of a couple $(X,Y)$ will be the tensor product of their probability distributions (which always make sense for distributions).
A: In analysis and probability, Dirac's delta $\delta_c$ is commonly seen as a function defined on a space of functions. Here are two examples:

*

*In the Theory of distributions for example $\delta_c$ is defined as  $\delta_c:\mathcal{C}^\infty_c(\mathbb{R}^d)\rightarrow\mathbb{R}$ defined as $\delta_c\phi = \phi(c)$.  Clearly $\delta_c$ defines a linear function.


*In measure theory or the thoery of integration, the $\delta_c$ is defined as a measure on a measurable space $(X,\mathscr{B})$ such that $\delta_c(A)=1$ if $c\in A$, $0$ otherwise. It is easily seen that $\delta_c$ induces a linear function on the space of measurable functions on $(X,\mathscr{B})$ be setting $\delta_cf=f(c)$.
your expressions (1)-(4) are not properly defined unless there is some additional context. For instance, if $\delta_c$ and $\mu$ are measures,  it make sense to talk about the product measure $\delta_c\otimes\mu$:
$$
\int f(x,y)\delta_c(dx)\mu(dy)=\int f(c,y)\mu(dy)$$
If  $\delta_c$ is though of as a  distributions, and may talk about $\delta_c*\phi$, for any $\phi\in C^\infty_c(\mathbb{R}^d)$
