Proofs of dirac delta property How would I formally prove this property of dirac delta?
$$\int \delta(a-x) \delta(x-b) \,dx = \delta(a-b) $$
I attempted to use the definition of a dirac delta
$$\int f(x)\delta(a-x)\delta(x-b)\,dx=\int f(x)\delta(a-b)\,dx$$
let $g(x)=f(x)\delta(a-x)$
$$\int f(x)\delta(a-x)\delta(x-b)\,dx=\int g(x)\delta(x-b)\,dx$$
let $u=x-b,\:x=u+b, \:du=dx$ 
$$\int g(x)\delta(x-b)\,dx=\int g(u+b)\delta(u)\,du=g(0+b)=g(b)$$
but I couldn't resolve $\frac{d\,g(x)}{dx}$ when I did the u sub
I also couldn't solve the integral of $\int f(x)\delta(a-b)\,dx=\delta(a-b)\int f(x)\, dx$
any hint would be appreciated

To be completely rigorous, note that in general the product of two measures/distributions/generalised functions is not well-defined; but the convolution is, which is what we have in this case (the convolution of $\delta(x)$ with $\delta(x-b)$ "evaluated" at $a$). Since the result is a measure, the equality just means that the integral against any function is the same, so we check this:
$$\int f(a) \int \delta(a-x) \delta(x-b)\ dx\ da = \int\left(\int f(a)\delta(a-x) da\right) \delta(x-b) dx \\ =\int f(x) \delta(x-b) dx=f(b)=\int f(a)\delta(b-a) da.$$
The interchange of integration here is essentially the definition of the convolution of measures - $\int \delta(a-x) \delta(x-b) dx$ is defined to be the measure such that this manipulation is valid.
