My professor of statistical mechanics stated briefly that the probability density of the "microcanonical ensemble" is given by the Dirac delta function $\delta (E-E_0)$ where $E_0$ is the total energy of the given system.
I don't understand why. Can anyone elaborate on this point please?
P.S. I can see that the normalization condition is satisfied due to the identity $$\int\limits_{-\infty}^{\infty}\delta (E-E_0)dE=1$$
Your professor was probably looking at the microcanonical distribution from the standpoint of the canonical distribution.
Consider the following expectation value of a variable A in the canonical distribution: $$\langle A(\beta,X)\rangle = \frac{\int A(E,X)\Omega(E,X)exp(-\beta E)dE}{\int \Omega(E,X)exp(-\beta E)dE}.$$ The denominator is the canonical partition function $Z(\beta,X)$, which is the $E \to \beta$ Laplace transform of the microcanonical distribution $\Omega(E,X)$. The X variable represents extensive variables other than E.
The expectation value can also be written as an integral with respect to a probability density function $p(\beta,E,X)$ as follows: $$\langle A(\beta,X)\rangle = \int A(E,X)p(\beta,E,X)dE,$$where, from the first equation, p is given by:$$ p(\beta,E,X) = \frac{\Omega(E,X)exp(-\beta E)}{\int \Omega(E,X)exp(-\beta E)dE}.$$ Now introduce the delta function $\delta(E-E_o)$ into the integrals. This selects for a particular energy $E_o$. As a result, the weights $\Omega(E_o,X)exp(-\beta E_o)$ in the first equation cancel out.
In particular, the reduced canonical pdf can simply be written $ p(\beta,E,X) = \delta(E-E_o)$, which, as asserted by your professor, leads to the (fixed) value of A in the microcanonical distribution:$$A(E_o,X)=\int A(E,X)\delta(E-E_o)dE.$$
