The entropy functional over the phase-space is given by:

$$
S[\rho(\Gamma)] = - \int \rho(\Gamma) \ln (\rho(\Gamma)) d\Gamma
$$
where $\rho(\Gamma)$ is the probability density of the system over phase-space $\Gamma$.

To impose the constraint that $\rho(\Gamma)$ is normalized, we introduce a Lagrange multiplier $\lambda$ and form the modified functional $ \tilde{S}[\rho(\Gamma)] $:
$$
\tilde{S}[\rho(\Gamma)] = S[\rho(\Gamma)] + \lambda \left( \int \rho(\Gamma) d\Gamma - 1 \right)
$$
Here, the term $\lambda \left( \int \rho(\Gamma) d\Gamma - 1 \right)$ enforces the normalization condition:
$$
\int \rho(\Gamma) d\Gamma = 1
$$


The functional derivative is defined according to
$$
\int d\Gamma \frac{\delta \tilde S\left[\rho(\Gamma)\right]}{\delta \rho(\Gamma)} \eta(\Gamma) d\Gamma := \frac{d \tilde S\left[\rho(\Gamma) + h \eta(\Gamma) \right]}{dh}\Bigg|_{h=0}
$$
where $\eta(\Gamma)$ is any arbitrary test function.

The condition of **Maximum Entropy** requires finding the probability density $\rho_e(\Gamma)$ which extremises the entropy. Therefore, it can be mathematically represented as,

$$
\frac{\delta \tilde S\left[\rho(\Gamma) = \rho_e(\Gamma)\right]}{\delta \rho(\Gamma)}  = 0
$$

From the above equation, we can calculate that 
$$
\frac{\delta \tilde S\left[\rho(\Gamma)\right]}{\delta \rho(\Gamma)} = \ln \{ \rho(\Gamma) \} + \lambda + 1
$$

Setting the above value to zero, at $\rho(\Gamma) = \rho_e(\Gamma)$, we get
$\rho_e(\Gamma) = e^{-(\lambda + 1)}$. The main highlight is that the probability is independent of the phase-space point $\Gamma$ and therefore will be a constant uniform distribution for all the possible states. Further applying the normalization condition, we can show that 

$$
\rho_e(\Gamma) = \frac{1}{\int d\Gamma} = \frac{1}{\Omega}
$$

where $\Omega$ is the volume of the phase-space or the total number of microstates.


**Question**

In principle, the microcanonical ensemble is constrained to have an energy, E and therefore, the probability density function is given according to 

$$
\rho_{mc}(\Gamma) = \frac{1}{\Omega}\delta\left(H(\Gamma) - E\right)
$$

This means that there is an additional restriction in the probability density that it can only take non-zero value when the energy of the system is equal to E.
**How can I impose this constraint to match the $\rho_e(\Gamma)$ with $\rho_{mc}(\Gamma)$?**


**EDIT 1 (A possible approach) - Transformation of random variables**


A possible approach is to consider the essential role of Liouville's theorem on equilibrium probability density, which suggests that the $\rho(\Gamma)$ depends on the phase-space explicitly only through the Hamiltonian ($H(\Gamma)$) of the system. 

From the basis of probability theory, this can be considered as a transformation from the random variable $\Gamma$ to $E = H(\Gamma)$, such that the probability distribution, $\rho(\Gamma)$ and $\rho_{mc}(E)$ are according to,

$$
\rho_{mc}(E) = \int d\Gamma \rho(H(\Gamma) - E)
$$

where we impose that the system has energy $E$ is imposed through the Dirac-delta function.

This approach particulalry emphasizes that the random variable of interest is **not** ***phase-space coordinates***, but it's the ***total energy of the system*** (which is a constant of the motion)