The question is already posted here : https://math.stackexchange.com/questions/2530202/prove-that-mle-does-not-depend-on-the-dominating-measure
Let $(X,\mathbb{F})$ be a measurable space, $\left\{P_{\theta}:\theta \in \Theta\right\}$ be a family of distributions on $\mathbb{F},$ which is dominated by a $\sigma$-finite measure $\mu_0$ on $\mathbb{F}$ (i.e. all $P_{\theta}'$s are absolutely continuous w.r.t. $\mu_0$). Let $p(x,\theta)=\frac{dP_{\theta}}{d\mu_o}(x)$ be the density (radon nicodym derivative) of $P_{\theta}$ w.r.t. $\mu_0$.
Given a sample $(x_1,\ldots,x_n)$ from $P_{\theta}$ (unknown $\theta$) the maximum likelihood estimator (MLE) of $\theta$ is defined by $$\hat{\theta}_{MLE}=\arg max_{\theta \in \Theta} ~\left\{\prod_{i=1}^n p(x_i,\theta)\right\}$$
To show that $\hat{\theta}_{MLE}$ DOES NOT depend on the choice of the dominating measure $\mu_0,$ although the definition of $\hat{\theta}_{MLE}$ is connected to $\mu_0$ through the density $p$.