Consider domain $X$, label set $ Y=\{0,1\}$ and the zero-one loss.Given any probability distribution D over $ X\times \{0,1\} $, we've defined the Bayes classifier $ f_D $ to be-$$ f_{D}(x)=\begin{cases}1 & \text{if }\mathbb{P}[y=1|x]\geq\tfrac{1}{2}\\0 & \text{otherwise.}\end{cases} $$I wish to prove that, for any classifer $ g\colon X\rightarrow\{0,1\}$,$ L_D(f_D)\leq L_D(g)$, which means that $ f_D$ is optimal.$L_D(h) $ is defined to be the true error of the classifier $h$. That is,  $L_D(h)=D\{(x,y)\mid h(x)\not = y\}$.I'm having some hard time proving this given the definitions above, and some hints/intuition will be appreciated.
Showing that Bayes classifier is optimal
machine learning;classification;statistics
The true error of a classifier $h$ is$$\begin{align*}L_D(h) &= \sideset{\mathbb{E}}{}{}_{x,y \sim D} \Pr[h(x) \neq y] \\ &=\sideset{\mathbb{E}}{}{}_{x,y \sim D}\begin{cases}\Pr[y \neq 0|x] & \text{if } h(x) = 0, \\\Pr[y \neq 1|x] & \text{if } h(x) = 1.\end{cases}\end{align*}$$(All probabilities are with respect to $D$.)The optimal classifier is thus the one that minimizes the loss function$$\phi(x) = \begin{cases}\Pr[y \neq 0|x] & \text{if } h(x) = 0, \\\Pr[y \neq 1|x] & \text{if } h(x) = 1\end{cases}$$for all $x$. We can rewrite the loss function as$$\phi(x) = \begin{cases}\Pr[y = 1|x] & \text{if } h(x) = 0, \\1-\Pr[y = 1|x] & \text{if } h(x) = 1\end{cases}$$So if $\Pr[y=1|x] < 1-\Pr[y=1|x]$ we should choose $h(x) = 0$, and if $\Pr[y=1|x] > 1-\Pr[y=1|x]$ we should choose $h(x) = 1$; if $\Pr[y=1|x] = 1-\Pr[y=1|x]$ then the choice doesn't matter.Finally, $\Pr[y=1|x] < 1-\Pr[y=1|x]$ is the same as $\Pr[y=1|x] < 1/2$, and this explains the formula for $f_D(x)$.