I am having a hard time to understand the concept of the detailed balance; mostly because of the intermingled notation most of the resources use; which involves constant usage of random and state variables interchangeably.
Assuming a continuous state space, I wonder whether the following derivation of "detailed balance implies stationary distribution" true:
Let's assume we have the transition function $T(Z,Z') = p_{Z'\mid Z}(Z'\mid Z)$. $p_Z(Z=a)T(Z=a,Z'=b) = p_Z(Z=b)T(Z=b,Z'=a)$ as the result of detailed balance. Then I want to show that $p_Z(\cdot)$ is the stationary distribution of the Markov chain.
I set $$p_{Z'}(Z'=b) = \int p_Z(Z)p_{Z'\mid Z}(Z'=b\mid Z)\,dZ = \int p_Z(Z)T(Z,Z'=b) \, dZ$$
Then due to the detailed balance, we have $p_Z(Z)T(Z,Z'=b) = p_Z(Z=b)T(Z=b,Z')$ whenever the right side $Z'$ takes the same value as the left side $Z$. So, I replace the righthand side into integral and since it should be $Z'=Z$ and $dZ'/dZ=1$ I apply change of variables:
$$\int p_Z(Z=b)T(Z=b,Z')\,dZ' = p_Z(Z=b) = p_{Z'}(Z'=b)$$
Since this is true for all $Z'$ $p_{Z}(\cdot)$ is the stationary distribution.
Are my reasoning and calculations correct here?
[This is a comment, but I write it here because of the large formulas ...]
Your argument is correct, but your writing is not rigorous. I strongly advise that you distinguish in your notation between a random variable and its possible values. For example, I would write $T(Z=a,Z'=b)=p_{Z'|Z}(Z'=b\,|\,Z=a)$ for the conditional probability density function of $Z'$ given $Z$.
Then, the starting point would be $$p_{Z'}(Z'=b) = \int_x p_Z(Z=x)p_{Z'\mid Z}(Z'=b\mid Z=x)\,\mathrm{d}x = \int_x p_Z(Z=x)T(Z=x,Z'=b) \, \mathrm{d}x\;.$$ By the detailed balance, this can be written as $$\int_x p_{Z}(Z=b)T(Z=b,Z'=x)\,\mathrm{d}x = p_Z(Z=b)\int_x T(Z=b,Z'=x)\,\mathrm{d}x = p_Z(Z=b) \;.$$