Let $p\in (0,1)$ and define a kernel on $(0,1]$ as $$p(x,\cdot) =p \lambda(\cdot) + (1-p) \frac{\lambda(\cdot \cap [0,x])}{x} $$ where $\lambda$ is lebesgue measure on $[0,1]$. What is the invariant measure?
I tried to solve $\mu = \mu p$: For all measurable $B$,
$\mu(B) = \int \mu(dx)p(x,B) = \int \mu(dx)(p \lambda(B) + (1-p) \frac{\lambda(B\cap [0,x])}{x} )$
but I have difficulty with $\int \frac{\lambda(B\cap [0,x])}{x}\mu(dx)$.
Another idea is to find the limit $\lim_n p^n(x,\cdot)$ for an initial state $x$, but again I am struggling with $\int \frac{\lambda(B\cap [0,x])}{x}\lambda(dx)$.
To start write $\mu=\mu p$ as $$ \mu(t,1]=(1-t)p+(1-p)\int_t^1\left(\int_s^1{\mu(dx)\over x}\right) ds,\qquad 0<t\le 1. $$ This shows that $\mu$ has a density $m$ with respect to $\lambda$ satisfying $$ m(t) = p+(1-p)\int_t^1{m(x)\over x} dx,\qquad 0<t\le 1. $$ This in turn implies that $m$ is $C^1$ on $(0,1]$ and satisfies the ODE $$ m'(t) = -(1-p){m(t)\over t},\qquad 0<t\le 1. $$ Solving this and normalizing to be a probability density you get $$ m(t) =pt^{-(1-p)},\qquad 0<t\le 1. $$