Let $k \in L^1((0,1) \times (0,1))$ be non-negative and such that $$ \int_0^1 k(x,y) dy = 1, \quad x \in (0,1). $$ Let $A\colon L^1(0,1) \to L^1(0,1)$ be defined by $$ Af(y) = \int_0^1 k(x,y) f(x) dx, \quad y \in (0,1). $$ Assume that there exists a unique $f \in L^1(0,1)$ such that
- $f$ is positive a.e.,
- $\int_0^1 f(x) dx = 1$,
- $Af = f$.
Is it true that $A$ is an irreducible operator?
By irreducible I mean
For every measurable $\Omega \subset (0,1)$ with measure $0 < \mu(\Omega) < 1$ ($\mu$ - Lebesgue measure), the set $$ I = \{f \in L^1(0,1)\colon f_{|\Omega} = 0 \} $$ i not $A$-invariant, i.e., $$ A(I) \not\subset I $$ holds.
Edit: It is clear the the uniqueness of $f$ is crucial. If $k$ is $2$ times the characteristic function of $$(0,1/2] \times (0,1/2) \cup (1/2,1) \times (1/2,1),$$ then $A$ is redducible, however there are also many $f$'s such that $Af = f$.