If $A$ is "small," then the chain probabilistically regenerates/forgets its past once it enters $A$. Also, smallness has nothing to do with its measure or anything like that; there are many examples of chains with the entire state space that’s small.
Background: if a transition kernel is of the form $ p^{(1)}(x,B) = \nu(B) \quad{(*)} $ for all $x$ and $B$, then the chain elements are independent. This is because the transition doesn't depend on where it's coming from: $x$.
With a small set, we're only looking at $x \in A$. If $A$ is $k=1$ small, then $p^{(1)}(x,\cdot) \ge \delta \nu(\cdot)$ for all $x \in A$, and we can write the transition kernel as follows: \begin{align*} p^{(1)}(x,\cdot) &= \delta \nu(\cdot) + p^{(1)}(x,\cdot) -\delta \nu(\cdot)\\ &= \delta \nu(\cdot) + (1-\delta) \frac{p^{(1)}(x,\cdot) -\delta \nu(\cdot)}{1-\delta} \\ &= \delta \nu(\cdot) + (1-\delta)K(x, \cdot). \end{align*}
This is a discrete mixture, so with probability $\delta$ you're transitioning with $\nu$ and forgetting the past, and with probability $1-\delta$, you're transitioning with something that takes into account where you're coming from and not forgetting the past. The smallness property gives us the nonnegativity of $K$.
As @Did mentions, "$\delta$ is used to evaluate the rate of the loss of memory of the initial state by the chain." You can see that if $\delta = 1$, then we get equation $(*)$.
Other things: If $k > 1$, then it takes longer to forget, and if we’re talking about “petiteness” then it’s the same idea but with a random number of steps.