Membership problem of Recursive languages are decidable.
Let $L$ be a recursive language and $M$ be the Turing Machine that accepts it. For string $w,$ if $w ∈ L,$ then $M$ halts in final state. If $w ∉ L,$ then $M$ halts in non-final state. (halts always!). That's why Recursive languages are decidable for Membership problem
My question is in same logic, why finiteness, emptiness is undecidable? Don't want any concrete proof. I just want brief concepts like my approach.