# How to check $L$ is regular or not [duplicate]

If $$L=\{w \in \Sigma^*\mid w=uv,\text{ number of occurnce a's in u equal to number of occurrence b's in v}\}.$$

I think $$L=\Sigma^*$$ because for any string in $$\Sigma^*$$, we can split it to $$uv$$ such that it contain equal number of a's and b's. But I can't prove it or maybe that language isn't regular. Any help for prove that is a regular language or not are welcome.

Indeed $$L = \Sigma^\ast$$. Here is a proof.
Clearly, $$L \subseteq \Sigma^\ast$$, so it suffices to show $$\Sigma^\ast \subseteq L$$. Let $$w \in \Sigma^\ast$$. We start with an arbitrary decomposition $$w = uv$$. If $$|u|_a = |v|_b$$ we are done. Otherwise, wlog let $$|u|_a < |v|_b$$. Then by moving the cutpoint one step to the right, i.e. considering the decomposition $$uv_1, v_2 \ldots v_m$$, we increase either the number of $$a$$s in $$u$$ (if $$v_1 = a$$) or decrease the number of $$b$$s in $$v$$ (if $$v_1 = b$$), so the absolute value of the difference $$||v|_b - |u|_a|$$ is reduced by $$1$$ after such a step. Since this difference is bounded from below by $$0$$ the procedure terminates after finitely many steps.
(If $$\Sigma$$ contains more symbols than $$a, b$$ the argument works the same way but the difference might stay the same in some steps.)