If two languages together cover all words and one is regular, is the other one as well? If $L_1$$\subseteq$ $\Sigma^*$, $L_2$$\subseteq$ $\Sigma^*$ , $L_1$ is regular and $L_1$$\cup$ $L_2$ = $\Sigma^*$ then is $L_2$ necessarily regular?

I think that the answer is yes, but I'm not sure on my proof.

The reason that I think that $L_2$ is regular is because surely $L_2$ just accepts all the words in the language that $L_1$ doesn't? So, to me, that suggests that $L_2$ must be regular as well, I just don't know where to begin on a formal proof.

Any guidance would be appreciated.
In fact, the answer is no. 

If $L_1$$\subseteq$ $\Sigma^*$, $L_2$$\subseteq$ $\Sigma^*$ , $L_1$ is regular and $L_1$$\cup$ $L_2$ = $\Sigma^*$ then is $L_2$ is not necessarily regular.

We can prove this through counter-example.

If we let $L_1$ = $\Sigma^*$, then we can choose any non-regular language in $\Sigma$  for $L_2$.

If we take $\Sigma$ = {a,b} and then let $L_2$ = $a^n$$b^n$ (A non-regular language) then  $L_1$$\cup$ $L_2$ = $\Sigma^*$ and $L_2$ is non-regular as required.