$g \notin L^2$ $\implies$ $f+g \notin L^2$

Question

Suppose $f\in L^2[0,1]$ and $g\notin L^2[0,1]$. How can I prove that $f+g\notin L^2[0,1]$?

So basically we need to show that $\int_0^1 |f+g|^2 \, dx = \int_0^1 |f^2 + 2fg + g^2| \, dx$ is unbounded. I tried splitting the integral up depending on whether the integrand is positive or negative. But it didn't get me anywhere.

Answer 1

$V:=L^2[0,1]$ is a vector space. If $f+g \in V$ and $f \in V$, then

$g=(f+g)-f \in V$, a contradiction.