Cauchy's functional equation -- additional condition

Consider the function $f:R \to R$

$$f(x+y)=f(x)+f(y)$$
which is known as Cauchy's functional equation. I know that if $f$ is monotonic, continuous at one point, bounded, then the only solutions are $f(x)=cx,~ \forall c$ But once I successfully derived Cauchy's functional equation from a problem, and I know it's an involution, bijective, as well as an odd function. How can I derive (if possible) that $f(x)=x$ is the only solution?