Given a decreasing function s.t. $\int_0^\infty f(x)\,dx<\infty,$ prove $\sum_{n=1}^\infty f(na)$ converges 
Let $f\in C([0,\infty))$ be a decreasing function such that $\int_0^\infty f(x)\,dx$ converges.
Prove $\sum_{n=1}^\infty f(na)$ converges, $\forall a>0$

My attempt:
By the Cauchy criterion, there exists $M>0,$ such that for $t-1>M:$
$$f(t)=\int_{t-1}^t f(t) \, dx \leq \int_{t-1}^t f(x)\,dx\xrightarrow{t \to \infty} 0$$
Hence, $f$ is non-negative.
$f$ is decreasing $\implies f(na)\leq f(a), \forall a>0, n\in \mathbb{N}.$
By integral monotonicity and non-negativity of $f$:
$$\int_1^\infty f(nx)\,dx \leq \int_1^\infty f(x)\,dx \leq \int_0^\infty f(x)\,dx$$
Hence $\int_1^\infty f(nx)\,dx$ converges and therefore $\sum_{n=1}^\infty f(na)$ converges.
Is that correct? If so, why is continuity necessary ? Is there a simpler way to prove it?
Any help appreciated.

This is the integral test for convergence of series. Do a change of variable x --- to --- ax. The integral converges so does series g(n)=f(na).
