Show that $f(x) = O(1/x)$

Question

Let $f:\mathbb{R}\to\mathbb{R}$ such that $f\ge 0$, monotonically decreasing and $\int_0^\infty f(x) \ dx < \infty$. Prove that $f(x) = O(1/x)$.

So basically, both $f(x)$ and $1/x$ are non-negative and monotonically decreasing. We know that $\int_0^\infty \frac{1}{x} = \infty$ but $\int_0^\infty \lt \infty $ - so there must be some $N$ such that for every $x>N$: $f(x) \le \frac{1}{x}$ which implies $f(x) = O(\frac{1}{x})$.

Is that valid/rigorous enough?

Thanks.

Answer 1

For $t > 0$,

$$tf(t) = \int_0^{t}f(t)\, dx \le \int_0^t f(x)\, dx \le \int_0^\infty f(x)\, dx.$$

Thus $0 \le f(t) \le M/t$ for all $t > 0$, where $M = \int_0^\infty f(x)\, dx$. So $f(x) = O(1/x)$.