Is there a strictly monotone, integrable function $f: \mathbb{R} \rightarrow [0,\infty)$?

Question

Im not sure about the above question. Im guessing that there is none, else the question would probably not be asked that way, but i can't really pinpoint where the contradiction lies.

Answer 1

Hint: Assume without lost of generality $f:\mathbb R\to[0,\infty)$ is strictly increasing. Then, we have: $\int_{\mathbb R} f(x) \; \mathrm dx \ge \int_0^\infty f(x) \; \mathrm dx \ge \int_0^\infty f(0) \; \mathrm dx$.

Solution:

From $0\le f(-1) < f(0)$ follows $\int_0^\infty f(0) \mathrm dx = \infty$. Thus, $f$ is not integrable.

Answer 2

Hint:

Strict monotonicity implies injectivity on the domain.