Let $f:[0,\infty)\to [0,\infty)$ be an increasing function satisfying
$$\int\_0^\infty f(x)\frac{dx}{1+x^2}=\infty.$$


Can we find a continuous increasing function $F$ on $[0,\infty)$ satisfying
$$\int\_0^\infty F(x)\frac{dx}{1+x^2}=\infty$$
and $F(x)\leq f(x)?$



Given an increasing function $f$, to find a continuous function satisfying properties of $f$
Yes. Since $f$ is increasing, it is almost everywhere continuous and in particular locally integrable. So we can define $F$ by $F(t)=0$ for $t<1$ and $F(t)=\int\_{t-1}^tf(x)dx$ for $t\geq1$. This function will be continuous and increasing and satisfies $f(x-1)\leq F(x)\leq f(x)$. From the first inequality you can shown that the integral $\int\_0^\infty F(x)\frac{dx}{1+x^2}$ diverges.


