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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 $F(x)\leq f(x)?$

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  • $\begingroup$ You allow $g$ to be constantly zero? In that case, the second condition is vacuous. $\endgroup$
    – user130903
    Commented Dec 23, 2021 at 4:48
  • $\begingroup$ I edited the question to make it less ambiguous. The function $g$ is not in my hand but have to be same for both $f,F.$ $\endgroup$
    – Wilderness
    Commented Dec 23, 2021 at 12:13

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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.

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  • $\begingroup$ Thanks. I also get the idea graphically. $\endgroup$
    – Wilderness
    Commented Dec 23, 2021 at 13:07
  • $\begingroup$ From 1st property of $f$, clearly it is not integrable. Although it is locally integrable and that is what we require to define $F$. Isn't it? $\endgroup$
    – Wilderness
    Commented Dec 23, 2021 at 13:33
  • $\begingroup$ Yeah that's right will edit. $\endgroup$
    – Squala
    Commented Dec 23, 2021 at 13:59

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