Can an integral be proved to have a finite value if an upper bound of the integrand has a finite value for improper integrals?

Question

Can we say $ \int_{0}^{\infty} f(x) \text{dx} < \infty$ if

$\exists \quad g(x) : \quad g(x)\geq f(x)\; \forall x \in \mathbb{R}$

and $ \int_{0}^{\infty} g(x) \text{dx}$ is finite.

If yes, what theorem guarantees this result?

PS: We do not have the condition that $f(X)\geq 0$ but it is known that it is bounded below as well. EDIT: What if I have the condition that $|f(x)|\leq g(x)$ ? Does that help?

Answer 1

This is not true if we don't assume that $f(x)\ge0$. For example, let $g(x)=0$ and $f(x)=-1$.

If you have that $|f(x)|\le g(x)$ (which is true if $f(x)\ge0$), then define $$ E_N=\{x\in[0,N]:f(x)\ge0\} $$ and $$ F_N=\{x\in[0,N]:f(x)\lt0\} $$ Dominated Convergence says that each of the following limits exist and $$ A=\lim_{N\to\infty}\int_{E_N}f(x)\,\mathrm{d}x\le\lim_{N\to\infty}\int_{E_N}g(x)\,\mathrm{d}x $$ $$ B=\lim_{N\to\infty}\int_{F_N}-f(x)\,\mathrm{d}x\le\lim_{N\to\infty}\int_{F_N}g(x)\,\mathrm{d}x $$ Therefore, $|A-B|\le\int_0^\infty g(x)\,\mathrm{d}x$ and $$ \int_0^\infty f(x)\,\mathrm{d}x=A-B $$

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In the version of the Dominated Convergence Theorem cited in the linked article, we don't need to separate into positive functions. In that version, as long as $|f(x)|\le g(x)$, we have that $$ \lim_{N\to\infty}\int_0^Nf(x)\,\mathrm{d}x=\int_0^\infty f(x)\,\mathrm{d}x $$ and $$ \left|\int_0^\infty f(x)\,\mathrm{d}x\right|\le\int_0^\infty|f(x)|\,\mathrm{d}x\le\int_0^\infty g(x)\,\mathrm{d}x $$

Answer 2

If $\int_0^t f(x)dx$ exists for all $t \ge 0$, and if $|f(x)| \le g(x)$ for all $x \ge 0$, then $\int_0^{\infty} f(x)dx$ exists, and $|\int_0^{\infty} f(x)dx| \le \int_0^{\infty} g(x)dx$.

This is a consequence of Lebesgue's Dominated Convergene Theorem, with the sequence $(f_n)$ defined as $$f_n(x) = f(x) \text{ if } x \le n$$ $$f_n(x) = 0 \text{ otherwise}$$