Convergence of an improper intergral I have two functions- $f(x)$ and $g(x)$ ,that are continuous and differentiable for all $x\in(-\infty,+\infty)$. Additionally, $0<f(x)<g(x)$ for all $x$ and we know that $\int_{-\infty}^{+\infty}g(x)dx$ converges. Can i say that $\int_{-\infty}^{+\infty}f(x)dx$  converges as well?
It seems to me that $f(x)$ should converge as well. If $f(x)<g(x)$ for all $x$ and they are both non-negative for all $x$, then $\int_{-\infty}^{+\infty}f(x)dx<\int_{-\infty}^{+\infty}g(x)dx$. Thus the term on the left side should be finite as well when the term on the right side is finite. But I have not been able to find any source where they discuss the convergence of the integral when the lower limit tends to $-\infty$.
 A: By definition,$$\int_{-\infty}^\infty f(x)\,\mathrm dx=\int_{-\infty}^0f(x)\,\mathrm dx+\int_0^\infty f(x)\,\mathrm dx.$$Since $\int_{-\infty}^0g(x)\,\mathrm dx$ converges and since $(\forall x\in\Bbb R):0\leqslant f(x)\leqslant g(x)$, $\int_{-\infty}^0f(x)\,\mathrm dx$ converges; this follows from the fact that the map$$\begin{array}{ccc}(-\infty,0]&\longrightarrow&\Bbb R\\M&\mapsto&\int_M^0f(x)\,\mathrm dx\end{array}$$is increasing and bounded. By the same argument, $\int_0^\infty f(x)\,\mathrm dx$ converges.
A: Yes you can.
One way to see this is to consider $I_f(a,b)=\int^b_a f$ and $I_g(a,b)=\int^b_ag$. Since $0\leq f<g$,

*

*$0\leq I_f(a,b)\leq I_g(a,b)$.

*For fixed $a$ ($b$) both $I_f$ and $I_g$ are monotone increasing in $b$) (rest $a$).

*By monotonicity $\lim_{(a,b)\rightarrow(-\infty,\infty)}I_f$ and $\lim_{(a,b)\rightarrow(-\infty,\infty)}I_g$ exists in $\mathbb{R}\cup\{\infty\}$.

Your assumption that $\lim_{(a,b)\rightarrow(-\infty,\infty)}I_g <\infty$ implies that $\lim_{(a,b)\rightarrow(-\infty,\infty)}I_f<\infty$ too, and
$$I_f(-\infty,\infty)=\int^\infty_{-\infty}f\leq \int^\infty_{-\infty}g= I_g(-\infty,\infty)$$
