Comparison Test in $\Bbb R^n$

Question

Comparison Test in $\Bbb R^n$:

Let $A$ be open in $\Bbb R^n$; let $f,g : A \to \Bbb R$ be continuous; suppose that $|f(x)| < g(x)$ for $x \in A$. Show that if $\int_A g$ exists, so does $\int_A f$.

Can someone provide me some reference from where I can get the proof?

Also Hints for proving the statement are welcome.

Thank You.

Answer 1

Since $f$ is continuous, the integral $\int_X f$ is defined for all $X$ compact. So the only issue is boundedness of the (possibly improper) integral at $A$.

Since $A$ is open, we may write $A=\bigcup A_n$, an increasing union of compact sets. Then we would define $$ \int_A f=\lim_n \int_{A_n}f, $$ if the limit exists. The fact that $\int_A g$ exists, tells us that $$\int_A g=\lim_n\int_{A_n}g.$$ Thus $$ \lim_n \int_{A\setminus A_n} g=0.$$ Then, if $A_n\subset A_m$, $$ \left|\int_{A_m}f-\int_{A_n}f\right|=\left|\int_{A_m\setminus A_n}f\right|\leq\int_{A_m\setminus A_n}|f| \leq\int_{A\setminus A_n}|f|\leq \int_{A\setminus A_n}g\to0. $$ So the sequence $\{\int_{A_n}f\}$ is Cauchy, and so $\int_A f$ exists.