I'm having trouble with this proof, can anyone please help me out?
If $\int_{a}^{\infty}f(x) dx$ diverges and $0 \le f(x) \le g(x)$ for all $ x\in[a, \infty)$, then $\int_{a}^{\infty}g(x)dx$ also diverges.
$0 \le f(x) \le g(x)$
$\int_{a}^{\infty} 0 dx \le \int_{a}^{\infty}f(x) dx \le \int_{a}^{\infty}g(x)dx$
$ 0 \le \lim_{A\to \infty}\int_{a}^{A}(f(x) \le g(x)) dx$
$\lim_{A\to \infty}(f(A)-f(a) \le g(A)-g(a))$
On
You can take the contrapositive of this statement to prove it, because of the fact following fact: Let P and Q be two sentences, such that P $\implies$ Q. This is equivalent to say $\neg$ (Q) $\implies$ $\neg$ (P). So now, you accept that the function g(x) converges in a defined interval - of course, supposing that 0 $\leq$ f(x) $\leq$ g(x) in that interval - and you use this to prove that f(x) also converges in that interval. The proof isn't long, but you have to use some basic properties of intergrals. Hope this will help you.
This is not true: $0\leq g(x)\leq 1$, $x\in[1,\infty)$, $g(x)=1/x^{2}$ but $\displaystyle\int_{1}^{\infty}g(x)dx<\infty$ and of course $\displaystyle\int_{1}^{\infty}1dx=\infty$.
But the following is true: $0\leq f(x)\leq g(x)$, $x\in[a,\infty)$ and $\displaystyle\int_{a}^{\infty}f(x)dx$ diverges, then so is $\displaystyle\int_{a}^{\infty}g(x)dx$:
Because we know that $\lim_{M\rightarrow\infty}\displaystyle\int_{a}^{M}g(x)dx=\sup_{M\geq a}\int_{a}^{M}g(x)dx$, if $\displaystyle\int_{a}^{\infty}g(x)dx$ converges, which means that the limit exists, then $\displaystyle\int_{a}^{M}f(x)dx\leq\int_{a}^{M}g(x)dx\leq\sup_{M\geq a}\int_{a}^{M}g(x)dx$, then $\sup_{M\geq a}\displaystyle\int_{a}^{M}f(x)dx\leq\sup_{M\geq a}\int_{a}^{M}g(x)dx<\infty$, so $\lim_{M\rightarrow\infty}\displaystyle\int_{a}^{M}f(x)dx=\sup_{M\geq a}\int_{a}^{M}f(x)dx<\infty$, which means that $\displaystyle\int_{a}^{\infty}f(x)dx$ converges, a contradiction.