Convergence of integral, that is absolutely convergent, proof Can you think of any proof on convergence of improper integral, that is absolutely convergent? It is so obvious, that I really don't know where to start. 
Triangle inequality gives us $$\Bigg|\int_a^{\infty}f(x)\ dx\Bigg| < \int_a^{\infty}|f(x)|\ dx$$ but I'm not sure if it isn't what is to be proven. Or maybe it's not true for improper integrals and the only way to use it is through Cauchy criterion? (Let assume $f:[a,\infty)\longrightarrow\mathbb{R}$)
 A: If we define the positive and negative parts of $f$ as follows:
$$f^+(x) = \max\{0, f(x)\}$$
$$f^-(x) = -\min\{0, f(x)\}$$
then $f(x) = f^+(x) - f^-(x)$ and $|f(x)| = f^+(x) + f^-(x)$.
Since $f^+$ and $f^-$ are both nonnegative and bounded above by $|f|$, integrability of $|f|$ implies integrability of $f^+$ and $f^-$, and this in turn implies integrability of $f = f^+ - f^-$.

Here are the details as applied to an integral over $[a,\infty)$.
Suppose that $\int_a^\infty |f(x)| dx = A < \infty$. If $a < b$ then we have
$$\begin{align}
0 \leq \int_{a}^{b} f^+(x) dx + \int_{a}^{b} f^-(x) dx &= \int_{a}^{b} |f(x)| dx \\
&\leq \int_{a}^{\infty} |f(x)| dx \\
&= A
\end{align}$$
Since $f^+$ and $f^-$ are nonnegative, the integrals 
$\int_a^b f^+(x) dx$ and $\int_a^b f^-(x) dx$ are increasing as $b$ increases, and they are both bounded above by the finite number $A$. Therefore, both limits
$$P = \int_a^\infty f^+(x) dx$$
and $$N = \int_a^\infty f^-(x) dx$$
exist and are finite. Moreover, the lmiit of a sum is the sum of the limits (provided both limits exist), so $A = P + N$.
Now $f(x) = f^+(x) - f^-(x)$, so
$$\int_a^b f(x) dx = \int_a^b f^+(x) dx - \int_a^b f^-(x) dx$$
and since the limit of a difference equals the difference of the limits (provided both limits exist), we conclude that
$$\int_a^\infty f(x) dx = \int_a^\infty f^+(x) dx - \int_a^\infty f^-(x) dx = P - N$$
A: An alternate argument is to use that $0\le f(x)+|f(x)|\le2|f(x)|$,  
so $\int_{a}^{\infty}|f(x)|dx$ converges $\implies\int_{a}^{\infty}2|f(x)|dx$ converges $\implies$
$\;\;\;\;\int_{a}^{\infty}[f(x)+|f(x)|]dx$ converges by
the Comparison Test, 
so $\int_{a}^{\infty}f(x)dx=\int_{a}^{\infty}[f(x)+|f(x)|]dx-\int_{a}^{\infty}|f(x)|dx$ converges. 
