Show that $\displaystyle \int_a^\infty f(x) \, dx$ exists if, and only if, for every $\varepsilon>0$ there exists $M_\varepsilon$ such that $$\left|\int_s^t f(x) \, dx\right|<\varepsilon$$ for all $s,t\geq M_\varepsilon$.
So far I've shown the reverse implication. But I can't show that $\displaystyle \int_a^\infty f(x) \, dx$ exists implies that for every $\varepsilon>0$ there exists $M_\epsilon$ such that $$\left|\int_s^t f(x) \, dx \right|< \varepsilon$$ for all $s,t\geq M_\varepsilon$. My main problem is the absolute value, it ruins every attempt I make.
Define $g(y)=\int_0^y f(x)dx$. $\lim_{y\to+\infty}g(y)$ exists and is finite iff for some $L$, given $\varepsilon>0$
$$\exists\;M_{\varepsilon}>0:y>M_{\varepsilon}\implies|g(y)-L|<{\varepsilon\over2}\\ \text{or,}\;\exists\;M_{\varepsilon}>0:t,s>M_{\varepsilon}\implies|g(t)-g(s)|<|g(t)-L|+|g(s)-L|<\varepsilon\\ \text{or,}\;\exists\;M_{\varepsilon}>0:t>s>M_{\varepsilon}\implies\left|\int_s^t f(x)dx \right|<\varepsilon$$