if $\int _a^{\infty }\left|f\left(x\right)\right|\:dx$ converges then $\int _a^{\infty }f\left(x\right)\:dx$ also converge.
I wonder what i missing. we proved in our course thar for every $g$ >$f$>0 if $\int _a^{\infty }f\left(x\right)\:dx$ converges than $\int _a^{\infty }g \left(x\right)\:dx$ also converges.
i can't just say $\left|f\right|\:>\:f$ and use this theorem?
Since $f$ is not necessarily non-negative you cannot use this theorem. However, notice that for any $A\lt B$, $$\left|\int_a^A f(x)dx-\int_a^Bf(x)dx\right|\leqslant \int_A^B|f(x)|dx$$ and by the convergence of $\int_a^{+\infty}|f(x)|dx$, we conclude that the limit $\lim_{A\to +\infty}\int_a^Af(x)dx$ exists.
If you want to use the result $0\leqslant f\leqslant g$, notice that $0\leqslant f^+\leqslant f$ and the same with $f^-$. We deduce that $\int_a^{+\infty}f^+(x)dx$ and $\int_a^{+\infty}f^-(x)dx$ are convergent and since $f=f^+-f^-$ that $\int_a^{+\infty}f(x)dx$ is convergent.