Proving $f$ is $L^1$ implies convergence of $\sum_{k \in \mathbb{Z}}{ 2^k*m(|f| > 2^k)}$ I need help proving that if f is Lebesgue integrable then
$$\sum_{k \in \mathbb{Z}}{ 2^k*m(|f| > 2^k)}$$
 converges. 
I proved the opposite direction but I don't know how to put any bounds on each term, I would appreciate hints/suggestions. 
 A: Hint: write $m\left(\left|f\right|\gt 2^k\right)=\sum_{j\geqslant k}m\left(2^j\lt \left|f\right|\leqslant 2^{j+1}\right)$, multiply by $2^k$, sum over $k$ and switch the sums. This gives 
$$\sum_{k\in \mathbb Z}2^km\left(\left|f\right|\gt 2^k\right)=
\sum_{k\in \mathbb Z}2^k\sum_{j\geqslant k}m\left(2^j\lt \left|f\right|\leqslant 2^{j+1}\right)=
\sum_{j\in \mathbb Z}\sum_{k=-\infty}^j2^km\left(2^j\lt \left|f\right|\leqslant 2^{j+1}\right)=\sum_{j=0}^{+\infty}\left(2^{j+1}+1\right)m\left(2^j\lt \left|f\right|\leqslant 2^{j+1}\right),$$
where the last inequality is ude to the computation 
$$\sum_{k=-\infty}^j2^k=\sum_{k=-\infty}^{-1}2^k+\sum_{k=0}^j2^k=
\sum_{k=1}^{+\infty}2^{-k}+2^{j+1}=1/(1-1/2)-1+2^{j+1}.$$
Use now the bound $2^jm\left(2^j\lt \left|f\right|\leqslant 2^{j+1}\right)\leqslant \int_{\{2^j\lt \left|f\right|\leqslant 2^{j+1}\}}|f|\mathrm dm$
to conclude.
A: Fixing the answer of Davide Giraudo:
First, we have that $m\left(\left|\,f\right|\gt 2^k\right)=\sum_{j\geqslant k}m\left(2^j\lt \left|\,f\right|\leqslant 2^{j+1}\right)$.
And hence
$$
\sum_{k\in \mathbb Z}2^km\left(\left|\,f\right|\gt 2^k\right)=
\sum_{k\in \mathbb Z}2^k\sum_{j\geqslant k}m\left(2^j\lt \left|\,f\right|\leqslant 2^{j+1}\right)=
\sum_{j\in \mathbb Z}\sum_{k=-\infty}^j2^km\left(2^j\lt \left|\,f\right|\leqslant 2^{j+1}\right).$$
Next use the fact that $\sum_{k=-\infty}^j 2^k=2^{j+1}$, for all $j\in\mathbb Z$, to obtain that
$$
\sum_{j\in \mathbb Z}\sum_{k=-\infty}^j2^km\left(2^j\lt \left|\,f\right|\leqslant 2^{j+1}\right)=\sum_{j\in \mathbb Z}2^{j+1} m\left(2^j\lt \left|\,f\right|\leqslant 2^{j+1}\right).
$$
Finally 
$$
\int_X\lvert\,f\rvert\,dm=\sum_{j=-\infty}^\infty\int_{m\left(2^j\lt \left|\,f\right|\leqslant 2^{j+1}\right)}\lvert\,f\rvert\,dm\ge
\sum_{j=-\infty}^\infty\int_{m\left(2^j\lt \left|\,f\right|\leqslant 2^{j+1}\right)}2^j\,dm=\sum_{j=-\infty}^\infty 2^j{m\left(2^j\lt \left|\,f\right|\leqslant 2^{j+1}\right)}
$$
Therefore
$$
\sum_{j=-\infty}^\infty 2^j\,m\big(\lvert\,f\rvert>2^j\big)\le 2\int_X\lvert\,f\rvert\,dm.
$$
Note. In the same way one can obtain that
$$
\int_X\lvert\,f\rvert\,dm\le\sum_{j=-\infty}^\infty 2^j\,m\big(\lvert\,f\rvert>2^j\big)\le 2\int_X\lvert\,f\rvert\,dm.
$$
Thus, $f$ is integrable if and only if $\sum_{j=-\infty}^\infty 2^j\,m\big(\lvert\,f\rvert>2^j\big)<\infty$.
