Proving $f$ is $L^1$ implies convergence of $\sum_{k \in \mathbb{Z}}{ 2^k*m(|f| > 2^k)}$

Question

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.

Answer 1

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.

Answer 2

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 $\sum_{j=-\infty}^\infty 2^j\,m\big(\lvert\,f\rvert>2^j\big)<\infty$.