Prove that a sum of measures is finite if $f\in L^1$ Let $(X,\mu)$ be a measure space and $\mu(X)<\infty$. Let $f:X\to \mathbb{C}$ be a measurable function and let $f\in L^1(\mu)$. Prove that
$$\sum_{n=1}^{\infty}2^n \mu\{|f|\geq2^n\}<\infty\tag{1}$$
If we split $|f(x)|$ into intervals as follows, we get:
\begin{align*}
\int_X |f|d\mu &\geq \sum_{n=-\infty}^{\infty} 2^n \mu\{2^n\leq |f|\leq 2^{n+1}\} \\ \\
&= \sum_{n=-\infty}^{\infty} 2^n \mu\{|f|\geq 2^{n}\}-\sum_{n=-\infty}^{\infty}2^n \mu\{|f|\geq 2^{n+1}\} \\ \\
\end{align*}
But I'm not sure how to use the fact that $\mu(X)<\infty$ so as to prove $(1)$?
EDIT:
Following Kavi Rama Murthy's comment:
\begin{align*}
\sum_{n\geq 1} 2^n\left( \mu\{|f|\geq 2^n\} - \mu\{|f|\geq 2^{n+1}\} \right) &=2\mu\{|f|\geq 2\}+\frac{1}{2}\sum_{n\geq 2} 2^n\mu\{|f|\geq 2^n\}
\end{align*}
but how can we know that
$$\sum_{n=-\infty}^{\infty} 2^n \left( \mu\{|f|\geq 2^{n}\}- \mu\{|f|\geq 2^{n+1}\}\right) \geq \sum_{n\geq 1} 2^n \left( \mu\{|f|\geq 2^{n}\}- \mu\{|f|\geq 2^{n+1}\}\right)$$
because we don't know whether $\mu$ is a positive measure?
 A: I am under the impression that something like this has been asked before in MSE, but I was not able to find the appropriate question. Here is a solution following the OP's ideas.

Again,  $\mu$ is finite and $f\in L_1(\mu)$. Suppose
$\{a_n:n\in\mathbb{Z}_+ \}$ is a positive increasing sequence with $c a_{n+1}\leq a_n\nearrow\infty$ for some $0<c<1$ (see comment at the end). The OP takes $a_n=2^n$, which satisfies this condition: $\frac12 a_{n+1}\leq a_n<a_{n+1}$.
For all $k\in\mathbb{N}$. we have
$$a_k\mu(\{a_k<|f|\leq a_{k+1}\})\leq\int_{\{a_k<|f|\leq a_{k+1}\}}|f|\,d\mu$$
Adding all terms we obtain that
$$\begin{align}
\sum^\infty_{k=0} a_k\mu(a_k<|f|\leq a_{k+1})\leq\sum^\infty_{k=0}\int_{\{a_k<|f|\leq a_{k+1}\}}|f|\,d\mu\leq\int|f|\,d\mu<\infty\tag{0}\label{zero}
\end{align}$$
Since $f\in L_1$, we get from Chebyshev-Markov's inequality and dominated convergence that
$$t \mu(|f|>t)\leq\int_{\{|f|>t\}}|f|\,d\mu\xrightarrow{t\rightarrow\infty}0$$
Let $D(t)=\mu(|f|>t)$. Then, applying summation by parts to the partial sums of the series on the left-hand side of \eqref{zero} gives
$$\begin{align}
S_n:=\sum^n_{k=0} a_k\mu(a_k<|f|\leq a_{k+1})&=\sum^n_{k=0} a_k\big(D(a_{k+1})-D(a_k)\big)\\
&=-a_nD(a_{n+1})+a_0D(a_0)+\sum^n_{k=1}D(a_k)\big(a_k-a_{k-1}\big)\tag{1}\label{one}
\end{align}
$$
The first term in \eqref{one} $a_nD(a_{n+1})\leq a_{n+1}D(a_{n+1})\xrightarrow{n\rightarrow\infty}0$. Since $S_n$ converges, we conclude that
$$\begin{align}
\sum^\infty_{k=1}D(a_k)\big(a_k-a_{k-1}\big) <\infty\tag{2}\label{two}
\end{align}
$$
The OP takes $a_k=2^k$, in which case $\frac{1}{2}\sum^\infty_{k=1}2^k\mu(|f|>2^k)) <\infty$

Observation: Nowhere in the proof I used the "tempered" growth factor $0<c<1$. Th  reason to include it is to show that under some conditions, the converse holds true. That is, if \eqref{two} converges and $a_n\mu(|f|>a_n)$ is bounded, then  $f\in L_1$. Indeed, under these assumptions, we get  from \eqref{one} that
$$\sum^{\infty}_{k=0} a_k\mu(\{a_k<|f|\leq a_{k+1}\})<\infty$$
Then, from
$$
\int_{\{a_k<|f|\leq a_{k+1}\}}|f|\,d\mu\leq a_{k+1}\,\mu(\{a_k<|f|\leq a_{k+1}\})\le  \frac{1}{c}a_k\,\mu(\{a_k<|f|\leq a_{k+1}\})
$$
we conclude that $\int_{\{a_0<|f|\}}|f|\,d\mu<\infty$. Since $\mu$ is finite, $\mu(|f|\leq|a_0|)<\infty$ and the integrability of $f$ follows.
