For the first function we have:
if $\mu(X)\lt\infty$ and $f\in L^+$ then $\int fd\mu\lt\infty \iff \sum_{n=0}^\infty 2^n\mu(\{x\in X:f(x)\ge2^n\})\lt\infty$
For the second one:
let $f\in L^+$ bounded, then $\int fd\mu\lt\infty \iff \sum_{n=0}^\infty 2^{-n}\mu(\{x\in X:f(x)\ge2^{-n}\})\lt\infty$
The two proofs are very similar. The idea is to use a (rather inexact) approximation by simple functions. Assume $f \ge 0$. Consider the first case where $\mu(X) < \infty$. Since $$ \int f \, d\mu = \int_{\{f < 1\}} f \, d\mu + \int_{\{f \ge 1\}} f \, d\mu $$ and $$ \int_{\{f < 1\}} f \, d\mu \le \mu(X), $$ you have $$ \int f \, d\mu < \infty \iff \int_{\{f \ge 1\}} f \, d\mu < \infty.$$ For every $n$ you have $2^n\chi_{\{2^n \le f < 2^{n+1}\}} \le f \chi_{\{2^n \le f < 2^{n+1}\}}\le 2^n\chi_{\{2^n \le f < 2^{n+1}\}}$ so that the Beppo Levi theorem implies $$ \sum_{n=0}^\infty 2^n \mu(\{2^n \le f < 2^{n+1}\}) \le \int_{\{1 \le f < \infty \}} f \, d\mu \le \sum_{n=0}^\infty 2^{n+1} \mu(\{2^n \le f < 2^{n+1}\}).$$
From this you can conclude that $f \in L^1(\mu)$ if and only if $$\mu(\{f = \infty)\} = 0 \quad \text{and} \quad \sum_{n=0}^\infty 2^n \mu(\{2^n \le f < 2^{n+1}\}) < \infty.$$
To connect this to the sum given in the problem use summation by parts: if $\{a_n\}$ and $\{b_n\}$ are sequences, then $$\sum_{n=0}^{k-1} (a_n - a_{n+1}) b_{n+1} - a_0 b_0 + a_k b_{k+1} = \sum_{n=0}^k a_n (b_{n+1} - b_n).$$ With $a_n = \mu(\{f \ge 2^n\})$ and $b_n = 2^n$ this becomes $$\sum_{n=0}^{k-1} 2^{n+1} \mu(\{ 2^n \le f < 2^{n+1}\}) - \mu(\{f \ge 1\}) + 2^{k+1} \mu(\{f \ge 2^k\} = \sum_{n=0}^k 2^n \mu(\{f \ge 2^n\}).$$
Suppose that $f \in L^1(\mu)$. It is a standard result that $t \mu(\{f \ge t\}) \to 0$ as $t \to \infty$. We may take $k \to \infty$ in the summation by parts formula to obtain $$\sum_{n=0}^{\infty} 2^{n+1} \mu(\{ 2^n \le f < 2^{n+1}\}) - \mu(\{f \ge 1\}) = \sum_{n=0}^\infty 2^n \mu(\{f \ge 2^n\}).$$ Since the sum on the left converges, so does the sum on the right.
Now suppose that $\displaystyle \sum_{n=0}^\infty 2^n \mu(\{f \ge 2^n\})$ converges. Then $2^n \mu(\{f \ge 2^n\}) \to 0$ so that $\mu(\{f = \infty\}) = 0$, and by taking limits in the summation by parts formula once again $$\sum_{n=0}^{\infty} 2^{n+1} \mu(\{ 2^n \le f < 2^{n+1}\}) - \mu(\{f \ge 1\}) = \sum_{n=0}^\infty 2^n \mu(\{f \ge 2^n\}).$$ Since the sum on the right converges so does the sum on the left. Along with the fact $\mu(\{f = \infty\}) = 0$ you conclude $f \in L^1(\mu)$.