Let $(X, S, \mu)$ a finite measure space, $f$ non negative function. Show:
$$\int f \,d\mu < +\infty$$ if only if $$\sum_{n=0}^\infty 2^{n}\mu(\{x\in X \mid f(x) \geq 2^n \}) < + \infty.$$
$(\implies)$ if $\int f \, d\mu = M < +\infty$, $2^n$ is non-negative, by Chebyshev inequality:
$$\mu(\{x\in X \mid f(x) \geq 2^n \} \leq \frac{M}{2^n}, \text{ where } \frac{M}{2^n} \rightarrow 0 \text{ when } n \rightarrow \infty.$$
$$\sum_{n=0}^\infty 2^n\mu(\{x\in X \mid f(x) \geq 2^n \}) \leq \sum_{n=0}^\infty 2^n \frac{M}{2^n} \rightarrow 0. $$
Then
$$\sum_{n=0}^\infty 2^n\mu(\{x\in X \mid f(x) \geq 2^n \}) < +\infty.$$
($\Leftarrow$) If $$\sum_{n=0}^\infty 2^n \mu(\{x\in X \mid f(x) \geq 2^n \}) < + \infty.$$
Is the first implication correct? I am trying the other implication also.
The first implication is not proven correctly. You've shown that $$ \sum_{n=0}^{\infty} 2^{n}\mu(\{x\in X | f(x) \geq 2^n \}) \leq \sum_{n=0}^{\infty} 2^{n} \frac{M}{2^n}=\sum_{n=0}^\infty M=\infty $$ which is not strong enoough. Instead proceed in the following way. Since $f$ is non-negative we may write $$ \int fd\mu=\int_{0}^\infty \mu(f>x) \, dx\tag{0} $$ (Here $(f>x)$ is shorthand for the set $\{t\mid f(t)>x\}$). This is the tail formula for expectation in probability theory. Now use the fact that $$ \int_{0}^\infty \mu(f>x)=\int_0^1 \mu(f>x)\, dx+\sum_{k=0}^\infty\int_{2^{k}}^{2^{k+1}}\mu(f>x)\, dx\tag{1} $$ But since $x\to \mu(f>x)$ is decreasing function $$ 2^k\mu(f>2^{k+1})\leq \int_{2^{k}}^{2^{k+1}}\mu(f>x)\leq 2^k\mu(f>2^k)\tag{2} $$ Equations $(0)$ and $(1)$ and $(2)$ (combined with some reasoning) yield the result. I leave it to you to fill in the details.