Equivalence of a Lebesgue Integrable function I have the following question:
Let $X$: $\mu(X)<\infty$, and let $f \geq 0$ on $X$.  Prove that $f$ is Lebesgue integrable on $X$ if and only if $\sum_{n=0}^{\infty}2^n \mu(\lbrace x \in X : f(x) \geq 2^n \rbrace) < \infty $.
I have the following ideas, but am a little unsure. For the forward direction:
By our hypothesis, we are taking $f$ to be Lebesgue integrable.  Assume $\sum_{n=0}^{\infty}2^n \mu(\lbrace x \in X : f(x) \geq 2^n \rbrace) = \infty $.  Then for any n, no matter how large, $\mu(\lbrace x \in X : f(x) \geq 2^n \rbrace)$ has positive measure.  Otherwise, the sum will terminate for a certain $N$, giving us $\sum_{n=0}^{N}2^n \mu(\lbrace x \in X : f(x) \geq 2^n \rbrace) < \infty $.  Thus we have $f$ unbounded on a set of positive measure, which in combination with $f(x) \geq 0$, gives us that $\int_E f(x) d\mu=\infty$.  This is a contradiction to $f$ being Lebesgue integrable.  So our summation must be finite.
For the reverse direction: 
We have that $\sum_{n=0}^{N}2^n \mu(\lbrace x \in X : f(x) \geq 2^n \rbrace) < \infty \\$.  Assume that $f$ is not Lebesgue integrable, then we have $\int_E f(x) d\mu=\infty$.  Since we are integrating over a finite set $X$, then this means that $f(x)$ must be unbounded on a set of positive measure, which makes our summation infinite, a contradiction.
Any thoughts as to the validity of my proof?  I feel as if there is an easier, direct way to do it.

Let $E_n = \lbrace x \in X : 2^n \leq f(x) < 2^{n+1} \rbrace$.
Let $F = \lbrace x \in X : f(x) < 1 \rbrace$.
Then $X$ is a disjoint union of $F$ and $E_n$, $n = 0,1,\dots$.
Hence $\int_X f(x) d\mu = \int_F f(x) d\mu + \sum_{n=0}^{\infty} \int_{E_n} f(x) d\mu$.
Let $G_n = \lbrace x \in X : f(x) \geq 2^n \rbrace$.
Then $E_n = G_n - G_{n+1}$.
Let $m_n = \mu(G_n)$.
Since $\mu(E_n) = \mu(G_n - G_{n+1}) = m_n - m_{n+1}$, $2^n(m_n - m_{n+1}) \leq \int_{E_n} f(x) d\mu \leq 2^{n+1}(m_n - m_{n+1})$.
Hence
$\int_F f(x) d\mu + \sum_{n=0}^{\infty} 2^n(m_n - m_{n+1}) \leq \int_X f(x) d\mu \leq \int_F f(x) d\mu + \sum_{n=0}^{\infty} 2^{n+1}(m_n - m_{n+1})$
Let $s = \sum_{n=0}^{\infty} 2^n m_n$.
Suppose $f$ is integrable.
By the above left inequality,
$\sum_{n=0}^{\infty} 2^n(m_n - m_{n+1}) < \infty$.
Since $\sum_{n=0}^{\infty} 2^n(m_n - m_{n+1}) = m_0 + \sum_{n=0}^{\infty} 2^n m_{n+1} = m_0 + (\sum_{n=0}^{\infty} 2^{n+1} m_{n+1})/2 < \infty$, $s < \infty$.
Conversely suppose $s< \infty$.
Note that $\int_F f(x) d\mu < \infty$.
By the above right inequality, $\int_X f(x) d\mu \leq \int_F f(x) d\mu + \sum_{n=0}^{\infty} 2^{n+1}(m_n - m_{n+1}) = \int_F f(x) d\mu + 2m_0 + \sum_{n=0}^{\infty} 2^{n+1} m_{n+1} = \int_F f(x) d\mu + m_0 + s < \infty$.
