Forward direction:
First we note that for all n $E_{n+1} \subset E_{n}$, thus $\mu(E_n-E_{n+1})=\mu(E_{n})-\mu(E_{n+1})$. We also note that $E_n-E_{n+1}=${$x∈ X|n}, thus it follows that $E_1-E_2, E_2-E_3, E_3-E_4,...$ are disjoint measurable sets. Now define a function $g:X→R$, such that $g(x)=n$ if $x\in E_n-E_{n+1}$ and 0 otherwise, it follows easily that $g(x)\leq(f(x))^2$, therefore: $\int_X g d \mu \leq \int_X f^2d \mu$ Hence: $\sum_{n=1}^{\infty} n^2(\mu(E_n)-\mu(E_{n+1})) \leq \int_X f^2d \mu...(1)$
Now define $h_n(x)=n^2$ if $x\in E_n$ and 0 otherwise. It follows that $h_n \leq f^2$, therefore: $\int_X h_n d \mu \leq \int_X f^2d \mu$ Hence, we have for all n: $n^2\mu(E_n) \leq \int_X f^2d \mu...(2)$
Now, we use Abel's identity to deduce that: $\sum_{n=1}^k (n+1/2)\mu(E_{n+1})=(k+1)^2\mu(E_{k+1})/2-\mu(E_1)-\sum_{n=1}^k (n^2/2)(\mu(E_{n+1})-\mu(E_n))$ From inequalities (1),(2) we find that the R.H.S. of the previous equation is bounded, thus the sequence of partial sums $\sum_{n=1}^k (n+1/2)\mu(E_{n+1})$ is bounded from above, hence it converges. Since, for all n $n/2, thus we deduce that the sequence $\sum_{n=1}^k (n/2)\mu(E_{n+1})$ converges.
Backward Direction:
Using the limit comparison test, we can show that $\sum_{n=1}^{\infty} (2n+3)\mu(E_{n+1})$ converges. Using Abel's identity we have: $\sum_{n=1}^k ((n+1)^2)(\mu(E_{n+1}-\mu(E_n)))=(k+2)^2\mu(E_{k+1})/2-4\mu(E_1)-\sum_{n=1}^k (2n+3)\mu(E_{n+1})$ Since the R.H.S is bounded therefore the series in L.H.S. converges. Now by letting $g(x)=(n+1)^2$ if $x\in E_n-E_{n+1}$ and 0 otherwise we can see that g is integrable. Since $g \leq f$, therefore $f$ is integrable.