8.$ (\Rightarrow) $ First we apply Tchebyshev to $ E_n $ and find that
$ (n-1)\left|\{x \in X \mid |f(x)| \geq n-1 \}\right| \leq \int_{E_n}|f| $
or rather
$ (n-1) m(E_n) \leq \int_{E_n}|f| $
Since we have that $ m(E_n) $ is finite we can move it to the other side of the inequality.
$ nm(E_n) \leq \int_{E_n}|f| + m(E_n) $
Since this is true for all $ n $ we take sums on both sides and note that the $ E_n $ are disjoint.
$ \sum_{n=1}^{\infty}nm(E_n) \leq \sum_{n=1}^{\infty}\int_{E_n}|f| + \sum_{n=1}^{\infty}m(E_n) $
or
$ \sum_{n=1}^{\infty}nm(E_n)\leq \int_{X}|f| + m(X) $
And we are in a finite measure space so $ m(X) < \infty $ and since $ f \in L^1 $ we have $ \int_{X}|f| < \infty $.
Thus we have that $ \sum_{n=1}^{\infty}nm(E_n) < \infty $.
$ (\Leftarrow) $ Since $ |f|< n $ in each $ E_n $ we have that
$ \int_X|f| = \sum_{n=1}^{\infty}\left(\int_{E_n}|f|\right) \leq \sum_{n=1}^{\infty}\left( n m(E_n)\right)< \infty $
In other words, $ f \in L^1 $.
